Countably compact groups and sequential order

TL;DR

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Contribution

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Abstract

We use $\diamondsuit$ to construct, for every $\alpha\leq\omega_1$ a sequential countably compact topological group of sequential order $\alpha$. This establishes the independence of the existence of sequential countably compact non Fr\'echet groups from the usual axioms of ZFC and answers several questions of D.~Shakhmatov.

Equations

$$(d+K)\cap [A\cup \{0\}{]}_{\beta }^{k_{\omega }(\mathcal{K})}=\varnothing ifd\in ⟨D⟩\setminus ⟨F_{K,\beta }⟩$$

$$(d+K)\cap \sum _{i<n}{\overline{⟨A_i⟩}}^{k_{\omega }(\mathcal{K})}=\varnothing ifd\in ⟨D⟩\setminus ⟨F_{K,n}⟩$$

$$d_n\notin \sum _{i<n}{K}_i+⟨\{d_i:i<n\}⟩+[A_0\cup \{0\}{]}_{{\sigma }_n^0}^{k_{\omega }(\mathcal{K})}$$

$$p_n\notin \sum _{i\le n}{K}_i+⟨\{p_i:i<n\}⟩+\sum _{i\le n}{\overline{⟨A_i⟩}}^{k_{\omega }(\mathcal{K})}$$

$$0\notin {\overline{P\cap (\sum _{i\le m}{\overline{⟨A_i⟩}}^{k_{\omega }(\mathcal{K})}+K)}}^{k_{\omega }(\mathcal{K})}$$

$$\mathrm{so}(G)\le \sup \{\min \sum _{\alpha \in I}\mathrm{so}(G_{\alpha }):I\subseteq \gamma ,|I|<\omega \}+1.$$

$$\eta =\max \{\mu ,\mathrm{so}(s_{\Omega },P,k_{\omega }({\mathcal{K}}^{\Omega }))\}\le \sigma -\Sigma ;$$

$${\eta }_n=\max \{{\mu }_n,\mathrm{so}(s_{{\gamma }_n},P_n,k_{\omega }({\mathcal{K}}^{{\gamma }_n}))\le {\sigma }_i-\Sigma ;$$

$$W_n(m)=\{(G^{\gamma },{\mathcal{K}}^{\gamma },{\mathcal{U}}^{\gamma },S_{\gamma },s_{\gamma }):{\gamma }_{n-1}(m)\le \gamma <{\gamma }_n(m)\}$$

$$C=\{(G_{\gamma },{\mathcal{K}}_{\gamma },{\mathcal{U}}_{\gamma }):\mathbf{m}(\alpha )\le \gamma \le {\gamma }^0\}$$

$$\cap \{U\in {\mathcal{U}}_{\mathbf{m}(\alpha +3)}:H(I)\subseteq U\}=H(I)$$

$$U^{\prime }\cap (\sum _{i\in I}{\overline{⟨S^i⟩}}^{k_{\omega }({\mathcal{K}}_{\mathbf{m}(\alpha +3)})})=\sum _{i\in I}(U\cap {\overline{⟨S^i⟩}}^{k_{\omega }({\mathcal{K}}_{\mathbf{m}(\alpha +3)})});$$

$$W=\{(G_{\gamma },{\mathcal{K}}_{\gamma },{\mathcal{U}}_{\gamma }):\mathbf{m}(\alpha +1)<\gamma \le \mathbf{m}(\alpha +1)+k=\mathbf{m}(\alpha +2)\}$$

$$\{(G_{\gamma },{\mathcal{K}}_{\gamma },{\mathcal{U}}_{\gamma }):\gamma (k-1)<\gamma \le \gamma (k)\}$$

$$p_{n(j)},d_{n(j)}^i\in {\cap }_{n<j}{U}_{n}and{p}_{n(j)}-\sum _{i\in I}{d}_{n(j)}^i\in K\cap {\cap }_{n<j}{U}_n$$

$$W_1=\{(G_{\gamma },{\mathcal{K}}_{\gamma },{\mathcal{U}}_{\gamma },S_{\gamma },s_{\gamma }):{\gamma }_s\le \gamma \le \gamma (i_1,k)\}$$

$$(D^{+}(J,i_1),G_{<{\gamma }_s},{\mathcal{K}}_{<{\gamma }_s},{\mathcal{U}}_{<{\gamma }_s},{\delta }_{i_1})$$

$$s_{\gamma (i_1,k)}\in [\{d_n(i_1)+d_n^{\prime }:n\in J\}{]}_{{\eta }^{i_1}(k)}^{k_{\omega }({\mathcal{K}}_{\gamma (i_1,k)})}$$

$$\mathrm{so}(s_{\gamma (i_1,k)},⟨S^i⟩\setminus ⟨2S^i⟩,k_{\omega }({\mathcal{K}}_{\gamma (i_1,k)}))={\delta }_{i_1}+{\eta }^{i_1}(k)\ge {\sigma }_k^{i_1}.$$

$$W_j=\{(G_{\gamma },{\mathcal{K}}_{\gamma },{\mathcal{U}}_{\gamma },S_{\gamma },s_{\gamma }):\gamma (i_{j-1},k)<\gamma \le \gamma (i_j,k)\}$$

$$(D^{+}(J,i_j),G_{\gamma (i_{j-1},k)},{\mathcal{K}}_{\gamma (i_{j-1},k)},{\mathcal{U}}_{\gamma (i_{j-1},k)},{\delta }_{i_j})$$

$$\sum _{l\le j}{s}_{\gamma (i_l,k)}\in [\{\sum _{l\le j}{d}_n(i_l)+d_n^{\prime }:n\in J\}{]}_{{\eta }^{i_j}(k)}^{k_{\omega }({\mathcal{K}}_{\gamma (i_j,k)})}$$

$$\mathrm{so}(s_{\gamma (i_j,k)},⟨S^i⟩\setminus ⟨2S^i⟩,k_{\omega }({\mathcal{K}}_{\gamma (i_j,k)}))={\delta }_{i_j}+{\eta }^{i_j}(k)\ge {\sigma }_k^{i_j}.$$

$$\{(G_{\gamma },{\mathcal{K}}_{\gamma },{\mathcal{U}}_{\gamma }):\gamma (\omega )<\gamma \le \gamma (\omega )+s={\gamma }_t\}$$

$$\{\gamma \in {\omega }_1:\gamma =M\cap {\omega }_1,X\in M,M\le H(\theta )\}$$

$$d_n\notin \sum _{i<n}{K}_i+⟨\{d_i:i<n\}⟩+\sum _{m<n}{\overline{⟨S^m⟩}}^{k_{\omega }({\mathcal{K}}_{<\gamma })}$$

$$W_j=\{(G_{\gamma },{\mathcal{K}}_{\gamma },{\mathcal{U}}_{\gamma },S_{\gamma },s_{\gamma }):{\gamma }_j<\gamma \le {\gamma }_{j+1}\}$$

$$\mathrm{so}(s_{{\gamma }_{j+1}},⟨S^{i^{\prime }}⟩\setminus ⟨2S^{i^{\prime }}⟩,k_{\omega }({\mathcal{K}}_{{\gamma }_j}))\ge {\sigma }_{j+1}^{i^{\prime }}$$

$$\mathrm{so}(g,⟨S^{i^{\prime }}⟩\setminus ⟨2S^{i^{\prime }}⟩,k_{\omega }({\mathcal{K}}_{{\gamma }_{n+1}}))\ge \mathrm{so}(s_{{\gamma }_{j+1}},⟨S^{i^{\prime }}⟩\setminus ⟨2S^{i^{\prime }}⟩,k_{\omega }({\mathcal{K}}_{{\gamma }_{n+1}}))\ge {\sigma }_n^{i^{\prime }}$$

$$d(i,n)\in {\overline{⟨S^i⟩}}^{k_{\omega }({\mathcal{K}}_{\mathbf{m}(\alpha +1)})}\cap \bigcap \{{\overline{U}}^{\tau ({\mathcal{U}}_{\mathbf{m}(\alpha +1)})}:U\in {\mathcal{U}}_{<\mathbf{m}(\alpha )}\}$$

Full text

1 Introduction and notation.

The standard definition of a topological space contains no reference to convergence, sequential or otherwise. In common mathematical practice, however, convergence often appears as a valuable tool in many application areas for topology, such as analysis. In an effort to formalize the relationship between the sequential convergence and topology, the class of Fréchet spaces was defined in which the closure operator is directly described in terms of limits of convergent sequences. Later, a more general class of sequential spaces was introduced in [4] to encompass all the topological spaces in which convergent sequences fully describe the topology (see below for the definitions of these and other concepts used in the introduction).

The addition of an algebraic structure appropriately coupled with the topology (by requiring the operations to be continuous) imposes a number of restrictions on both the topology and the convergence. Thus separation axioms $T_{1}$–$T_{3{1\over 2}}$ are equivalent in topological groups, as are metrizability and the first countability axiom (see [1]). The investigation into sequential topological groups was started by V. Malykhin, P. Nyikos, I. Protasov, E. Zelenyuk, and others (see [23], [13] and the references therein) in the 1970s and 1980s. In the following decade, a number of questions about sequential topological groups (more generally, about ‘sequential phenomena’ in groups) had been stated, that have guided the future development of this field.

The survey paper [14] (see also [15]) presents a fairly complete overview of the state of the art in the study of convergence in the presence of an algebraic structure. One of the problems posed in [14] (Question 7.5) is the existence of a countably compact sequential group that is not Fréchet. This question is stated for precompact and pseudocompact sequential groups, as well. It was shown in [17] that it is consistent with the usual axioms of ZFC that there are no such countably compact groups. The goal of this paper is to establish the consistency of the existence of such groups thus showing that such existence is independent of ZFC. Note that every compact sequential (or even countably tight) topological group is metrizable (see [1]).

Another question in [14] deals with the measure of (sequential) complexity of the closure operator in sequential groups, called the sequential order (see below for a definition). Namely, Question 7.4 asks if countably compact (pseudocompact, precompact) sequential groups of arbitrary sequential orders exist, consistently, or otherwise.

A similar question was asked by P. Nykos in [13] about the class of all sequential groups. The existence of sequential groups with nontrivial ($\not\in\{0,1,\omega_{1}\}$) sequential orders was shown to be independent of ZFC in [17] and [18]. The example in this paper (see Theorem 1) thus establishes that the answer to Question 7.4(iii) from [14] (existence of countably compact groups of arbitrary sequential orders) is likewise independent of ZFC (for the classes of pseudocompact and precompact groups it is still an open question whether it is consistent that there are no such groups of sequential order $\omega_{1}$).

We use the standard set-theoretic notation, see [10] and [1]. All spaces are assumed to be regular unless stated otherwise.

Abusing the notation somewhat we write $\sigma-1=\sigma^{\prime}$ where $\sigma\geq 1$ is a successor ordinal and $\sigma^{\prime}+1=\sigma$ or $\sigma=0$ and $\sigma^{\prime}=-1$ (we do not go as far as call $-1$ an ordinal though). For brevity, we use the term increasing to mean non decreasing, and use strictly increasing when a stronger condition is assumed. Define $\alpha\leq_{L}\beta$ as $\alpha<\beta$ for a limit $\beta$ and $\alpha\leq\beta$ otherwise. Note that if $\alpha$ is a successor $\alpha\leq\beta$ is equivalent to $\alpha\leq_{L}\beta$.

Our notation for various ordinal invariants is more detailed than customary as we must frequently keep track of several topologies on the same space. Whenever the topology is clear from the context we omit it from the notation as well. The following definition is the starting point for most arguments about convergence.

Definition 1**.**

Let $(X,\tau)$ be a topological space, $A\subseteq X$. Define the sequential closure of $A$, $[A]^{\tau}_{0}=[A]^{\tau}={\{\,x\in X:\exists S\subseteq A,S\to x\,\}}$. Given an ordinal $\sigma>0$, define $[A]_{\sigma}^{\tau}=[[A]_{\sigma-1}^{\tau}]^{\tau}$ if $\sigma$ is a successor and $[A]_{\sigma}^{\tau}=\cup_{\sigma^{\prime}<\sigma}[A]_{\sigma^{\prime}}^{\tau}$ otherwise.

Sequential topological spaces may be defined as exactly those $(X,\tau)$ in which for every $A\subseteq X$ there exists a $\sigma$ such that $\overline{A}^{\tau}=[A]_{\sigma}^{\tau}$. It is a quick argument to show that in such spaces $\sigma\leq\omega_{1}$.

Definition 2**.**

Let $(X,\tau)$ be a sequential space, $A\subseteq X$, and $x\in X$. Define $\mathop{\mathfrak{so}}(x,A,\tau)=\inf{\{\,\sigma:x\in[A]_{\sigma}^{\tau}\,\}}\cup\{\omega_{1}\}$.

As a quick observation, if $x\in\overline{A}^{\tau}$ then $\mathop{\mathfrak{so}}(x,A,\tau)=\sigma<\omega_{1}$ is a successor ordinal, and whenever $\sigma>0$ there are $x_{n}\in A$ such that $x_{n}\to x$ in $\tau$ and $\mathop{\mathfrak{so}}(x_{n},A,\tau)=\sigma_{n}$ is an increasing (non decreasing) sequence of ordinals such that $\sigma_{n}<\sigma$ and $\sigma_{n}\to\sigma-1$.

The central ordinal invariant in the study of convergence can now be defined as follows.

Definition 3**.**

Let $(X,\tau)$ be a sequential space. Define the sequential order of $X$, $\mathop{\mathfrak{so}}(X,\tau)=\sup{\{\,\mathop{\mathfrak{so}}(x,A,\tau):A\subseteq X,x\in\overline{A}^{\tau}\,\}}$.

The construction below depends heavily on the algebraic properties of the underlying group. Recall that a group $G$ is called boolean if $a+a=0_{G}$ for any $a\in G$. All such groups are abelian and may be viewed as vector spaces over ${\mathbb{F}}_{2}$. One can thus consider (linearly) independent subsets of $G$ in the usual sense. If $A\subseteq G$ by ${\langle{A}\rangle}$ we denote the span of $A$ in $G$. A convenient property of boolean groups that is used without mentioning below is that $a+b=a-b$ for any $a,b\in G$ for a boolean $G$.

The following construction will be used often.

Definition 4**.**

Let $G$ be a boolean group, $S\subseteq G$ be an independent subset. Define the even subgroup of ${\langle{S}\rangle}$ (relative to $S$) as ${\langle{2S}\rangle}={\langle{S+S}\rangle}$.

Under most circumstances, the set $S$ will be clear from the context and the ‘relative to $S$’ part will be omitted. We assume below that all groups are abelian unless stated otherwise, although a number of statements hold under more general assumptions.

Lemma 1**.**

Let $G$ be a topological group, $A,B\subseteq G$. Let $a\in[A]_{\alpha}$ and $b\in[B]_{\beta}$ for some $\alpha,\beta<\omega_{1}$. Then $a+b\in[A+B]_{\max\{\alpha,\beta\}}$.

Proof.

We may assume that both $\alpha$ and $\beta$ are successor ordinals. If $\max\{\alpha,\beta\}=0$ then $a\in A$ and $b\in B$ so $a+b\in A+B=[A+B]_{0}$.

Suppose the Lemma holds for all successor $\alpha^{\prime},\beta^{\prime}$ such that $\max\{\alpha^{\prime},\beta^{\prime}\}<\gamma$. Let $\mathop{\mathfrak{so}}(a,A)=\alpha$ and $\mathop{\mathfrak{so}}(b,B)=\beta$, $\max\{\alpha,\beta\}=\gamma$. Pick $b_{n}\to b$ and $a_{n}\to a$ such that $\mathop{\mathfrak{so}}(a_{n},A)=\alpha_{n}$, $\mathop{\mathfrak{so}}(b_{n},B)=\beta_{n}$ where $\alpha_{n}<\gamma$, $\beta_{n}<\gamma$ are such that $\alpha_{n}\to\alpha-1$ and $\beta_{n}\to\beta-1$. By the hypothesis $a_{n}+b_{n}\in[A+B]_{\max\{\alpha_{n},\beta_{n}\}}=[A+B]_{\gamma_{n}}$ where $\gamma_{n}<\gamma$. Thus $a+b\in[A+B]_{\gamma}$. ∎

Lemma 2**.**

Let $G$ be a topological group and $K\subseteq G$ be a sequentially compact subspace. Let $P\subseteq D+K$. Then for any $\alpha<\omega_{1}$ $[P]_{\alpha}\subseteq[D]_{\alpha}+K$.

Proof.

Suppose the lemma holds for all $\alpha^{\prime}<\alpha$ and let $\mathop{\mathfrak{so}}(x,P)=\alpha$. If $\alpha=0$ the argument is trivial, otherwise, there are $x_{n}=d_{n}+g_{n}$ such that $x_{n}\to x$, $d_{n}\in D$, $g_{n}\in K$ and $\mathop{\mathfrak{so}}(x_{n},P)=\alpha_{n}<\alpha$. By the hypothesis, we may assume that $x_{n}=d_{n}^{\prime}+g_{n}^{\prime}$ where $\mathop{\mathfrak{so}}(d_{n}^{\prime},D)\leq\alpha_{n}$ and $g_{n}^{\prime}\in K$. Using the sequential compactness of $K$ we may assume (after passing to a subsequence if necessary) that $g_{n}^{\prime}\to g$ for some $g\in K$. Thus $d_{n}^{\prime}\to d=x-g$ and $\mathop{\mathfrak{so}}(d,D)=\alpha^{\prime}\leq\alpha$. ∎

Corollary 1**.**

Let $G$ be a topological group and $K\subseteq G$ be a sequentially compact subspace. Let $P={\{\,p_{n}:n\in\omega\,\}}$ and $p_{n}=a_{n}+d^{n}+d_{n}$ where $a_{n}\in K$, $d_{n}\to d$, and $g\in[{\{\,d^{i}:i\in\omega\,\}}]_{\sigma}$.

Then there exists a $p\in[P]_{\max\{\sigma,1\}}$ such that $p\in g+d+K$. If $a_{n}\to a$ one may assume that $p=a+g+d$.

Proof.

Put $D={\{\,d^{n}+d_{n}:n\in\omega\,\}}$ and apply Lemma 1 to show that $g+d\in[D]_{\max\{\sigma,1\}}$. Now note that $D\subseteq P+(-K)$ and use Lemma 2 (with the roles of $P$ and $D$ reversed) to find a $p\in[P]_{\max\{\sigma,1\}}$ and an $a\in K$ such that $p=g+d+a$.

If $a_{n}\to a$ note that $p=a+g+d\in[P]_{\max\{\sigma,1\}}$ by Lemma 1. ∎

The next concept is used to build approximations of the sequential group topology.

Definition 5**.**

Let $(X,\tau)$ be a topological space. Then $(X,\tau)$ is called $\mathop{k_{\omega}}$ if there exists a countable family ${\cal K}$ of subspaces of $X$ such that $U\in\tau$ if and only if $U\cap K$ is relatively open in $K$ for every $K\in{\cal K}$. We say that $\tau$ is determined by ${\cal K}$ and write $\tau=\mathop{k_{\omega}}({\cal K})$.

A rich source of $\mathop{k_{\omega}}$ topologies on $X$ is provided by the following well known construction. Let ${\cal K}$ be a countable family of subsets of $X$ such that each $K\in{\cal K}$ is endowed with a compact topology $\tau_{K}$. Define a new topology $\tau=\mathop{k_{\omega}}({\cal K})={\{\,U:\forall K\in{\cal K}\;(U\cap K)\in\tau_{K}\,\}}$. It is easy to see that such $\tau$ is automatically $\mathop{k_{\omega}}$. The construction above makes sense for uncountable ${\cal K}$ as well. We will use $\mathop{k_{\omega}}({\cal K})$ to denote the appropriate topology even though for an uncountable ${\cal K}$, the topology $\mathop{k_{\omega}}({\cal K})$ is not necessarily $\mathop{k_{\omega}}$.

Note that for an arbitrary ${\cal K}$ such $\tau$ is not guaranteed to be Hausdorff (although it is always $T_{1}$ provided each $\tau_{K}$ is), nor does $\tau_{K}$ necessarily coinside with the topology inherited by $K$ from $\tau$. Both of these properites are readily ensured by starting with a countable family ${\cal K}$ of compact subspaces of $X$ in some (not necessarily $\mathop{k_{\omega}}$) topology $\tau^{\prime}$ on $X$ and taking $\tau_{K}$ to be the appropriate subspace topology.

Group $\mathop{k_{\omega}}$ topologies have been well studied and are useful building blocks for various examples of sequential groups (see, for example [14] and the references therein).

The following simple lemma demonstrates a straightforward way to build $\mathop{k_{\omega}}$ group topologies (see [21], Lemma 4 for a proof of a more general statement).

Lemma 3**.**

Let $G$ be a boolean topological group and ${\cal K}$ be a countable family of compact subspaces of $G$ closed under finite sums. Then $\mathop{k_{\omega}}({\cal K})$ is a Hausdorff group topology on $G$. Moreover, $\mathop{k_{\omega}}({\cal K})$ is the finest group topology on $G$ in which each $K\in{\cal K}$ remains compact.

2 Basic definitions.

Most known constructions of nontrivial sequential spaces use a technique that separates the analysis of convergence from that of the topology. Among the tools used to deal with convergence, the standard test spaces (such as the sequential fan $S(\omega)$, Arens’ space $S_{2}$, Archangel’skii-Franklin space $S_{\omega}$, etc.) feature prominently. Additional ordinal invariants, such as the Cantor-Bendixson index, scatteredness rank (see [24], [18]), etc. are often used to bound the sequential order of ‘intermediate spaces’ in the construction.

To outline the reasons why such methods are of limited utility for the problem in this paper consider the following argument.

Lemma 5 in [21] states that given a $\mathop{k_{\omega}}$ boolean group $G$ and a closed discrete subset $D\subseteq G$ one can find a coarser $\mathop{k_{\omega}}$-topology on $G$ in which $D$ has a limit point. This suggests the following brute force strategy for building a countably compact sequential group that is not Fréchet.

Consider a subgroup $G$ of $2^{\mathfrak{c}}$ (algebraically) generated by a subspace homeomorphic to a compact sequential space $K$ of sequential order $\geq 2$ (for example, the one point compactification of the well-known Mrowka’s space $\psi^{*}$). Endow $G$ with the natural $\mathop{k_{\omega}}$ topology determined by the family ${\cal K}$ of all the iterated sums of $K$.

Recursively (using an appropriate set-theoretic principle such as $\diamondsuit$, see [21] for details of similar constructions) add new convergent sequences using the lemma from [21] mentioned above to make $G$ countably compact. The topology determined by all the added compact subspaces together with the original family ${\cal K}$ will be sequential and countably compact providing the desired example.

It is instructive to see why the naive approach above fails. If such a group topology $\tau$ on $G$ existed it would be easy to find a quotient $G^{\prime}$ of $G$ such that the quotient topology on $G^{\prime}$ has a countable pseudocharacter. As was shown in [21] $G^{\prime}$ will then necessarily be countable (at least with the $K$ chosen above) thus yielding a contradiction, since $G^{\prime}$ must be countably compact. In fact, it is still an open question whether it is consistent with the axioms of ZFC that a countably compact sequential group may contain a compact subspace of sequential order $\geq 2$.

A more detailed analysis of countably compact (boolean) groups helps to reveal the main source of difficulties with the approach above.

Recall that a topological group $G$ is called precompact (or totally bounded) if it can be embedded as a subgroup in some compact group. The following lemma follows from Pontryagin’s duality for compact abelian groups and the well-known characterization of precompact groups. Since this result will never be used directly, its proof is omitted.

Lemma 4**.**

Every countably compact group is precompact. Every precompact (thus every countably compact) boolean group has a linear topology, i.e. has a base of neighborhoods of [math] consisting of (clopen) subgroups of (necessarily) finite index.

The construction in [21] that ‘forces’ $D$ to aquire a limit point adds a single convergent sequence to the original topology of $G$ along with all of the iterated sums of such sequence. Unless some precautions with the choice of the new convergent sequence (as well as its limit) are taken, it may easily destroy the precompactness of any topology on $G$ compatible with the $\mathop{k_{\omega}}$-topology.

The construction in this paper uses two separate topologies, one to deal with the convergence, and the other to ensure that the limit space is a topological group: a $\mathop{k_{\omega}}$ topology and a coarser precompact first countable metrizable topology (see Definition 7 below), respectively. In order to keep the topologies compatible throughout the construction, instead of using the standard test spaces to estimate the sequential order of points, estimates of the sequential order of some homogeneous countable subspaces are used instead (the odd subspace, see Definitions 4 and 11).

In the arguments below it will be convenient to consider $\mathop{k_{\omega}}$ groups together with the countable family of compact subsets that determine the topology. We therefore introduce the following shortcut.

Definition 6**.**

Call $(G,{\cal K})$ a $\mathop{k_{\omega}}$-pair (with respect to $\tau$) if $(G,\tau)$ is a boolean topological group with the $\mathop{k_{\omega}}$ topology $\tau$ and ${\cal K}$ is a countable family of compact subspaces of $G$ closed under finite sums and intersections such that $\tau=\mathop{k_{\omega}}({\cal K})$ and $\cup{\cal K}=G$.

In almost every case, the existence of the Hausdorff topology $\tau$ will be clear from the context and will not be discussed while referring to a $\mathop{k_{\omega}}$-pair.

As was noted above, any inductive construction of a countably compact group must provide a mechanism for ensuring the precompactness of the final topology. The following definition is used throughout and forms one of the building blocks of the construction.

Definition 7**.**

Call ${(G,{\cal K},{\cal U})}$ a convenient triple if $G$ is a boolean group, ${\cal U}$ is a countable family of subgroups closed under finite intersections that forms an open base of neighborhoods of [math] in some Hausdorff precompact topology $\tau({\cal U})$ on $G$, and ${\cal K}$ is a countable family of compact (in $\tau({\cal U})$) subgroups of $G$ closed under finite sums and intersections such that $\cup{\cal K}=G$.

Trivially, if ${(G,{\cal K},{\cal U})}$ is a convenient triple then $(G,{\cal K})$ is a $\mathop{k_{\omega}}$-pair (with respect to $\mathop{k_{\omega}}({\cal K})$) and every $K\in{\cal K}$ is metrizable.

A number of arguments involve translating various subsets by compact subspaces. The definition below lists a few ordinal invariants that measure the effects of such shifts.

Definition 8**.**

Let $(G,{\cal K})$ be a $\mathop{k_{\omega}}$-pair, $A\subseteq G$. Suppose $D\subseteq G$ is countable and there exist an $\alpha<\omega_{1}$ and $K\in{\cal K}$ such that $D\subseteq[A]_{\alpha}+K$. Call the smallest $\alpha$ with this property the ${\cal K}$-depth of $D$ over $A$ and write ${\bf h}_{A}^{{\cal K}}({D})=\alpha$. If no such $\alpha$ exists ${\bf h}_{A}^{{\cal K}}({D})$ is defined to be $\omega_{1}$.

Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of subsets of $G$. Call $m\in\omega$ the ${\cal K}$-depth of $D$ over ${\cal C}$ and write ${\bf h}_{{\cal C}}^{{\cal K}}({D})=m$ if $m\in\omega$ is the smallest such that $D\subseteq\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}+K$ for some $K\in{\cal K}$. If no such $m$ exists ${\bf h}_{{\cal C}}^{{\cal K}}({D})=\omega$.

Let $0\in\overline{P}^{\mathop{k_{\omega}}({\cal K})}$ for some $P\subseteq G$. Call $m\in\omega$ the asymptotic ${\cal K}$-depth of $P$ over ${\cal C}$ and write ${{}^{\omega}{\bf h}_{{\cal C}}^{{\cal K}}({P})}=m$ if $m\in\omega$ is the smallest such that $0\in\overline{P\cap(\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}+K)}$ for some $K\in{\cal K}$. If no such $m$ exists ${{}^{\omega}{\bf h}_{{\cal C}}^{{\cal K}}({P})}=\omega$.

The following property is an immediate corollary of the definition above.

Lemma 5**.**

Let $(G,{\cal K})$ be a $\mathop{k_{\omega}}$-pair, $A\subseteq G$, and $P,D\subseteq G$ be countable subsets. Let $P\subseteq D+K$ for some $K\in{\cal K}$. Then ${\bf h}_{A}^{{\cal K}}({P})\leq{\bf h}_{A}^{{\cal K}}({D})$.

Note that adding or removing finitely many points does not change the ${\cal K}$-depth of a set. Combined with the lemma above it follows that $D\subseteq^{*}D^{\prime}$ implies ${\bf h}_{A}^{{\cal K}}({D})\leq{\bf h}_{A}^{{\cal K}}({D^{\prime}})$.

We now define some minimality properites of sets with respect to their ${\cal K}$-depth.

Definition 9**.**

Let $(G,{\cal K})$ be a $\mathop{k_{\omega}}$-pair, $A\subseteq G$, ${\cal A}={\{\,A_{i}:i\in\omega\,\}}$ be a family of subsets of $G$, and $D\subseteq G$ be a countable subset. Let ${\bf h}_{A}^{{\cal K}}({D})=\delta\leq\omega_{1}$. Define {labeling}${\mathbb{M}}(A,{\cal K})$

for any $\beta<\delta$, $K\in{\cal K}$ there exists a finite $F_{K,\beta}\subseteq D$ such that

[TABLE]

{labeling}

${\mathbb{M}}(A,{\cal K})$**

for any $n\in\omega$, $K\in{\cal K}$ there exists a finite $F_{K,n}\subseteq D$ such that

[TABLE]

By choosing $F_{K,\beta}=F_{K,n}$ above for any $K\in{\cal K}$ and any $\beta<\omega_{1}$ one shows that ${\mathbb{M}}({\cal A},{\cal K})$ implies ${\mathbb{M}}({\langle{A_{n}}\rangle},{\cal K})$ and ${\bf h}_{{\langle{A_{n}}\rangle}}^{{\cal K}}({D})=\omega_{1}$ for every $n\in\omega$. Also note that both properties imply that ${\langle{D}\rangle}\cap K$ is finite for every $K\in{\cal K}$ so ${\langle{D}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K})$.

The following lemma will be part of most ‘thinning out’ arguments below.

Lemma 6**.**

Let $(G,{\cal K})$ be a $\mathop{k_{\omega}}$-pair, $D^{\prime}\subseteq G$ be infinite, closed and discrete in $\mathop{k_{\omega}}({\cal K})$, and let ${\cal A}={\{\,A_{i}:i\in\omega\,\}}$ be a family of subsets of $G$.

There exists an infinite independent $D\subseteq D^{\prime}$ such that ${\langle{D}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ and $D$ satisfies ${\mathbb{M}}(A_{i},{\cal K})$ for every $i\in\omega$. If ${\bf h}_{{\cal A}}^{{\cal K}}({D^{\prime}})=\omega$ then $D$ can be chosen to satisfy ${\mathbb{M}}({\cal A},{\cal K})$.

Proof.

Passing to a subset if necessary, assume that $D^{\prime}$ is countable. Let ${\bf h}_{A_{0}}^{{\cal K}}({D^{\prime}})=\sigma_{0}\leq\omega_{1}$. Suppose $0<\sigma_{0}\leq\omega_{1}$ and pick $\sigma_{n}^{0}\to\sigma_{0}$ if $\sigma_{0}<\omega_{1}$ is a limit ordinal and $\sigma_{n}^{0}=\sigma_{0}-1$ otherwise. If $\sigma_{0}=\omega_{1}$ let $\sigma_{n}^{0}=\omega_{1}$.

Select points $d_{n}\in D^{\prime}$ so that

[TABLE]

If the recursion terminates at some $n\in\omega$ then $D^{\prime}\subseteq[A_{0}\cup\{0\}]_{\sigma_{n}^{0}}+K$ for some $K\in{\cal K}$. If $\sigma_{n}^{0}<\omega_{1}$ then ${\bf h}_{A_{0}}^{{\cal K}}({D^{\prime}})\leq\sigma_{n}^{0}<\sigma_{0}$ contradicting the choice of $\sigma_{0}$. If $\sigma_{n}^{0}=\omega_{1}$ then $\sigma_{0}=\omega_{1}$. Since $D^{\prime}$ is countable there is a $\sigma^{\prime}<\omega_{1}$ such that $D^{\prime}\subseteq[A_{0}]_{\sigma^{\prime}}^{\mathop{k_{\omega}}({\cal K})}+K$ contradicting ${\bf h}_{A_{0}}^{{\cal K}}({D^{\prime}})=\omega_{1}$.

Let $D_{0}={\{\,d_{n}:n\in\omega\,\}}$ and $D\subseteq^{*}D_{0}$. Let $K=K_{i^{\prime}}\in{\cal K}$, $\beta<\sigma_{0}$, $\sigma_{m}^{0}\geq\beta$. By picking $i\geq i^{\prime}$ large enough if necessary we may assume that ${\langle{D\setminus D_{0}}\rangle}+K\subseteq\sum_{i^{\prime}<i}K_{i^{\prime}}$. Let $n>\max\{i,m\}$. Put $F_{K}={\{\,d_{j}:j<n\,\}}$ and let $d\in{\langle{D}\rangle}\setminus{\langle{F_{K}}\rangle}$. Then $d=d_{n^{\prime}}+d^{\prime}+a$ where $n^{\prime}\geq n$, $a\in{\langle{D\setminus D_{0}}\rangle}$, and $d^{\prime}\in{\langle{{\{\,d_{j}:j<n^{\prime}\,\}}}\rangle}$. Now $d_{n^{\prime}}\not\in\sum_{i^{\prime}<i}K_{i^{\prime}}+d^{\prime}+[A_{0}\cup\{0\}]_{\beta}^{\mathop{k_{\omega}}({\cal K})}$ so $d\not\in K+[A_{0}\cup\{0\}]_{\beta}^{\mathop{k_{\omega}}({\cal K})}$. Thus ${\langle{D}\rangle}\cap K$ is finite for every $K\in{\cal K}$ so ${\langle{D}\rangle}$ is closed in ${\mathop{k_{\omega}}({\cal K})}$, ${\bf h}_{A_{0}}^{{\cal K}}({D})=\sigma_{0}$, and ${\mathbb{M}}(A_{0},{\cal K})$ holds for every infinite $D\subseteq^{*}D_{0}$. The argument for $\sigma_{0}=0$ is similar, replacing $[A_{0}\cup\{0\}]_{\sigma_{n}^{0}}^{\mathop{k_{\omega}}({\cal K})}$ with $\{0\}$. If ${\bf h}_{{\cal A}}^{{\cal K}}({D})=\omega$ replace $[A_{0}\cup\{0\}]_{\sigma_{n}^{0}}^{\mathop{k_{\omega}}({\cal K})}$ with $\sum_{i\leq n}\overline{{\langle{A_{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$.

Repeatedly using the construction above construct $D_{0}\supseteq D_{1}\supseteq\cdots\supseteq D_{n}\supseteq\cdots$ such that ${\langle{D_{i}}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ for every $i\in\omega$ and ${\mathbb{M}}(A_{i},{\cal K})$ holds for every infinite $D\subseteq^{*}D_{i}$. Let $D\subseteq^{*}D_{i}$ for every $i\in\omega$. If $D_{0}$ satisfies ${\mathbb{M}}({\cal A},{\cal K})$ put $D=D_{0}$. ∎

The remark after Definition 9 now gives the following corollary (after selecting a trivial $A=\{0\}$).

Corollary 2**.**

Let $(G,{\cal K})$ be a $\mathop{k_{\omega}}$-pair, $D\subseteq G$ be a countable closed and discrete subspace of $G$. Then there exists an infinite $D^{\prime}\subseteq D$ such that ${\langle{D^{\prime}}\rangle}$ is closed and discrete in $G$.

Lemma 7**.**

Let $(G,{\cal K},{\cal U})$ be a convenient triple and let ${\cal A}={\{\,A_{i}:i\in\omega\,\}}$ be a family of subsets of $G$. Let $P\subseteq G$ be a countable subset such that $0\in\overline{P}^{\tau({\cal U})}$ and ${\bf h}_{{\cal A}}^{{\cal K}}({P\cap U})=\omega$ for every $U\in{\cal U}$ (in particular, if ${{}^{\omega}{\bf h}_{{\cal A}}^{{\cal K}}({P})}=\omega$).

Then there exists an $S\subseteq P$ such that $S\to 0$ in $\tau({\cal U})$ and $S$ satisfies ${\mathbb{M}}({\cal A},{\cal K})$.

Proof.

Let ${\cal K}={\{\,K_{n}:n\in\omega\,\}}$ and ${\cal U}={\{\,U_{n}:n\in\omega\,\}}$. Select points $p_{n}\in P\cap\cap_{i\leq n}U_{i}$ so that

[TABLE]

Since ${\bf h}_{{\cal A}}^{{\cal K}}({P\cap U})=\omega$ for every $U\in{\cal U}$ the choice of $p_{n}$ is always possible. Verifying that $S={\{\,p_{n}:n\in\omega\,\}}\to 0$ in $\tau({\cal U})$ is routine. The proof that $S$ satisfies ${\mathbb{M}}({\cal A},{\cal K})$ is is the same as in Lemma 6.

To see that the condition of the lemma follows from ${{}^{\omega}{\bf h}_{{\cal A}}^{{\cal K}}({P})}=\omega$, let $m\in\omega$, $K\in{\cal K}$, and $U\in{\cal U}$. Then ${{}^{\omega}{\bf h}_{{\cal A}}^{{\cal K}}({P})}=\omega$ implies

[TABLE]

therefore $P\cap U\not\subseteq\sum_{i\leq m}\overline{{\langle{A_{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}+K$. Thus ${\bf h}_{{\cal A}}^{{\cal K}}({P\cap U})=\omega$. ∎

3 ‘Convexity’ and extensions in boolean groups

Given a convenient triple ${(G,{\cal K},{\cal U})}$, a common operation is to extend $G$ and ${\cal K}$ by adding points from the compact completion of $(G,\tau({\cal U}^{\prime}))$ where ${\cal U}^{\prime}$ extends ${\cal U}$. The convergence properties of the new convenient triple ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ will depend on the precompact topology whose extension produces $G^{\prime}$. This section lists a few results to help control these properties.

The following lemma may be viewed as a boolean $\mathop{k_{\omega}}$ version of the classical Hahn-Banach theorem.

Lemma 8**.**

Let $(G,{\cal K})$ be a $\mathop{k_{\omega}}$-pair such that every $K\in{\cal K}$ is a (compact) subgroup of $G$, let $H$ be a subgroup of $G$ closed in ${\mathop{k_{\omega}}({\cal K})}$, and $g\in G\setminus H$. Then there exists a subgroup $U\subseteq G$ of finite index open in ${\mathop{k_{\omega}}({\cal K})}$ such that $H\subseteq U$ and $g\not\in U$.

Proof.

Let ${\cal K}={\{\,K_{i}:i\in\omega\,\}}$. Build by induction a sequence of closed subgroups $H\subseteq H_{0}\subseteq\cdots\subseteq H_{n}\subseteq\cdots$ such that $H_{n+1}=H_{n}+K_{n+1}^{\prime}$ for some subgroup $K_{n+1}^{\prime}\subseteq K_{n+1}$ clopen in $K_{n+1}$, and $g\not\in H_{n}$. If $H_{n}$ has been built, $g\not\in H_{n}$ so there exists a clopen subgroup $K_{n+1}^{\prime}\subseteq K_{n+1}$ such that $g\not\in H_{n+1}=H_{n}+K_{n+1}^{\prime}$. Then $H_{n+1}$ is a closed subgroup of $G$ and $H_{n+1}\cap K_{i}$ is relatively open in $K_{i}$ for every $i\leq n+1$.

Let $H^{\prime}=\cup_{n\in\omega}H_{n}$. Now the intersection $H^{\prime}\cap K$ is relatively open in $K$ for every $K\in{\cal K}$ so $H^{\prime}$ is an open subgroup of $G$ in ${\mathop{k_{\omega}}({\cal K})}$. Let $G^{\prime}\subseteq G$ be a subgroup such that $G^{\prime}\cap{\langle{H^{\prime}\cup\{g\}}\rangle}=\{0\}$ and $G^{\prime}+{\langle{H^{\prime}\cup\{g\}}\rangle}=G$. Then $g\not\in U=H^{\prime}+G^{\prime}$ is open in ${\mathop{k_{\omega}}({\cal K})}$ and of finite index. ∎

In the statement of the next lemma we abuse the notation to use $\tau({\cal U}_{0})$ for both the topology of the compact completion of $G$, as well as the topology on $G$ generated by ${\cal U}_{0}$.

Lemma 9**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, let $H$ be a subgroup of $G$ closed in ${\mathop{k_{\omega}}({\cal K})}$. Then there exists a countable family of open (in $\mathop{k_{\omega}}({\cal K})$) subgroups of finite index ${\cal U}_{0}\supseteq{\cal U}$ such that $\overline{H}^{\tau({\cal U}_{0})}\cap G=\cap{\{\,U\in{\cal U}_{0}:H\subseteq U\,\}}=H$.

Proof.

Let ${\cal K}={\{\,K_{n}:n\in\omega\,\}}$. For each $K\in{\cal K}$ let ${\{\,U_{n}^{K}\subseteq K:n\in\omega\,\}}$ be a countable base of neighborhoods of [math] in $K$ consisting of clopen subgroups of finite index. The family ${\cal K}_{n}^{K}={\{\,U_{n}^{K}+a:a\in K\,\}}$ is finite so ${\cal K}^{+}=\cup_{m,n\in\omega}{\cal K}_{n}^{K_{m}}$ is countable. For each $K\in{\cal K}^{+}$, let $U(K)\subseteq G$ be an open in ${\mathop{k_{\omega}}({\cal K})}$ subgroup of finite index such that $H\subseteq U(K)$ and $H\cap K=\varnothing$ if such $U(K)$ exists. Let $U(K)=G$ otherwise. Let ${\cal U}_{0}$ be the closure of ${\cal U}\cup{\{\,U(K):K\in{\cal K}^{+}\,\}}$ under finite intersections.

Let $g\in G\setminus H$. By Lemma 8 there exists an open in ${\mathop{k_{\omega}}({\cal K})}$ subgroup $U$ of finite index such that $H\subseteq U$ and $g\not\in U$. Let $K^{\prime}\in{\cal K}$ be such that $g\in K^{\prime}$ and let $U_{n}^{K^{\prime}}$ be such that $U_{n}^{K^{\prime}}\subseteq K^{\prime}\cap U$. Then $K=g+U_{n}^{K^{\prime}}\in{\cal K}^{+}$ and $U\cap K=\varnothing$ so $H\subseteq U(K)$, $U(K)\cap K=\varnothing$. Since $U(K)\in{\cal U}_{0}$ is clopen, $g\not\in\overline{H}^{\tau({\cal U}_{0})}$. ∎

Lemma 10**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, let $S$ be a countable independent subset of $G$, closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. Then there exist a countable family of open subgroups of finite index ${\cal U}_{0}\supseteq{\cal U}$ such that $\overline{{\langle{S^{\prime}}\rangle}}^{\tau({\cal U}_{0})}\cap G={\langle{S^{\prime}}\rangle}$ for any $S^{\prime}\subseteq S$ and $\overline{{\langle{2S}\rangle}}^{\tau({\cal U}_{0})}\cap\overline{{\langle{S}\rangle}\setminus{\langle{2S}\rangle}}^{\tau({\cal U}_{0})}=\varnothing$.

Proof.

Note that ${\langle{S^{\prime}}\rangle}=\cap{\{\,{\langle{S\setminus\{s\}}\rangle}:s\in S\setminus S^{\prime}\,\}}$ and $s\not\in{\langle{2S}\rangle}$ for any $s\in S$. Now apply Lemma 9 repeatedly to find ${\cal U}_{0}\supseteq{\cal U}$ such that $\overline{{\langle{S\setminus\{s\}}\rangle}}^{\tau({\cal U}_{0})}\cap G={\langle{S\setminus\{s\}}\rangle}$ for every $s\in S$ and $\overline{{\langle{2S}\rangle}}^{\tau({\cal U}_{0})}\cap G={\langle{2S}\rangle}$. ∎

We now define a basic extension operation used in the construction.

Definition 10**.**

Let $(G,{\cal K},{\cal U})$ and $(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})$ be convenient triples and an independent $D\subseteq G$ be such that ${\langle{D}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. Call $(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})$ a primitive sequential extension (pse for short) of $(G,{\cal K},{\cal U})$ over $D$ if the following conditions hold:

1. (1)

$G\subseteq G^{\prime}$, $\overline{U}^{\tau({\cal U}^{\prime})}\in{\cal U}^{\prime}$ for every $U\in{\cal U}$, and ${\cal K}^{\prime}$ is the closure of ${\cal K}\cup\{L\}$ under finite sums where $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$ (thus $L$ is compact in $\mathop{k_{\omega}}({\cal K}^{\prime})$); 2. (2)

$\overline{{\langle{D\setminus F}\rangle}}^{\tau({\cal U}^{\prime})}\cap G={\langle{D\setminus F}\rangle}$* for any $F$;*

Abusing the notation we will also use $\tau({\cal U}^{\prime})$ to refer to the topology on $G$ induced by $\tau({\cal U}^{\prime})$.

If the new points added to $G$ are ‘sufficiently far’ from some set $A\subseteq G$, the convergence properties at the points ‘near $A$’ may not be affected.

Lemma 11**.**

Let $(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})$ be a pse of $(G,{\cal K},{\cal U})$ over $D\subseteq G$. Let $A\subseteq G$, ${\bf h}_{A}^{{\cal K}}({D})=\delta\leq\omega_{1}$, and let $D$ satisfy ${\mathbb{M}}(A,{\cal K})$. If $\mathop{\mathfrak{so}}(a,A,\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma\leq_{L}\delta$ then $a\in G$ and $\mathop{\mathfrak{so}}(a,A,{\mathop{k_{\omega}}({\cal K})})=\sigma$.

Proof.

If $a\in[A]_{0}$ then $a\in A\subseteq G$. Suppose the Lemma has been proved for all $\sigma^{\prime}<\sigma$. Pick $a_{n}\to a$ (in $\mathop{k_{\omega}}({\cal K}^{\prime})$) such that $\mathop{\mathfrak{so}}(a_{n},A,\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma_{n}$ for some increasing $\sigma_{n}\to\sigma-1$. Then by the inductive hypothesis $a_{n}\in G$ and $\mathop{\mathfrak{so}}(a_{n},A,{\mathop{k_{\omega}}({\cal K})})=\sigma_{n}$.

Since $a_{n}\to a$ in $\mathop{k_{\omega}}({\cal K}^{\prime})$ there is a $K^{\prime}\in{\cal K}$ such that ${\{\,a_{n}:n\in\omega\,\}}\subseteq K^{\prime}+L$ where $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$. By thinning out and reindexing we may assume that $a_{n}=a^{n}+d^{n}$ where $a^{n}\in K^{\prime}$, $a^{n}\to a^{\prime}\in K^{\prime}$ and $d^{n}\in L$, $d^{n}\to d\in L$. Since $a^{n},a_{n}\in G$, by (2) $d^{n}\in{\langle{D}\rangle}$.

Let $\beta=\sigma$ if $\delta$ is limit or $\beta=\delta-1$ otherwise. Then $\beta<\delta$ and $\sigma_{n}\leq\beta$ for every $n\in\omega$. Using ${\mathbb{M}}(A,{\cal K})$ pick a finite $F\subseteq D$ such that $(d+K^{\prime})\cap[A]_{\beta}=\varnothing$ for every $d\not\in{\langle{F}\rangle}$. Then $(d+K^{\prime})\cap[A]_{\sigma_{n}}=\varnothing$ for every $n\in\omega$ and $d\not\in{\langle{F}\rangle}$.

Now $d^{n}\in{\langle{F}\rangle}$ for every $n\in\omega$. We may assume that $d^{n}=d$ for every $n\in\omega$ so $a_{n}=a^{n}+d\in K^{\prime\prime}$ for some $K^{\prime\prime}\in{\cal K}$ and $a_{n}\to a$ in ${\mathop{k_{\omega}}({\cal K})}$. It follows that $a\in G$ and $\mathop{\mathfrak{so}}(a,A,{\mathop{k_{\omega}}({\cal K})})\leq\sigma$. Since $\mathop{k_{\omega}}({\cal K})\subseteq\mathop{k_{\omega}}({\cal K}^{\prime})$, $\mathop{\mathfrak{so}}(a,A,{\mathop{k_{\omega}}({\cal K})})=\sigma$. ∎

4 Parity and separation

While Lemma 11 provides one way for preserving the sequential order at some points, it is not always possible to expect a given set to be far from a fixed witness to the sequential order. A different mechanism is needed, introduced in this section.

The next definition is a convenient way to set a lower bound on the sequential order in a $\mathop{k_{\omega}}$ group. Note that it is not required that $0\in[S]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ (or $0\in\overline{S}^{\mathop{k_{\omega}}({\cal K})}$).

Definition 11**.**

Let $(G,{\cal K},{\cal U})$ be a convenient triple and $S\subseteq G$. Let $\sigma$ be a successor. Say that ${\mathbb{S}}({\cal K},S,\sigma)$ holds if for every $K\in{\cal K}$ and every $\sigma^{\prime}<\sigma-1$ there exists a clopen subgroup ${\mathbb{U}}_{K}({\cal K},S,\sigma^{\prime})\subseteq K$ such that ${\mathbb{U}}_{K}({\cal K},S,\sigma^{\prime})\cap[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma^{\prime}}=\varnothing$.

Lemma 12**.**

Let $(G,{\cal K},{\cal U})$ be a convenient triple and $S\subseteq G$ be a countable independent subset. Let $\sigma<\omega_{1}$ be a successor ordinal. Then ${\mathbb{S}}({\cal K},S,\sigma)$ holds if and only if $\mathop{\mathfrak{so}}(0,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\mathop{k_{\omega}}({\cal K})})\geq\sigma$.

Proof.

Suppose $\mathop{\mathfrak{so}}(0,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\mathop{k_{\omega}}({\cal K})})=\sigma^{\prime}<\sigma$. Then there exist $s_{n}\to 0$ in $\mathop{k_{\omega}}({\cal K})$ such that $\mathop{\mathfrak{so}}(s_{n},{\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\mathop{k_{\omega}}({\cal K})})=\sigma_{n}\leq\sigma^{\prime}-1$. By taking a subsequence if necessary we may assume that $s_{n}\in K$ for some $K\in{\cal K}$. Since $\sigma_{n}\leq\sigma^{\prime}-1<\sigma^{\prime}\leq\sigma-1$, by ${\mathbb{S}}({\cal K},S,\sigma)$ there exists a clopen subgroup $U={\mathbb{U}}_{K}({\cal K},S,\sigma^{\prime}-1)\subseteq K$ such that $U\cap[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma^{\prime}-1}=\varnothing$. Since $0\in U$ and $s_{n}\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma^{\prime}-1}$, this contradicts $s_{n}\to 0$. Hence ${\mathbb{S}}({\cal K},S,\sigma)$ implies $\mathop{\mathfrak{so}}(0,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\mathop{k_{\omega}}({\cal K})})\geq\sigma$.

The converse will not be used so its proof is omitted. Its proof uses the property that each $K\in{\cal K}$ is first countable. ∎

The algebraic tool used to control the convergence properties in the extension is given by the following parity homomorphism.

Definition 12**.**

Let $(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})$ be a pse of $(G,{\cal K},{\cal U})$ over $D\subseteq G$, and let $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$. Define ${\bf p}^{S}:[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}\to 2$ and ${\bf p}^{L}:L\to 2$ by letting ${\bf p}^{S}(a)=0$ if $a\in[{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ and ${\bf p}^{S}(a)=1$ if $a\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$. If $d\in L$ put ${\bf p}^{L}(d)=0$ if $d\in\overline{{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}$ and ${\bf p}^{L}(d)=1$ otherwise. If $b=a+d$ for some $a\in[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ and $d\in L$ put ${\bf p}(b)={\bf p}^{S}(a)+{\bf p}^{L}(d)$.

The next lemma shows that the parity homomorphism is well defined under some conditions.

Lemma 13**.**

Let $(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})$ be a pse of $(G,{\cal K},{\cal U})$ over $D\subseteq[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ where $S\subseteq G$ is an independent set, and let $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$. If $\overline{{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}\cap\overline{{\langle{D}\rangle}\setminus{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}=\varnothing$ and ${\mathbb{S}}({\cal K},S,\sigma+1)$ holds then ${\bf p}:[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}+L\to 2$ is a well defined homomorphism, continuous on $L$.

Proof.

Since ${\langle{S}\rangle}={\langle{2S}\rangle}\cup({\langle{S}\rangle}\setminus{\langle{2S}\rangle})$ if $a\in[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ for some $\sigma<\omega_{1}$ then either $a\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ or $a\in[{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$. If $a\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}\cap[{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ then $a\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma^{\prime}}^{\mathop{k_{\omega}}({\cal K})}\cap[{\langle{2S}\rangle}]_{\sigma^{\prime}}^{\mathop{k_{\omega}}({\cal K})}$ for some successor ordinal $\sigma^{\prime}\leq\sigma$. Thus there are $a_{n}^{0}\to a$ and $a_{n}^{1}\to a$ such that $a_{n}^{0}\in[{\langle{2S}\rangle}]_{\sigma_{n}^{0}}$ and $a_{n}^{1}\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma_{n}^{1}}$ for some $\sigma_{n}^{0},\sigma_{n}^{1}\leq\sigma^{\prime}-1<\sigma$. Now $a_{n}^{0}+a_{n}^{1}\to 0$ and $a_{n}^{0}+a_{n}^{1}\in[({\langle{S}\rangle}\setminus{\langle{2S}\rangle})+{\langle{2S}\rangle}]_{\sigma^{\prime}-1}^{\mathop{k_{\omega}}({\cal K})}=[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma^{\prime}-1}^{\mathop{k_{\omega}}({\cal K})}$ by Lemma 1. Pick a $K\in{\cal K}$ such that ${\{\,a_{n}^{0}+a_{n}^{1}:n\in\omega\,\}}\subseteq K$. Then ${\mathbb{U}}_{K}({\cal K},S,\sigma^{\prime}-1)\cap{\{\,a_{n}^{0}+a_{n}^{1}:n\in\omega\,\}}=\varnothing$ contradicting $a_{n}^{0}+a_{n}^{1}\to 0$.

Thus ${\bf p}^{S}$ is well defined on $[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$. A similar argument involving Lemma 1 and $({\langle{S}\rangle}\setminus{\langle{2S}\rangle})+({\langle{S}\rangle}\setminus{\langle{2S}\rangle})={\langle{2S}\rangle}$ shows that ${\bf p}^{S}$ is a homomorphism.

It follows from the definition of ${\bf p}^{L}$ and the choice of $D$ that ${\bf p}^{L}$ is a continuous homomorphism on $L$.

If $d\in L\cap[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ then $d\in{\langle{D}\rangle}$ by (2) so ${\bf p}^{S}(d)={\bf p}^{L}(d)$ by Lemma 1 and $D\subseteq[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$. Let $a+d=a^{\prime}+d^{\prime}\in[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}+L$ where $a,a^{\prime}\in[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ and $d,d^{\prime}\in L$. Then $a+a^{\prime}=d+d^{\prime}$ so $d+d^{\prime}\in G$ and thus $d+d^{\prime}\in{\langle{D}\rangle}\subseteq[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ by (2) and Lemma 1. Now ${\bf p}^{S}(a)+{\bf p}^{S}(a^{\prime})={\bf p}^{S}(a+a^{\prime})={\bf p}^{S}(d+d^{\prime})={\bf p}^{L}(d+d^{\prime})={\bf p}^{L}(d)+{\bf p}^{L}(d^{\prime})$ so ${\bf p}^{S}(a)+{\bf p}^{L}(d)={\bf p}^{S}(a^{\prime})+{\bf p}^{L}(d^{\prime})$ and ${\bf p}$ is well defined. ∎

While the parity homomorphism is unlikely to be continuous on its domain (even if an appropriate topology is agreed upon) it satisfies the following weak continuity property.

Lemma 14**.**

Let $(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})$ be a pse of $(G,{\cal K},{\cal U})$ over $D\subseteq G$. Let $D$ satisfy ${\mathbb{M}}({\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\cal K})$ and $\overline{{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}\cap\overline{{\langle{D}\rangle}\setminus{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}=\varnothing$, and let $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$. Let ${\bf h}_{{\langle{S}\rangle}\setminus{\langle{2S}\rangle}}^{{\cal K}}({D})=\delta\leq\omega_{1}$ and $D\subseteq[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\delta}^{\mathop{k_{\omega}}({\cal K})}$ if $\delta<\omega_{1}$. Let $\alpha$ be a successor and ${\mathbb{S}}({\cal K},S,\alpha)$ hold. If $\delta<\alpha$, and $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma<\alpha$ then $g\in[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}+L$, ${\bf p}(g)=1$ where ${\bf p}:[{\langle{S}\rangle}]_{\alpha-1}^{\mathop{k_{\omega}}({\cal K})}+L\to 2$ is the homomorphism in Definition 12.

Proof.

Since $\delta<\alpha$ the homomorphism ${\bf p}:[{\langle{S}\rangle}]_{\alpha-1}^{\mathop{k_{\omega}}({\cal K})}+L\to 2$ is well defined by Lemma 13.

Suppose the Lemma has been proved for all $g^{\prime}\in G$, $\sigma^{\prime}<\sigma$ such that $\mathop{\mathfrak{so}}(g^{\prime},{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma^{\prime}$ and let $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma$.

If $\sigma\leq_{L}\delta$ then $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\mathop{k_{\omega}}({\cal K})})=\sigma$ by Lemma 11. Since $\sigma$ is a successor, assume below that $\sigma>\delta$ and let $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\prime})$ be such that $\mathop{\mathfrak{so}}(g_{n},{\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\cal K}^{\prime})=\sigma_{n}$ where $\sigma_{n}\to\sigma-1$ and $\sigma_{n}<\sigma$.

Applying the inductive hypothesis each $g_{n}\in[{\langle{S}\rangle}]_{\sigma_{n}}^{\mathop{k_{\omega}}({\cal K})}+L$, and ${\bf p}(g_{n})=1$ so there exist $a_{n}\in[{\langle{S}\rangle}]_{\sigma_{n}}^{\mathop{k_{\omega}}({\cal K})}$ and $d_{n}\in L$ such that $g_{n}=a_{n}+d_{n}$. Since $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\prime})$ there are (possibly after thinning out and reindexing) a $K^{\prime}\in{\cal K}$, $a^{n}\in K^{\prime}$, and $d^{n}\in L$ such that $a^{n}\to a$, $d^{n}\to d$ and $g_{n}=a^{n}+d^{n}$. Now $a^{n}-a_{n}=d_{n}-d^{n}=d_{n}^{\prime}$ so $d_{n}^{\prime}\in{\langle{D}\rangle}$ by (2). Thus $a^{n}=a_{n}+d_{n}^{\prime}$ where $d_{n}^{\prime}\in[{\langle{S}\rangle}]_{\delta}^{\mathop{k_{\omega}}({\cal K})}$. Therefore $a^{n}\in[{\langle{S}\rangle}]_{\sigma_{n}^{\prime}}^{\mathop{k_{\omega}}({\cal K})}$ for some $\sigma_{n}^{\prime}<\sigma$ by $\delta<\sigma$ and Lemma 1.

After picking a subsequence and reindexing, assume that ${\bf p}(a^{n})$ and ${\bf p}(d^{n})$ are constant. Then ${\bf p}(a)={\bf p}(a^{n})$, ${\bf p}(d)={\bf p}(d^{n})$ by the definition of ${\bf p}$ and Lemma 13 so ${\bf p}(g)={\bf p}(a+d)={\bf p}(a_{n}+d_{n})={\bf p}(g_{n})=1$ by Lemma 13. ∎

The main reason the parity homomorphism was defined is the proof of the following lemma that states the conditions under which the sequential order of some points is preserved across primitive sequential extensions.

Lemma 15**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, $S\subseteq G$ be such that ${\mathbb{S}}({\cal K},S,\alpha)$ holds for some successor ordinal $\alpha<\omega_{1}$. Let ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ be a pse of ${(G,{\cal K},{\cal U})}$ over $D\subseteq G$ that has the following properties.

1. (1)

$D$* satisfies ${\mathbb{M}}({\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\cal K})$, ${\bf h}_{{\langle{S}\rangle}\setminus{\langle{2S}\rangle}}^{{\cal K}}({D})=\delta$ and either $\delta\geq\alpha-1$ or $D\subseteq[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\delta}^{\mathop{k_{\omega}}({\cal K})}$;* 2. (2)

if $\delta<\alpha-1$ then $\overline{{\langle{D}\rangle}\setminus{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}\cap\overline{{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}=\varnothing$;

Then ${\mathbb{S}}({\cal K}^{\prime},S,\alpha)$ holds so that ${\mathbb{U}}_{K}({\cal K}^{\prime},S,\sigma)={\mathbb{U}}_{K}({\cal K},S,\sigma)$ for every $K\in{\cal K}$ and $\sigma<\alpha-1$.

Proof.

Let $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$, $K\in{\cal K}$, and $\sigma<\alpha-1$.

Suppose $\sigma<\delta$. Pick a finite $F_{K}\subseteq D$ using ${\mathbb{M}}({\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\cal K})$ such that $(K+d)\cap[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}=\varnothing$ for every $d\in{\langle{D}\rangle}\setminus{\langle{F_{K}}\rangle}$. Put $U_{K+L}(\sigma)=K+\overline{{\langle{D\setminus F_{K}}\rangle}}^{\tau({\cal U}^{\prime})}$.

Suppose $U_{K+L}(\sigma)\cap[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K}^{\prime})}\not=\varnothing$. Let $u\in K$, $d\in\overline{{\langle{D\setminus F_{K}}\rangle}}^{\tau({\cal U}^{\prime})}$, and $a\in G^{\prime}$ be such that $u+d=a$ where $a\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$. By Lemma 11 $a\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ so $d\in G$. Then $d\in{\langle{D\setminus F_{K}}\rangle}$ by (2) and $(K+d)\cap[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}=\varnothing$ by the choice of $F_{K}$ contradicting the choice of $a$.

Suppose $\sigma\geq\delta$. Then $\delta<\alpha-1$. Put $U_{K+L}(\sigma)={\mathbb{U}}_{K}({\cal K},S,\sigma)+\overline{{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}$. Then $U_{K+L}(\sigma)$ is a compact subgroup of finite index in $K+L$ and thus clopen in $K+L$. Suppose $U_{K+L}(\sigma)\cap[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K}^{\prime})}\not=\varnothing$. Then there exist $u\in{\mathbb{U}}_{K}({\cal K},S,\sigma)$ and $d\in\overline{{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}$ such that $u+d=a$ for some $a\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$. By Lemma 13 and (2) ${\bf p}:[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}+L\to 2$ is well defined and $a=a^{\prime}+d^{\prime}$ where $a^{\prime}\in[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$, $d^{\prime}\in L$ and ${\bf p}(a^{\prime}+d^{\prime})=1$ by Lemma 14.

Now $u=a+d$ thus $a+d\in G$. Since ${\bf p}(a^{\prime}+d^{\prime})=1$ by Lemma 14 and ${\bf p}(d)=0$ by Definition 12, ${\bf p}(a^{\prime}+d^{\prime}+d)=1$. Since $a+d\in G$ and $u=a+d=a^{\prime}+d^{\prime}+d$, $d^{\prime}+d\in L\cap G={\langle{D}\rangle}\subseteq[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ by (2) and $\delta\leq\sigma$.

Since $a^{\prime}+d^{\prime}+d\in[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ by Lemma 1 and ${\bf p}^{S}(a^{\prime}+d^{\prime}+d)={\bf p}(a^{\prime}+d^{\prime}+d)=1$, $a^{\prime}+d^{\prime}+d\in[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ by the definition of ${\bf p}^{S}$ contradicting the choice of ${\mathbb{U}}_{K}({\cal K},S,\sigma)$.

Now for every $K\in{\cal K}^{\prime}$ pick a $p(K)\in{\cal K}$ such that $K\subseteq p(K)+L$ and $p(K)=K$ if $K\in{\cal K}$. Note that ${\mathbb{U}}_{p(K)}({\cal K},S,\sigma)\subseteq U_{p(K)+L}(\sigma)$ for $K\in{\cal K}$. Put ${\mathbb{U}}_{K}({\cal K}^{\prime},S,\sigma)=U_{p(K)+L}(\sigma)\cap K$ if $K\in{\cal K}^{\prime}\setminus{\cal K}$ and ${\mathbb{U}}_{K}({\cal K}^{\prime},S,\sigma)={\mathbb{U}}_{K}({\cal K},S,\sigma)$ otherwise. ∎

Let $\gamma$ be an ordinal. Suppose for every $\sigma<\gamma$ a convenient triple ${(H^{\sigma},{\cal K}^{\sigma},{\cal U}^{\sigma})}$ is defined so that the following conditions hold:

1. (1)

$H^{\sigma^{\prime}}\subseteq H^{\sigma}$, ${\cal K}^{\sigma^{\prime}}\subseteq{\cal K}^{\sigma}$, and ${\cal U}^{\sigma^{\prime}}\subseteq{\{\,U\cap H^{\sigma^{\prime}}:U\in{\cal U}^{\sigma}\,\}}$ if $\sigma^{\prime}\leq\sigma<\gamma$; 2. (2)

$H^{\sigma^{\prime}}$ is dense in $H^{\sigma}$ in $\mathop{k_{\omega}}({\cal K}^{\sigma})$ for every $\sigma^{\prime}\leq\sigma$;

Define ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ by taking $H^{<\gamma}=\cup_{\sigma<\gamma}H^{\sigma}$, ${\cal K}^{<\gamma}=\cup_{\sigma<\gamma}{\cal K}^{\sigma}$, ${\cal U}^{<\gamma}={\{\,\overline{U}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}:U\in{\cal U}^{\sigma},\sigma<\gamma\,\}}$.

Note that in the case of a successor $\gamma$, ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}={(H^{\gamma-1},{\cal K}^{\gamma-1},{\cal U}^{\gamma-1})}$.

Lemma 16**.**

The family ${\cal U}^{<\gamma}$ forms a base of clopen subgroups of finite index for a precompact group topology $\tau({\cal U}^{<\gamma})$ on $H^{<\gamma}$ and each $H^{\sigma}$, $\sigma<\gamma$ is dense in $H^{<\gamma}$ in $\mathop{k_{\omega}}({\cal K}^{<\gamma})$. If $\gamma<\omega_{1}$ then ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ is a convenient triple.

Proof.

Let $U\in{\cal U}^{\sigma^{\prime}}$ for some $\sigma^{\prime}<\gamma$. Then $U+F=H^{\sigma^{\prime}}$ for some finite $F\subseteq H^{\sigma^{\prime}}$. Thus $\overline{U}^{\mathop{k_{\omega}}({\cal K}^{\sigma})}+F=H^{\sigma}$ for any $\sigma^{\prime}\leq\sigma<\gamma$. Therefore, $\overline{U}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}+F=H^{<\gamma}$ so $\overline{U}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}$ is a clopen subgroup of finite index in $H^{<\gamma}$.

Furthermore, if $g\in H^{<\gamma}\setminus\{0\}$ then $g\in H^{\sigma^{\prime}}$ for some $\sigma^{\prime}<\gamma$ and there exists a $U^{\prime}\in{\cal U}^{\sigma^{\prime}}$ such that $g\not\in U^{\prime}$. If $\sigma\geq\sigma^{\prime}$ then by (1) $U^{\prime}=U\cap H^{\sigma^{\prime}}$ for some $U\in{\cal U}^{\sigma}$. Thus $g\not\in\overline{U^{\prime}}^{\mathop{k_{\omega}}({\cal K}^{\sigma})}$ for any $\sigma<\gamma$. Now $g\not\in\overline{U^{\prime}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}=\cup_{\sigma<\gamma}\overline{U^{\prime}}^{\mathop{k_{\omega}}({\cal K}^{\sigma})}$. Thus $\tau({\cal U}^{<\gamma})$ forms a base of a precompact $T_{1}$ group topology on $H^{<\gamma}$ that consists of clopen subgroups. The rest of the properties are routine. ∎

The lemma below shows that iterated extensions preserve the sequential order some points have when they are added to the group.

Lemma 17**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, $S\subseteq G$ be a countable independent subset, and $\delta<\omega_{1}$ be a successor ordinal. Suppose the family ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma<\alpha\,\}}$ where $\alpha<\omega_{1}$ has the following properties.

1. (1)

${(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}$* is a primitive sequential extension of ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ over some $D_{\gamma}\subseteq H^{<\gamma}$ for every $0\leq\gamma<\alpha$ and ${(H^{<0},{\cal K}^{<0},{\cal U}^{<0})}={(G,{\cal K},{\cal U})}$;* 2. (2)

if ${\bf h}_{{\langle{S}\rangle}\setminus{\langle{2S}\rangle}}^{{\cal K}^{<\gamma}}({D_{\gamma}})=\delta_{\gamma}$ then either $\delta_{\gamma}\geq\delta-1$ or $D_{\gamma}\subseteq[{\langle{S}\rangle}\setminus{\langle{2S}\rangle}]_{\delta_{\gamma}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}$, $D_{\gamma}$ satisfies ${\mathbb{M}}({\langle{S}\rangle}\setminus{\langle{2S}\rangle},{\cal K}^{<\gamma})$, and $\overline{{{\langle{D_{\gamma}}\rangle}\setminus{\langle{2{D_{\gamma}}}\rangle}}}^{\tau({\cal U}^{\gamma})}\cap\overline{{\langle{2{D_{\gamma}}}\rangle}}^{\tau({\cal U}^{\gamma})}=\varnothing$; 3. (3)

${\mathbb{S}}({\cal K}^{\gamma},S,\delta)$* holds for every $\gamma<\alpha$;*

Suppose $g\in H^{\gamma}\setminus H^{<\gamma}$ for some $\gamma<\alpha$ and $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma<\delta$. Then $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma$.

Proof.

Suppose the statement holds for all $g^{\prime}\in H^{<\alpha}$ such that $\mathop{\mathfrak{so}}(g^{\prime},{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma^{\prime}<\sigma$ for some $\sigma<\delta$, and let $g\in H^{\gamma}\setminus H^{<\gamma}$ be such that $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma$. Let $\gamma^{\prime}$ be the smallest ordinal with the following property. There exist $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\gamma^{\prime}})$ such that $\mathop{\mathfrak{so}}(g_{n},{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma_{n}<\sigma$ such that $\sigma_{n}\to\sigma-1$, $g_{n}\in H^{\gamma_{n}}\setminus H^{<\gamma_{n}}$, and $\gamma_{n}\leq\gamma^{\prime}$ is increasing.

Note that $g_{n}$ as above exist by the definition of $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))$. By $g_{n}\to g$ there exists a $K\in{\cal K}^{\beta}$ for some $\beta<\alpha$ such that $g_{n}\in K$ for every $n\in\omega$. Thus $\gamma_{n}\leq\beta$ and $\gamma^{\prime}\leq\beta<\alpha$ is well defined.

Suppose $\gamma^{\prime}>\gamma$. Then ${(H^{\gamma^{\prime}},{\cal K}^{\gamma^{\prime}},{\cal U}^{\gamma^{\prime}})}$ is a primitive sequential extension of ${(H^{<\gamma^{\prime}},{\cal K}^{<\gamma^{\prime}},{\cal U}^{<\gamma^{\prime}})}$ over some $D_{\gamma^{\prime}}\subseteq H^{<\gamma^{\prime}}$ such that ${\bf h}_{{\langle{S}\rangle}\setminus{\langle{2S}\rangle}}^{{\cal K}^{<\gamma^{\prime}}}({D_{\gamma^{\prime}}})=\delta_{\gamma^{\prime}}$.

If $\sigma\leq_{L}\delta_{\gamma^{\prime}}$ then by Lemma 11 $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\gamma^{\prime}}))=\sigma$. Otherwise, since ${\mathbb{S}}({\cal K}^{\gamma^{\prime}},S,\delta)$ holds and $\delta_{\gamma^{\prime}}<\sigma<\delta$, ${\bf p}:[{\langle{S}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K}^{<\gamma^{\prime}})}+\overline{\langle D_{\gamma^{\prime}}\rangle}\to 2$ is defined by Lemma 13. By Lemma 14 $g=a+s$ where $\mathop{\mathfrak{so}}(a,{\langle{S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\gamma^{\prime}}))\leq\sigma$, $s\in\overline{\langle D_{\gamma^{\prime}}\rangle}^{\tau({\cal U}^{<\gamma^{\prime}})}$, and ${\bf p}(a+s)=1$. Since $g\in H^{\gamma}\subseteq H^{<\gamma^{\prime}}$, $s\in H^{<\gamma^{\prime}}$ so by (2) $s\in\langle D_{\gamma^{\prime}}\rangle$. By Lemma 1 $\mathop{\mathfrak{so}}(a+s,{\langle{S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\gamma^{\prime}}))=\max\{\sigma,\delta_{\gamma^{\prime}}\}=\sigma$. Since ${\bf p}(a+s)=1$, $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\gamma^{\prime}}))=\sigma$.

Thus we may assume that $g_{n}$ are chosen so that $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{<\gamma^{\prime}})$. Let $K\in{\cal K}^{\beta}$ for some $\beta<\gamma^{\prime}$ be such that $g_{n}\in K$. Then $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\beta})$ and $\gamma_{n}\leq\beta$ contradicting the choice of $\gamma^{\prime}$.

Thus $\gamma^{\prime}\leq\gamma$ and $g_{n}\in H^{\gamma}$ for every $n\in\omega$ so by the hypothesis $\mathop{\mathfrak{so}}(g_{n},{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma}))\leq\sigma_{n}$ and $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma$. ∎

The next lemma shows that not only is the sequential order preserved, it is preserved in a ‘uniform’ way if certain conditions are met.

Lemma 18**.**

Let $(G,{\cal K},{\cal U})$ be a convenient triple and $S\subseteq G$ be a countable independent subset. Suppose the family ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma<\alpha\,\}}$, $\alpha\leq\omega_{1}$ has the following properties:

1. (1)

conditions (1) and (2) of Lemma 17 hold; 2. (2)

${\mathbb{S}}({\cal K}^{\gamma},S,\delta)$* holds so that ${\mathbb{U}}_{K}({\cal K}^{\gamma},S,\sigma)={\mathbb{U}}_{K}({\cal K}^{\beta},S,\sigma)$ for any $K\in{\cal K}^{\beta}$, $\sigma<\delta-1$, and $\beta\leq\gamma$;*

Then ${\mathbb{S}}({\cal K}^{<\alpha},S,\delta)$ holds by defining ${\mathbb{U}}_{K}({\cal K}^{<\alpha},S,\sigma)={\mathbb{U}}_{K}({\cal K}^{\gamma},S,\sigma)$ for any $\sigma<\delta-1$ and $K\in{\cal K}^{\gamma}$.

Proof.

Let $K\in{\cal K}^{\gamma}$, $\sigma<\delta-1$, and $g\in H^{\beta}\setminus H^{<\beta}$ where $\beta,\gamma<\alpha$. Suppose $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma^{\prime}\leq\sigma$. Then by Lemma 17 $\mathop{\mathfrak{so}}(g,{\langle{S}\rangle}\setminus{\langle{2S}\rangle},\mathop{k_{\omega}}({\cal K}^{\beta}))=\sigma^{\prime}$. If $\beta\leq\gamma$ then $g\not\in{\mathbb{U}}_{K}({\cal K}^{\gamma},S,\sigma)={\mathbb{U}}_{K}({\cal K}^{<\alpha},S,\sigma)$. If $\beta>\gamma$ then $g\not\in K$. ∎

If the new points are sufficiently ‘far’ from a given set, the next lemma shows that the sequential order of the points ‘near’ the set is not affected.

Lemma 19**.**

Let $\alpha<\omega_{1}$ and let ${(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}$ be a primitive sequential extension of ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ over some $D_{\gamma}$ for every $\gamma<\alpha$. Let $A\subseteq G=H^{-1}$, ${\bf h}_{A}^{{\cal K}^{<\gamma}}({D_{\gamma}})=\omega_{1}$, and $D_{\gamma}$ satisfy ${\mathbb{M}}(A,{\cal K}^{<\gamma})$.

Then $\mathop{\mathfrak{so}}(g,A,\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\mathop{\mathfrak{so}}(g,A,\mathop{k_{\omega}}({\cal K}^{<0}))$.

Proof.

Suppose the Lemma holds for all $g^{\prime}\in H^{<\alpha}$ such that $\mathop{\mathfrak{so}}(g^{\prime},A,\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma^{\prime}<\sigma$ for some $\sigma<\omega_{1}$ and let $\mathop{\mathfrak{so}}(g,A,\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma$. Let $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{<\alpha})$ so that $\mathop{\mathfrak{so}}(g_{n},A,\mathop{k_{\omega}}({\cal K}^{<\alpha}))<\sigma$. By the hypothesis $\mathop{\mathfrak{so}}(g_{n},A,\mathop{k_{\omega}}({\cal K}^{<0}))<\sigma$. There exists a $\gamma<\alpha$ such that ${\{\,g_{n}:n\in\omega\,\}}\subseteq K\in{\cal K}^{\gamma}$ so $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\gamma})$ and $\mathop{\mathfrak{so}}(g,A,\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma$. Let $\gamma$ be the smallest such and suppose $\gamma\geq 0$.

Since ${\bf h}_{A}^{{\cal K}^{<\gamma}}({D_{\gamma}})=\omega_{1}$ and $D_{\gamma}$ satisfies ${\mathbb{M}}(A,{\cal K}^{<\gamma})$, the sequential order $\mathop{\mathfrak{so}}(g,A,\mathop{k_{\omega}}({\cal K}^{<\gamma}))=\sigma$ by Lemma 11 so there are $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\gamma^{\prime}})$ for some $\gamma^{\prime}<\gamma$ and $\mathop{\mathfrak{so}}(g_{n},A,\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\mathop{\mathfrak{so}}(g_{n},A,\mathop{k_{\omega}}({\cal K}^{<0}))<\sigma$ contradicting the choice of $\gamma$. ∎

The constructions above are intended for the case when the sequential order is reflected by a single point in the group, i.e. when the sequential order desired is a successor ordinal. In the case when the group to be constructed must have a limit sequential order, some additional tools are required.

The following lemma presents a rough idea of how one might go about handling the limit case. Note that the sequentiality of the $\Sigma$-product of sequential spaces was proved in [8], Corollary 2.5 (one can also use an argument similar to the one below).

Lemma 20**.**

Let $G$ be a $\Sigma$-product of countably compact sequential spaces $G_{\alpha}$, $\alpha<\gamma$ for some ordinal $\gamma$. Then

[TABLE]

Here the $\min$ is taken over all the possible reorderings of $I$.

Sketch of proof.

Below we refer to the ‘center’ point of each $G_{\alpha}$, as well as the corresponding points in $\Sigma G_{\alpha}$ and finite products of $G_{\alpha}$ as [math]. Let $A\subseteq G$ be such that $\overline{A}\ni 0$. Introduce a finer topology $\tau$ on $G$ by making each $G_{\alpha}$ discrete. It is well known that $\tau$ is Féchet. Consider a finite $I\subseteq\gamma$ and a basic neighborhod $U_{I}={\{\,p\in G:p(\alpha)=0\mbox{ if }\alpha\in I\,\}}$ of [math] in $\tau$. Consider the natural projection $\pi_{I}(A)$ of $A$ into the finite product $\prod_{\alpha\in I}G_{\alpha}$. Note that $\overline{\pi_{I}(A)}\ni 0$ so by [12], Theorem 2.2 $0\in[\pi_{I}(A)]_{\sigma_{I}}$ where $\sigma_{I}\leq\min\sum_{\alpha\in I}\mathop{\mathfrak{so}}(G_{\alpha})$. Using induction on $\sigma_{I}$ (and the sequential compactness of $G$) one shows that there exists an $x\in U_{I}$ such that $x\in[A]_{\sigma_{I}}$. Thus $\overline{X}^{\tau}\ni 0$ where $X={\{\,x\in[A]_{\sigma_{I}}:I\subseteq\gamma,|I|<\omega\,\}}$. Since $\tau$ is Fréchet, this completes the proof. ∎

The lemma above exhibits two obstacles to obtaining a group of a sequential order that is a limit ordinal. Consider the case of $\omega+\omega$. If the product approach suggested by the lemma above is to be followed, it is clear that infinitely many factors are required, infinitely may of which must have the sequential order above $\omega+1$. The best estimate for the sequential order of a product of $n$ of such factors (obtained in [12]) will then exceed $n\omega>\omega+\omega$. Thus one must be concerned with providing a better growth rate for the sequential order of finite products of groups.

The other obstacle is the $+1$ part of the uppper bound estimate. Even if the sequential order of finite products can be made to grow slowly, the points in the ‘last sequence’ converging to a given point in the product might still have unbounded sequential orders thus leading to a successor sequential order for the group.

It is unclear to the authors whether the examples below can be given a product structure, thus a different approach was taken. The examples ‘look like’ products of groups, although the countable product structure is destoyed.

The lemma below is a parity neutral version of Lemma 14.

Lemma 21**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let $D\subseteq G$ be such that ${\bf h}_{{\langle{A}\rangle}}^{{\cal K}}({D})=\alpha_{\cal K}\leq\omega_{1}$. Suppose $D\subseteq\overline{{\langle{A}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ if $\alpha_{\cal K}<\omega_{1}$, and $D$ satisfies ${\mathbb{M}}({\langle{A}\rangle},{\cal K})$. Let ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ be a primitive sequential extension of ${(G,{\cal K},{\cal U})}$ over $D$.

If $g\in\overline{{\langle{A}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$ then $g=a+s$ where $a\in\overline{\langle A\rangle}^{\mathop{k_{\omega}}({\cal K})}$ and $s\in L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$.

Proof.

Let $\mathop{\mathfrak{so}}(g,{\langle{A}\rangle},\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma$. If $\sigma\leq_{L}\alpha_{\cal K}$ by Lemma 11 $\mathop{\mathfrak{so}}(g,{\langle{A}\rangle},{\mathop{k_{\omega}}({\cal K})})=\sigma$. Suppose the statement holds for all $g^{\prime}\in G$ such that $\mathop{\mathfrak{so}}(g^{\prime},{\langle{A}\rangle},\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma^{\prime}<\sigma$ where $\sigma\not\leq_{L}\alpha_{\cal K}$.

Pick $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\prime})$ such that $\mathop{\mathfrak{so}}(g_{n},{\langle{A}\rangle},\mathop{k_{\omega}}({\cal K}^{\prime}))=\sigma_{n}<\sigma$. By the hypothesis $g_{n}=g^{n}+s_{n}$ where $g^{n}\in\overline{{\langle{A}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ and $s_{n}\in L$. Since $g_{n}\to g$ one can find a $K\in{\cal K}$, $a_{n}\in K$, $s^{n}\in L$ such that $g_{n}=a_{n}+s^{n}$, $a_{n}\to a$, $s^{n}\to s$. Then $d_{n}=s_{n}+s^{n}=g^{n}+a_{n}\in G$ so by (2) $d_{n}\in{\langle{D}\rangle}$. Thus $a_{n}=g^{n}+d_{n}$ and $a_{n}\in\overline{{\langle{A}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ by Lemma 1. Thus $g=a+s$ where $a\in\overline{{\langle{A}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ and $s\in L$. ∎

The next definition itroduces a product-like structure into the group. The preservation of this structure is the subject of several lemmas that follow.

Definition 13**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of countable subsets of $G$. Call ${\cal C}$ discrete separated by ${\cal K}$ in $G$ if for any infinite, closed and discrete $D^{i}={\{\,d_{n}^{i}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$, $i\in I$ where $I\subseteq\omega$ is finite the set ${\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}\not\subseteq K$ for any $K\in{\cal K}$ (equivalently, ${\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}$ is infinite, closed, and discrete).

The following lemma follows from the definition of discrete separation.

Lemma 22**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of countable subsets of $G$ discrete separated by ${\cal K}$ in $G$.

Let $I\subseteq\omega$ be finite and let $A_{i}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ be closed (in ${\mathop{k_{\omega}}({\cal K})}$) subsets of $G$ for $i\in I$. Then $\sum_{i\in I}A_{i}$ is closed in ${\mathop{k_{\omega}}({\cal K})}$.

Lemma 23**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of countable subsets of $G$ discrete separated by ${\cal K}$ in $G$.

Let $D\subseteq G$ be a countable closed discrete subspace such that $D$ satisfies property ${\mathbb{M}}({\langle{S^{i}}\rangle},{\cal K})$ for every $i\in\omega$, and $D\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ for some $m\in\omega$. Let ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ be a primitive sequential extension of ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ over $D$. Then ${\cal C}$ is discrete separated by ${\cal K}^{\prime}$ in $G^{\prime}$.

Proof.

Suppose there exist infinite closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\prime})$ subspaces $D^{i}={\{\,d_{n}^{i}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$ where $i\in I$ for some finite $I\subseteq\omega$, and a $K^{\prime}\in{\cal K}^{\prime}$ such that ${\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}\subseteq K^{\prime}$.

Suppose $i\not=m$ and ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}}({D})<\omega_{1}$. Let $D={\{\,p_{n}:n\in\omega\,\}}$ and $p_{n}=a_{n}^{\prime}+p_{n}^{\prime}$ where $a_{n}^{\prime}\in K^{\prime\prime}$ for some $K^{\prime\prime}\in{\cal K}$ and $p_{n}^{\prime}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$. Passing to a subset, if necessary, we may assume that $p_{n}^{\prime}\not=p_{k}^{\prime}$ if $n\not=k$, so $D^{\prime}={\{\,p_{n}^{\prime}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ is infinite, closed, and discrete in $\mathop{k_{\omega}}({\cal K})$. Then ${\{\,p_{n}+p_{n}^{\prime}:n\in\omega\,\}}={\{\,a_{n}^{\prime}:n\in\omega\,\}}\subseteq K^{\prime\prime}$ contradicting the property that ${\cal C}$ is discrete separated by ${\cal K}$. Hence ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}}({D})=\omega_{1}$ for every $i\not=m$. By Lemma 11 $d_{n}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}\subseteq G$ for $i\not=m$.

Let $K^{\prime}\subseteq K+L$ where $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$ and $K\in{\cal K}$. Now $\sum_{i\in I}d_{n}^{i}=a_{n}+d_{n}$ where $a_{n}\in K$ and $d_{n}\in L$. If $m\not\in I$ then $d_{n}\in G$ by the argument in the preceeding paragraph, so by (2) and the choice of $D$, $d_{n}\in{\langle{D}\rangle}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$. By passing to a subset if necessary, we may assume that the set $D^{m}={\{\,d_{n}=d_{n}^{m}:n\in\omega\,\}}$ is either infinite, closed, and discrete, or is a subset of some $K^{\prime\prime}\in{\cal K}$. In the first case ${\{\,\sum_{i\in I\cup\{m\}}d_{n}^{i}:n\in\omega\,\}}\subseteq K$, while in the second case ${\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}\subseteq K+K^{\prime\prime}\in{\cal K}$, contradicting the property that ${\cal C}$ is discrete separated. Thus we may assume $m\in I$.

Since $L\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$, the set ${\{\,d_{n}^{m}+d_{n}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$ is also closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\prime})$ so we may assume, after replacing $D^{m}$, if necessary, that $\sum_{i\in I\setminus\{m\}}d_{n}^{i}+d_{n}^{m}=a_{n}\in G$. Then $d_{n}^{m}\in G$ and by Lemma 21 $d_{n}^{m}=d_{n}^{\prime}+s_{n}$ where $d_{n}^{\prime}\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ and $s_{n}\in L$. Since $d_{n}^{m}\in G$, $s_{n}\in{\langle{D}\rangle}$ by (2) so $D^{m}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ by Lemma 1. Now $D^{i}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ are closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ contradicting the assumption that ${\cal C}$ is discrete separated in $G$. ∎

Lemma 24**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of countable subsets of $G$ discrete separated by ${\cal K}$ in $G$. Let $D\subseteq G$ be a countable closed discrete subspace that satisfies ${\mathbb{M}}({\cal C},{\cal K})$.

Let ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ be a primitive sequential extension of ${(G,{\cal K},{\cal U})}$ over $D$. Then ${\cal C}$ is discrete separated by ${\cal K}^{\prime}$ in $G^{\prime}$.

Proof.

Note that ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}}({D})=\omega_{1}$ for every $i\in\omega$.

Suppose there exist infinite closed and discrete subspaces $D^{i}={\{\,d_{n}^{i}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$ such that ${\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}\subseteq K^{\prime}\in{\cal K}^{\prime}$. Let $K^{\prime}\subseteq K+L$ where $L=\overline{{\langle{D}\rangle}}^{\tau({\cal U}^{\prime})}$ and $K\in{\cal K}$. Now $\sum_{i\in I}d_{n}^{i}=a_{n}+d_{n}$ where $a_{n}\in K$ and $d_{n}\in L$.

By Lemma 11 $d_{n}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ for every $i\in I$, $n\in\omega$ so $d_{n}\in G$ and by (2) $d_{n}\in{\langle{D}\rangle}$. Use ${\mathbb{M}}({\cal C},{\cal K})$ to find a finite $F\subseteq D$ such that $(d+K)\cap(\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})})=\varnothing$ if $d\in{\langle{D}\rangle}\setminus{\langle{F}\rangle}$. Thus ${\{\,d_{n}:n\in\omega\,\}}\subseteq{\langle{F}\rangle}$ and ${\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}\subseteq K^{\prime\prime}$ for some $K^{\prime\prime}\in{\cal K}$ contradicting that ${\cal C}$ is discrete separated by ${\cal K}$ in $G$. ∎

It is not possible to preserve the sequential order of every new point (even if a uniform bound is imposed) in the extension so the next lemma deals with a much weaker property.

Lemma 25**.**

Let $\alpha<\omega_{1}$ and let ${(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}$ be a primitive sequential extension of ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ over some $D_{\gamma}\subseteq H^{<\gamma}$ for every $\gamma<\alpha$. Let $S^{m}\subseteq G=H^{<0}$, $m\in\omega$ be countable subsets. Let ${\bf h}_{{\langle{S^{m}}\rangle}}^{{\cal K}^{<\gamma}}({D_{\gamma}})=\delta_{\gamma}^{m}$. Suppose for every $m\in\omega$ the set $D_{\gamma}$ satisfies ${\mathbb{M}}({\langle{S^{m}}\rangle},{\cal K}^{<\gamma})$ and has the property that either $\delta_{\gamma}^{m}=\omega_{1}$ or $D_{\gamma}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}$.

If $g\in H^{\beta}\setminus H^{<\beta}$ for some $\beta<\alpha$ and $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})}$ then $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\beta})}$.

Proof.

Suppose the statement holds for all $g^{\prime}\in H^{<\alpha^{\prime}}$ such that $\mathop{\mathfrak{so}}(g^{\prime},{\langle{S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha^{\prime}})=\sigma^{\prime}$ where $(\alpha^{\prime},\sigma^{\prime})<(\alpha,\sigma)$. Let $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma$. If $\sigma=0$ then $g\in{\langle{S^{m}}\rangle}$, otherwise there are $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{<\alpha})$ such that $\mathop{\mathfrak{so}}(g_{n},{\langle{S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma_{n}<\sigma$. Since $g_{n}\to g$ there exists a $K\in{\cal K}^{\gamma}$ for some $\gamma<\alpha$ such that $g_{n}\in K$.

Suppose $\alpha$ is a successor ordinal. If $\beta=\alpha-1$ then ${\cal K}^{<\alpha}={\cal K}^{\beta}$ so the statement holds. Otherwise $\beta<\alpha-1$ and $g\in H^{<\alpha-1}$ so by Lemma 21 $g=a+s$ where $a\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha-1})}$ and $s\in\overline{{\langle{D_{\alpha-1}}\rangle}}^{\tau({\cal U}^{\alpha-1})}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\alpha-1})}$ if $\delta_{\alpha-1}^{m}<\omega_{1}$. If $\delta_{\alpha-1}^{m}=\omega_{1}$ then $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha-1})}$ by Lemma 11. Since $a,g\in H^{<\alpha-1}$, it follows that $s\in H^{<\alpha-1}$, so by (2) $s\in{\langle{D_{\alpha-1}}\rangle}$ and $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha-1})}$. Since $(\alpha-1,\sigma)<(\alpha,\sigma)$, the statement holds by the hypothesis.

If $\alpha$ is a limit ordinal $\gamma<\alpha^{\prime}$ for some $\alpha^{\prime}<\alpha$. Since $(\alpha,\sigma_{n})<(\alpha,\sigma)$ and $g_{n}\in H^{\gamma}$, by the hypothesis $g_{n}\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha^{\prime}})}$ so $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha^{\prime}})}$ and $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\beta})}$ by the hypothesis. ∎

Lemma 26**.**

Let $\alpha<\omega_{1}$ and let ${(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}$ be a primitive sequential extension of ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ over some $D_{\gamma}$ for every $\gamma<\alpha$. Let $S^{i}\subseteq G=H^{<0}$ be countable subsets such that ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ is discrete separated by ${\cal K}^{\gamma}$ in $H^{\gamma}$ for every $\gamma<\alpha$. Let ${\bf h}_{{\langle{S^{m}}\rangle}}^{{\cal K}^{<\gamma}}({D_{\gamma}})=\delta_{\gamma}^{m}$. Suppose for every $m\in\omega$ the set $D_{\gamma}$ satisfies ${\mathbb{M}}({\langle{S^{m}}\rangle},{\cal K}^{<\gamma})$ and has the property that whenever $\delta_{\gamma}^{m}<\omega_{1}$ the set $D_{\gamma}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}$.

Then ${\cal C}$ is discrete separated by ${\cal K}^{<\alpha}$ in $H^{<\alpha}$.

Proof.

Suppose the statement holds for all $\gamma<\alpha$. Let $d_{n}^{i}\in H^{\alpha_{n}^{i}}\setminus H^{<\alpha_{n}^{i}}$ and $K\in{\cal K}^{\beta}$, $\beta<\alpha$ be such that $D={\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}\subseteq K$, where $D^{i}={\{\,d_{n}^{i}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})}$ are closed discrete subspaces of $H^{<\alpha}$, $i\in I$, and $I\subseteq\omega$ is finite.

If $\gamma=\max\{{\{\,\sup\alpha_{n}^{i}:i\in I\,\}}\cup\{\beta\}\}<\alpha$ then by Lemma 25 $D^{i}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$ so ${\cal C}$ is not discrete separated in $\mathop{k_{\omega}}({\cal K}^{\gamma})$.

Suppose, say, $\alpha_{n}^{0}$ is unbounded in $\alpha$. Pick $n\in\omega$ so that $\alpha_{n}^{0}>\beta$ and let $\gamma=\max_{i\in I}\alpha_{n}^{i}$. By the inductive hypothesis, ${\cal C}$ is discrete separated by ${\cal K}^{<\gamma}$ in $H^{<\gamma}$. Since ${\langle{D_{\gamma}}\rangle}$ is closed and discrete, this implies that $\delta_{\gamma}^{m}<\omega_{1}$ for at most one $m\in\omega$ (note that $D_{\gamma}$ cannot be finite, otherwise $H^{\gamma}\setminus H^{<\gamma}=\varnothing$). Then $\gamma>\beta$ and ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}^{<\gamma}}({D_{\gamma}})=\omega_{1}$ for all $i\in I$ such that $i\not=m$ for some $m\in\omega$. By Lemma 11 $d_{n}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{{\cal K}^{<\gamma}}$ for every $i\not=m$ so $m\in I$ (otheriwse $\max_{i\in I}\alpha_{n}^{i}<\gamma$). Thus $\sum_{i\in I}d_{n}^{i}\in H^{\gamma}\setminus H^{<\gamma}$ contradicting $\sum_{i\in I}d_{n}^{i}\in K\subseteq H^{<\gamma}$. ∎

Putting the concepts introduced above together, the next definition introduces a basic ‘extension step’ of the construction.

Definition 14**.**

Let ${(G,{\cal K},{\cal U})}$ and ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ be convenient triples, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of countable subsets of $G$, and ${\{\,\sigma^{i}:i\in\omega\,\}}\subseteq\omega_{1}$ be a set of successor ordinals. Let $D\subseteq G$ be a countable independent closed and discrete subset of $G$ in ${\mathop{k_{\omega}}({\cal K})}$. Call ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ a fine primitive sequential extension (or fpse for short) of ${(G,{\cal K},{\cal U})}$ over $(D,{\cal C})$ (or $D$ if ${\cal C}$ is clear) if the following properties hold.

1. (1)

${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$* is a primitive sequential extension of ${(G,{\cal K},{\cal U})}$ over $D$, which satisfies ${\mathbb{M}}({\cal K},{\langle{S^{i}}\rangle})$ for every $i\in\omega$, and either $D$ satisfies ${\mathbb{M}}({\cal K},{\cal C})$ or $D\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ for some $m\in\omega$;* 2. (2)

if ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}}({D})=\delta_{D}^{m}<\sigma^{m}-1$ then $D\subseteq[{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}]_{\delta_{D}^{m}}^{\mathop{k_{\omega}}({\cal K})}$, $D$ satisfies ${\mathbb{M}}({\cal K},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle})$, and $\overline{{\langle{D}\rangle}\setminus{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}\cap\overline{{\langle{2D}\rangle}}^{\tau({\cal U}^{\prime})}=\varnothing$;

Definition 15**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of countable subsets of $G$. Call a family $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma<\alpha\,\}}$ of convenient triples an fpse-chain above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma<\alpha\,\}}$ relative to ${\cal C}$ if ${(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}$ is an fpse of ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ over $(D_{\gamma},{\cal C})$ where $D_{\gamma}\subseteq H^{<\gamma}$ for $\gamma>\gamma^{0}$ and ${(H^{\gamma^{0}},{\cal K}^{\gamma^{0}},{\cal U}^{\gamma^{0}})}$ is an fpse of ${(G,{\cal K},{\cal U})}$ over $(D_{\gamma^{0}},{\cal C})$ where $D_{\gamma^{0}}\subseteq G$. Call each $D_{\gamma}$ the free sequence at $\gamma$.

Lemma 27**.**

Let $\alpha<\omega_{1}$ and let ${(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}$ be a primitive sequential extension of ${(H^{<\gamma},{\cal K}^{<\gamma},{\cal U}^{<\gamma})}$ over some $D_{\gamma}\subseteq H^{<\gamma}$ for every $\gamma^{0}<\gamma<\alpha$. Let $S^{i}\subseteq H^{\gamma^{0}}$, $i\in\omega$ be such that for every $\gamma<\alpha$ there is at most one $i\in\omega$ such that ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}^{<\gamma}}({D_{\gamma}})<\omega_{1}$, $D_{\gamma}$ satisfies ${\mathbb{M}}({\langle{S^{i}}\rangle},{\cal K}^{<\gamma})$ for every $\gamma^{0}<\gamma<\alpha$ and every $i\in\omega$, and $D_{\gamma}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}$ whenever ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}^{<\gamma}}({D_{\gamma}})<\omega_{1}$.

Let $g=\sum_{i\in I}g_{i}$ where $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})}$ for some finite $I\subseteq\omega$. If $g\in H^{\gamma}\setminus H^{<\gamma}$ and $g_{i}\in H^{\gamma_{i}}\setminus H^{<\gamma_{i}}$ then $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma_{i}})}$ and $\gamma=\max_{i\in I}\gamma_{i}$.

Proof.

By Lemma 25 $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma_{i}})}$. Let $\gamma^{\prime}=\max_{i\in I}\gamma_{i}$. If $\gamma^{\prime}=\gamma^{0}$ then $\gamma_{i}=\gamma^{0}$ for every $i\in I$ so $g\in H^{\gamma^{0}}$. Otherwise ${(H^{\gamma^{\prime}},{\cal K}^{\gamma^{\prime}},{\cal U}^{\gamma^{\prime}})}$ is a primitive sequential extension of ${(H^{<\gamma^{\prime}},{\cal K}^{<\gamma^{\prime}},{\cal U}^{<\gamma^{\prime}})}$ over $D_{\gamma^{\prime}}\subseteq H^{<\gamma^{\prime}}$. If ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}^{<\gamma^{\prime}}}({D_{\gamma^{\prime}}})=\omega_{1}$ then $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma^{\prime}})}$ by Lemma 11 so $\gamma_{i}<\gamma^{\prime}$. Thus there is a unique $j\in I$ such that $\gamma_{j}=\gamma^{\prime}$ and $\gamma_{i}<\gamma^{\prime}$ for $i\not=j$. Now $g=\sum_{i\in I}g_{i}$ so $\gamma^{\prime}=\gamma$. ∎

Certain discrete separated families behave like direct sums algebraically.

Lemma 28**.**

Let ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ be a pse of ${(G,{\cal K},{\cal U})}$ over $D\subseteq G$. Let $S^{i}\subseteq G$, $i\in\omega$ be such that $\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}\cap(\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})})=\{0\}$ for any finite $I\subseteq\omega\setminus\{m\}$. Suppose there is at most one $i\in\omega$ such that ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}}({D})<\omega_{1}$ and if such $i$ exists $D\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$. Let $D$ satisfy ${\mathbb{M}}({\cal K},{\langle{S^{i}}\rangle})$ for every $i\in\omega$.

Then $\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}\cap(\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})})=\{0\}$ for any finite $I\subseteq\omega\setminus\{m\}$.

Proof.

Let $g=\sum_{i\in I}g_{i}$ where $I\subseteq\omega$ is finite, $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$ for $i\in I$, $m\not\in I$, and $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\prime})}$.

Suppose ${\bf h}_{{\langle{S^{m}}\rangle}}^{{\cal K}}({D})=\omega_{1}$. Then $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ by Lemma 11 and $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ by Lemma 27 so $g=0$ by the property of ${\cal C}$.

If ${\bf h}_{{\langle{S^{m}}\rangle}}^{{\cal K}}({D})<\omega_{1}$ and $g\in G^{\prime}\setminus G$ then ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}}({D})=\omega_{1}$ and $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ for $i\in I$. Thus $g=\sum_{i\in I}g_{i}\in G$ contradicting the assumption on $g$. Thus $g\in G$ and by Lemma 21 $g=a+s$ where $a\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ and $s\in\overline{D}^{\tau({\cal U}^{\prime})}$. Since $g\in G$, $s\in G$ and by (2) $s\in{\langle{D}\rangle}$. Thus $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ and $g=0$ by the condition on ${\cal C}$. ∎

It is convenient to have an upper bound measure for the sequential order of the construction.

Definition 16**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be a family of countable subsets of $G$ and $\xi={\{\,\sigma^{i}:i\in\omega\,\}}\subseteq\omega_{1}$ be a set of successor ordinals. Call $({\cal C},\xi)$ a sequential scale in ${(G,{\cal K},{\cal U})}$ if the following conditions are satisfied.

1. (1)

${\cal C}$* is discrete separated by ${\cal K}$ in $G$;* 2. (2)

$\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}\cap\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}=\{0\}$* for every finite $I\subseteq\omega\setminus\{m\}$;* 3. (3)

${\mathbb{S}}({\cal K},S^{i},\sigma^{i})$* holds for every $i\in\omega$;*

A trivial observation shows that a collection ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ of subsets will form a sequential scale for any $\xi\subseteq\omega_{1}$ provided $\cup_{i\in\omega}S^{i}$ is independent, and ${\langle{\cup_{i\in\omega}S^{i}}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$.

Lemma 29**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi\subseteq\omega_{1}$. Let $P\subseteq G$ be an infinite subset closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. Suppose $P\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ or ${\bf h}_{{\cal C}}^{{\cal K}}({P})=\omega$.

Then there exists an fpse ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ of ${(G,{\cal K},{\cal U})}$ such that $S\to s$ for some infinite $S\subseteq P$ and $s\in G^{\prime}\setminus G$.

Proof.

Using the observation immediately following Lemma 5 and passing to a subset if neccessary, assume that ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}}({P^{\prime}})=\sigma$ for every infinite $P^{\prime}\subseteq P$. Let $P={\{\,p_{n}:n\in\omega\,\}}$, $D={\{\,d_{n}:n\in\omega\,\}}\subseteq[{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}]_{\sigma}^{\mathop{k_{\omega}}({\cal K})}$ if $\sigma<\sigma^{m}-1$, $K\in{\cal K}$, and $a_{n},a\in K$ be such that $p_{n}=d_{n}+a_{n}$ and $a_{n}\to a$. If $\sigma\geq\sigma^{m}-1$ put $D=P$. Using Lemma 6 find an infinite $J\subseteq\omega$ such that the set $Q={\{\,d_{n}:n\in J\,\}}$ satisfies ${\mathbb{M}}({\langle{S^{i}}\rangle},{\cal K})$ for $i\in\omega$ and satisfies ${\mathbb{M}}({\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},{\cal K})$ if $\sigma<\sigma^{m}-1$. Note that ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}}({Q})=\sigma$ by the assumption on $P$ and Lemma 5.

Use Lemma 10 to find a family of clopen subgroups ${\cal U}_{0}\supseteq{\cal U}$ of finite index in $G$ such that $\overline{{\langle{Q^{\prime}}\rangle}}^{\tau({\cal U}_{0})}\cap G={\langle{Q^{\prime}}\rangle}$ for every infinite $Q^{\prime}\subseteq Q$ and $\overline{{\langle{2{Q}}\rangle}}^{\tau({\cal U}_{0})}\cap\overline{{{\langle{Q}\rangle}\setminus{\langle{2{Q}}\rangle}}}^{\tau({\cal U}_{0})}=\varnothing$ if $\sigma<\sigma^{m}-1$ (all closures with respect to $\tau({\cal U}_{0})$ are assumed to be taken in the appropriate compact completion of $G$). By passing to a subset if necessary assume $Q\to q$ in $\tau({\cal U}_{0})$ for some $q$ in the compact completion of $G$. Put $L=\overline{{\langle{Q}\rangle}}^{\tau({\cal U}_{0})}$, let $G^{\prime}=G+L$, and let ${\cal K}^{\prime}$ be the closure of ${\cal K}\cup\{L\}$ under finite sums and intersections. Let ${\cal U}^{\prime}$ be the trace of ${\cal U}_{0}$ on $G^{\prime}$. Now (1), (2), (1), and (2) follow from the construction and $S={\{\,d_{n}+a_{n}:n\in J\,\}}\to q+a=s$ in $\tau({\cal U}^{\prime})$ so ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ is the desired fpse.

If ${\bf h}_{{\cal C}}^{{\cal K}}({P})=\omega$ use Lemma 6 to find an infinite $S\subseteq P$ that satisfies ${\mathbb{M}}({\cal C},{\cal K})$ and use an argument similar to the one above to construct ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$. ∎

Lemma 30**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let $I\subseteq\omega$ be finite, and let $U\in{\cal U}$.

Then there exists a $U^{\prime}\subseteq U$ such that $(\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})})\cap U^{\prime}=\sum_{i\in I}(\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}\cap U)$ and $U^{\prime}$ is a clopen subgroup of finite index in $(G,\mathop{k_{\omega}}({\cal K}))$.

Proof.

The group $H=\sum_{i\in I}(\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}\cap U)$ is closed in ${\mathop{k_{\omega}}({\cal K})}$ by Lemma 22 and is of finite index in $\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$. Let $g_{1},\ldots,g_{k}\in G\setminus H$ be such that $H+{\{\,g_{i}:i\leq k\,\}}\supseteq\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$. Use Lemma 8 to construct an open subgroup of finite index $U_{i}$, $i=1,\ldots,k$ such that $H\subseteq U_{i}$ and $g_{i}\not\in U_{i}$. Put $U^{\prime}=\cap_{i\leq k}U_{i}\cap U$. ∎

To keep the sequential order low one must be able to construct sequences converging to some points that are already present in the group.

Lemma 31**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ and $\xi={\{\,\sigma^{i}:i\in\omega\,\}}\subseteq\omega_{1}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$. Let $I\subseteq\omega$ be finite, $D^{i}={\{\,d_{n}^{i}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ be such that $D^{i}\to 0$ in $\tau({\cal U})$ for every $i\in I$, and ${\bf h}_{{{\langle{S^{i^{\prime}}}\rangle}\setminus{\langle{2{S^{i^{\prime}}}}\rangle}}}^{{\cal K}}({D^{i^{\prime}}})\geq\sigma^{i^{\prime}}-1>0$ for some $i^{\prime}\in I$.

Then there exists an infinite $J\subseteq\omega$ and an fpse ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ of ${(G,{\cal K},{\cal U})}$ over $D={\{\,d_{n}^{i^{\prime}}:n\in J\,\}}$ such that $D\to 0$ in $\mathop{k_{\omega}}({\cal K}^{\prime})$ and ${\{\,d_{n}^{i}:n\in J\,\}}\to 0$ in $\tau({\cal U}^{\prime})$ for every $i\in I$.

Proof.

Pick an infinite $J\subseteq\omega$ such that either $D(i)={\{\,d_{n}^{i}:n\in J\,\}}\to 0$ or ${\langle{D(i)}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ and $D(i^{\prime})$ satisfies ${\mathbb{M}}({\langle{S^{i}}\rangle},{\cal K})$ for every $i\in\omega$. Since ${\bf h}_{{{\langle{S^{i^{\prime}}}\rangle}\setminus{\langle{2{S^{i^{\prime}}}}\rangle}}}^{{\cal K}}({D^{i^{\prime}}})>0$ the set ${\langle{D(i^{\prime})}\rangle}$ is infinite, closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. By taking a subset of $J$ if necessary assume that $D(i^{\prime})$ satisfies ${\mathbb{M}}({\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},{\cal K})$ and the set $\cup_{i\in I}D(i)$ is independent (using (2)). Since ${\cal C}$ is discrete separated and $D(i)\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$, the group ${\langle{\cup_{i\in I^{\prime}}D(i)}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ where $I^{\prime}={\{\,i\in I:{\langle{D(i)}\rangle}\hbox{ is closed discrete in }{\mathop{k_{\omega}}({\cal K})}\,\}}$.

Use Lemma 8 and the argument in Lemma 9 to find a ${\cal U}^{\prime}\supseteq{\cal U}$ such that for every $U\in{\cal U}^{\prime}$ and every $i\in I$ the set $D(i)\setminus U$ is finite and $\overline{{\langle{\cup_{i\in I^{\prime}}D(i)\setminus F_{i}}\rangle}}^{\tau({\cal U}^{\prime})}\cap G={\langle{\cup_{i\in I^{\prime}}D(i)\setminus F_{i}}\rangle}$ for any finite $F_{i}\subseteq G$. Then $D(i)\to 0$ in $\tau({\cal U}^{\prime})$.

Put $L=\overline{D}^{\tau({\cal U}^{\prime})}$ and let ${\cal K}^{\prime}$ be the closure of ${\cal K}\cup\{L\}$ under finite intersections and sums. Now $D(i^{\prime})\to 0$ in $\mathop{k_{\omega}}({\cal K}^{\prime})$, (1) and (2) hold, and, since ${\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}}({D})\geq\sigma^{i^{\prime}}-1$, properties (1)–(2) hold so ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ is an fpse of ${(G,{\cal K},{\cal U})}$ over $D$. ∎

The next lemma presents the basic structure of the example.

Lemma 32**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ and $\xi={\{\,\sigma^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$. Let ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma<\alpha\,\}}$ be an fpse-chain over ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma<\alpha\,\}}$ relative to ${\cal C}$.

Then $({\cal C},\xi)$ is a sequential scale in ${(H^{<\alpha},{\cal K}^{<\alpha},{\cal U}^{<\alpha})}$ and the following properties hold.

1. (1)

${\mathbb{U}}_{K}({\cal K}^{<\alpha},S^{i},\sigma)={\mathbb{U}}_{K}({\cal K}^{\gamma^{\prime}},S^{i},\sigma)={\mathbb{U}}_{K}({\cal K}^{\gamma},S^{i},\sigma)$* for every $K\in{\cal K}^{\gamma}$, $\gamma\leq\gamma^{\prime}<\alpha$, and $\sigma<\sigma^{i}-1$;* 2. (2)

if $g\in H^{\gamma}\setminus H^{<\gamma}$ for $\gamma<\alpha$ and $g\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})}$ then $g\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$; 3. (3)

if $g\in H^{\gamma}\setminus H^{<\gamma}$ for some $\gamma<\alpha$ and $\mathop{\mathfrak{so}}(g,{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\sigma<\sigma^{i}$ then $\mathop{\mathfrak{so}}(g,{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma$;

Proof.

Suppose the statement has been proved for all $\alpha^{\prime}<\alpha$. Suppose $\alpha$ is a successor ordinal such that $\alpha^{\prime}=\alpha-1$. Then by the hypothesis $({\cal C},\xi)$ is a sequential scale in ${(H^{<\alpha^{\prime}},{\cal K}^{<\alpha^{\prime}},{\cal U}^{<\alpha^{\prime}})}$ and ${(H^{<\alpha},{\cal K}^{<\alpha},{\cal U}^{<\alpha})}={(H^{\alpha^{\prime}},{\cal K}^{\alpha^{\prime}},{\cal U}^{\alpha^{\prime}})}$ is an fpse of ${(H^{<\alpha^{\prime}},{\cal K}^{<\alpha^{\prime}},{\cal U}^{<\alpha^{\prime}})}$ over some $D_{\alpha^{\prime}}\subseteq H^{<\alpha^{\prime}}$. Now (1) and (2) hold by Lemmas 23, 24, and 25, and (1). Property (2) follows from Lemma 28 and (1). Property (3) follows from Lemma 15 applied to every $S^{i}$ and property (2). Moreover, by Lemma 15 one may choose ${\mathbb{U}}_{K}({\cal K}^{<\alpha},S^{i},\sigma)={\mathbb{U}}_{K}({\cal K}^{\alpha^{\prime}},S^{i},\sigma)={\mathbb{U}}_{K}({\cal K}^{\gamma},S^{i},\sigma)$ for any $K\in{\cal K}^{\gamma}$, $\sigma<\sigma^{i}-1$, $i\in\omega$, $\gamma<\alpha$ so (1) holds. Now (3) follows from Lemma 17, (1), and (2).

Let $\alpha$ be a limit ordinal. Then (1) holds by Lemma 26, the hypothesis, and (1). Property (2) holds by Lemma 25 and (1).

Let $I\subseteq\omega$ be finite and $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})}$ for $i\in I$, $m\not\in I$ and $g_{m}\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})}$. Then $g_{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$ where $\gamma=\max_{i\in I\cup\{m\}}\gamma_{i}$ and $g_{i}\in H^{\gamma_{i}}\setminus H^{<\gamma_{i}}$ for $i\in I\cup\{m\}$. Then $\gamma<\alpha$ and property (2) holds by the hypothesis. Properties (3) and (1) hold by Lemma 18, (2), and the hypothesis. Property (3) holds by Lemma 17. ∎

5 Stability

A number of arguments below require that new sequences are added in a strict order (both to make the recursion work, as well as to keep the convergence structure intact) depending on the sequential order of the points involved.

The sequential order may change as the new sequences are added, however, so a mechanism to keep the order fixed is needed. One such mechanism is described below.

While the results below hold for any type of fpse-chains, the proofs are more involved and the additional generality is not required in the constructions that follow. We thus introduce a narrow subclass of fpse-chains.

Definition 17**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Call an fpse-chain $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ relative to $({\cal C},\xi)$ of finite type if $D_{\gamma}\subseteq\overline{{\langle{S^{i(\gamma)}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}$ for every $\gamma$.

If $i(\gamma)=m$ for every $\gamma$ say that $C$ is close to ${\langle{S^{m}}\rangle}$. Otherwise, if $i(\gamma)\not=m$ for every $\gamma$ say that $C$ is away from ${\langle{S^{m}}\rangle}$.

The constructions below add new sequences to several ‘summands’ of the ‘direct sum’. The next construction presents a way to rearrange the order in which new sequences are added.

Definition 18**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ be an fpse-chain of finite type above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ relative to $({\cal C},\xi)$.

Let $m\in\omega$ and $\gamma^{(m)}={\{\,\gamma^{0}\leq\gamma\leq\gamma^{1}:D_{\gamma}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}\,\}}$. Let $\mu<\omega_{1}$ be such that $h:\mu\to\gamma^{(m)}$ is the unique 1-to-1 monotone map. Define a chain $C^{(m)}(\mu^{\prime})={\{\,{(H^{\lambda}_{(m)},{\cal K}^{\lambda}_{(m)},{\cal U}^{\lambda}_{(m)})}:\lambda<\mu^{\prime}\leq\mu\,\}}$ above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{h(\lambda)}:\lambda<\mu^{\prime}\,\}}$ recursively as follows.

Let $C^{(m)}(\varnothing)=\varnothing$. Suppose $C^{(m)}(\mu^{\prime\prime})$ has been defined for all $\mu^{\prime\prime}<\mu^{\prime}$. If $\mu^{\prime}$ is a limit ordinal, put $C^{(m)}(\mu^{\prime})=\cup_{\mu^{\prime\prime}<\mu^{\prime}}C^{(m)}(\mu^{\prime\prime})$. Otherwise $\mu^{\prime}=\mu^{\prime\prime}+1$ for some $\mu^{\prime\prime}<\omega_{1}$. Put $L_{\mu^{\prime}}=\overline{D_{h(\mu^{\prime})}}^{\tau({\cal U}_{h(\mu^{\prime})})}$. Let $H^{\mu^{\prime}}_{(m)}=H^{\mu^{\prime\prime}}_{(m)}+L_{\mu^{\prime}}$ and let ${\cal K}^{\mu^{\prime}}_{(m)}$ be the closure of ${\cal K}^{\mu^{\prime\prime}}_{(m)}\cup\{L_{\mu^{\prime}}\}$ under finite sums and intersections. Let ${\cal U}^{\mu^{\prime}}_{(m)}={\{\,U\cap H^{\mu^{\prime}}_{(m)}:U\in{\cal U}^{h(\mu^{\prime})}\,\}}$. In the definition above, $H^{\lambda}_{(m)}\subseteq H^{\gamma^{1}}$ and ${\cal K}^{\lambda}_{(m)}\subseteq{\cal K}^{\gamma^{1}}$. It is an easy argument to see that each ${(H^{\lambda}_{(m)},{\cal K}^{\lambda}_{(m)},{\cal U}^{\lambda}_{(m)})}$ is a convenient triple.

Call $C^{(m)}=C^{(m)}(\mu)$ the $m$-reduction of $C$.

It is not immediately clear that the $m$-reduction is an fpse-chain. This is the subject of Lemma 38.

The lemma below holds for any pse-chains, thus the sequential scale is only needed to narrow its statement to fpse-chains and plays no other role in the proof.

Lemma 33**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma\leq\gamma^{0}\,\}}$ be an fpse-chain above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma\leq\gamma^{0}\,\}}$ relative to $({\cal C},\xi)$. Let $L_{\gamma}=\overline{D_{\gamma}}^{\tau({\cal U}_{\gamma})}$.

Let $K\subseteq H^{\gamma}$ be compact in $\mathop{k_{\omega}}({\cal K}^{\gamma})$. Then $K\subseteq K^{\prime}+\sum_{\lambda\in F}L_{\lambda}$ where $K^{\prime}\in{\cal K}$ and $F\subseteq\gamma+1$ is finite.

Proof.

Induction on $\gamma^{0}$. ∎

Further similarity with the direct sum is provided by the following ‘projection’ result.

Lemma 34**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let $K\subseteq G$ be compact in $\mathop{k_{\omega}}({\cal K})$, $I\subseteq\omega$ be finite and $m\in\omega$. Then the set $\pi_{I,m}(K)={\{\,d\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}:d+d^{\prime}\in K\hbox{ for some }d^{\prime}\in\sum_{i\in I\setminus\{m\}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}\,\}}$ is compact in ${\mathop{k_{\omega}}({\cal K})}$.

Proof.

Suppose $\pi_{I,m}(K)$ is not compact. Then there exists an infinite closed and discrete subset $D^{m}={\{\,d_{n}^{m}:n\in\omega\,\}}$ such that for some $d_{n}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ and finite $I\subseteq\omega$, $I\ni m$, and $D={\{\,\sum_{i\in I}d_{n}^{i}:n\in\omega\,\}}\subseteq K$. After thinning out and reindexing we may assume that each $D^{i}={\{\,d_{n}^{i}:n\in\omega\,\}}$ is either closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ or $D^{i}\to d^{i}$ for some $d^{i}\in G$. Let $I^{\prime}\subseteq I$ be the set of all $i\in I$ such that $D^{i}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. Then $m\in I^{\prime}$ and ${\{\,\sum_{i\in I^{\prime}}d_{n}^{i}:n\in\omega\,\}}\subseteq K^{\prime}$ for some $K^{\prime}\in{\cal K}$ contradicting the assumption that ${\cal C}$ is discrete separated. ∎

Lemma 35**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ be an fpse-chain of finite type above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ relative to $({\cal C},\xi)$.

Let $m\in\omega$ and $\gamma^{(m)}={\{\,\gamma^{0}\leq\gamma\leq\gamma^{1}:D_{\gamma}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}\,\}}$. Let $\mu<\omega_{1}$ be such that $h:\mu\to\gamma^{(m)}$ is the unique 1-to-1 monotone map. Let $C^{(m)}={\{\,{(H^{\lambda}_{(m)},{\cal K}^{\lambda}_{(m)},{\cal U}^{\lambda}_{(m)})}:\lambda<\mu\,\}}$ be the $m$-reduction of $C$ and let $\mathop{\mathfrak{so}}(g,D,\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma<\omega_{1}$ where $D\subseteq{\langle{S^{m}}\rangle}$.

Then $g\in H^{\lambda}_{(m)}$ and $\mathop{\mathfrak{so}}(g,D,\mathop{k_{\omega}}({\cal K}^{\lambda}_{(m)}))=\sigma$ where $\lambda=\sup{\{\,\lambda^{\prime}:h(\lambda^{\prime})\leq\gamma\,\}}$. If $\lambda$ is a limit ordinal and $\gamma\not=h(\lambda)$, $\lambda$ may be replaced with with $<\lambda$.

Proof.

Suppose the satement holds for all $g^{\prime}\in G^{\gamma}$ such that $\mathop{\mathfrak{so}}(g^{\prime},D,\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma^{\prime}<\sigma$ and let $\mathop{\mathfrak{so}}(g,D,\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma$. If $\sigma=0$ then the proof is immediate so assume that $\sigma>0$ and there are $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\gamma})$ such that $\mathop{\mathfrak{so}}(g_{n},D,\mathop{k_{\omega}}({\cal K}^{\gamma}))=\sigma_{n}<\sigma$.

By the hypothesis $\mathop{\mathfrak{so}}(g_{n},D,\mathop{k_{\omega}}({\cal K}^{\lambda}_{(m)}))=\sigma_{n}$ where $\lambda$ is defined as in the statement of the Lemma. Since $g_{n}\to g$ in $\mathop{k_{\omega}}({\cal K}^{\gamma})$ there exists a compact $K\in{\cal K}^{\gamma}$ such that $g_{n}\in K$ and by Lemma 33 $K\in K^{\prime}+\sum_{\nu\in F}L_{\nu}$ for some finite $\nu\subseteq\gamma+1$. Thus $g_{n}=g^{n}+\sum_{\nu\in F}g_{n}^{\nu}$ where $g^{n}\in K^{\prime}\subseteq G$, $g_{n}^{\nu}\in L_{\nu}$, and $L_{\nu}=\overline{D_{\nu}}^{\tau({\cal U}^{\nu})}\subseteq\overline{{\langle{S^{i(\nu)}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\nu})}$.

Let $I={\{\,i(\nu):\nu\in F\,\}}\cup\{m\}$. By Lemma 1 $g^{n}\in\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$ so $g^{n}=\sum_{i\in I}g^{n}(i)$ where $g^{n}(i)\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$. Since $g^{n}\in G$ by Lemma 27 $g^{n}(i)\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ for every $i\in I$.

Since $({\cal C},\xi)$ is a sequential scale $g^{n}(i)+\sum_{i(\nu)=i}g_{n}^{\nu}=0$ for every $i\not=m$ by (2) so $g_{n}=g^{n}(m)+g_{n}^{\nu}$ where $g^{n}(m)\in\pi_{I,m}(K^{\prime})$ and $g_{n}^{\nu}\in\sum_{i(\nu)=m}L_{\nu}$. If $i(\nu)=m$ then $L_{\nu}\in{\cal K}^{\nu}_{(m)}$ where $\nu\leq\gamma$ so $\nu=h(\lambda^{\prime})$ for some $\lambda^{\prime}\leq\lambda$. Thus $g_{n}\in K^{\prime\prime}\in{\cal K}^{\lambda}_{(m)}$ and $\mathop{\mathfrak{so}}(g,D,\mathop{k_{\omega}}({\cal K}^{\lambda}_{(m)}))=\sigma$. ∎

Lemma 36**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ be an fpse-chain of finite type above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ relative to $({\cal C},\xi)$.

Let $m\in\omega$ and $\gamma^{(m)}={\{\,\gamma^{0}\leq\gamma\leq\gamma^{1}:D_{\gamma}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}\,\}}$. Let $\mu<\omega_{1}$ be such that $h:\mu\to\gamma^{(m)}$ is the unique 1-to-1 monotone map. Let $C^{(m)}={\{\,{(H^{\lambda}_{(m)},{\cal K}^{\lambda}_{(m)},{\cal U}^{\lambda}_{(m)})}:\lambda<\mu\,\}}$ be the $m$-reduction of $C$ and let $D\subseteq{\langle{S^{m}}\rangle}$, $P_{0}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$ .

Then ${\bf h}_{D}^{{\cal K}^{\gamma}}({P_{0}})={\bf h}_{D}^{{\cal K}^{\lambda}_{(m)}}({P_{0}})$ where $\lambda=\sup{\{\,\lambda^{\prime}:h(\lambda^{\prime})\leq\gamma\,\}}$. If $\lambda$ is a limit ordinal and $\gamma\not=h(\lambda)$, $\lambda$ may be replaced with $<\lambda$.

Proof.

Let $P_{1}\subseteq[D]_{\delta}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$ and $K\in{\cal K}^{\gamma}$ be such that $P_{0}\subseteq P_{1}+K$. By Lemma 35 $P_{1}\subseteq[D]_{\delta}^{\mathop{k_{\omega}}({\cal K}^{\lambda}_{(m)})}$ and by Lemma 33 $K=K^{\prime}+\sum_{\nu\in F}L_{\nu}$ where $F\subseteq\gamma+1$ is finite, $K^{\prime}\in{\cal K}$, and $L_{\nu}=\overline{D_{\nu}}^{\tau({\cal U}^{\nu})}\subseteq\overline{{\langle{S^{i(\nu)}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\nu})}$. Let $g=p_{1}-p_{0}\in K$ for some $p_{1}\in P_{1}$ and $p_{0}\in P_{0}$. Then $g\in\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$ by Lemma 1 and $g=g^{\prime}+\sum_{\nu\in F}g_{\nu}$ where $g_{\nu}\in L_{\nu}\subseteq\overline{{\langle{S^{i(\nu)}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$ and $g^{\prime}\in K^{\prime}$.

As in the proof of Lemma 35 $g=g^{\prime\prime}+g_{m}$ where $g^{\prime\prime}\in\pi_{I,m}(K^{\prime})$ and $g_{m}\in\sum_{i(\nu)=m}L_{\nu}$ where $I={\{\,i(\nu):\nu\in F\,\}}\cup\{m\}$. Thus $p_{1}-p_{0}\in\pi_{I,m}(K^{\prime})+\sum_{i(\nu)=m}L_{\nu}\subseteq K^{\prime\prime}\in{\cal K}^{\lambda}_{(m)}$ so $P_{0}\subseteq P_{1}+K^{\prime\prime}$. ∎

Lemma 37**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let $m\in\omega$ and $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ be an fpse-chain of finite type above ${(G,{\cal K},{\cal U})}$ away from ${\langle{S^{m}}\rangle}$. Let $D\subseteq{\langle{S^{m}}\rangle}$ and $P\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{\gamma})}$.

Then $P\subseteq G$ and ${\bf h}_{D}^{{\cal K}^{\gamma}}({P})={\bf h}_{D}^{{\cal K}}({P})$. If $P$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ then $P$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\gamma})$.

Proof.

Note that $C^{(m)}=\varnothing$. Applying Lemma 35 to each $p\in P$ one shows that $P\subseteq G$. Let ${\bf h}_{D}^{{\cal K}^{\gamma}}({P})=\delta$. It follows from Lemma 36 and $C^{(m)}=\varnothing$ that ${\bf h}_{D}^{{\cal K}}({P})\leq\delta$. To see that $P$ is closed and discrete note that otherwise for some infinite $P^{\prime}\subseteq P$ there exists a compact $K\in{\cal K}^{\gamma^{1}}$ such that $P^{\prime}\subseteq K$. Just as in the proof of Lemma 36 we may assume that $K\in{\cal K}$ contradicting the assumption that $P$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. ∎

Lemma 38**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple and ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$ for some $\xi$. Let $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ be an fpse-chain of finite type above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{\gamma}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ relative to $({\cal C},\xi)$.

Then the $m$-reduction $C^{(m)}={\{\,{(H^{\lambda}_{(m)},{\cal K}^{\lambda}_{(m)},{\cal U}^{\lambda}_{(m)})}:\lambda<\mu\,\}}$ is an fpse-chain above ${(G,{\cal K},{\cal U})}$ along ${\{\,D_{h(\lambda)}:\lambda<\mu\,\}}$ relative to $({\cal C},\xi)$ close to ${\langle{S^{m}}\rangle}$.

Proof.

Let $m\in\omega$ and $\gamma^{(m)}={\{\,\gamma^{0}\leq\gamma\leq\gamma^{1}:D_{\gamma}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\gamma})}\,\}}$. Let $\mu<\omega_{1}$ be such that $h:\mu\to\gamma^{(m)}$ is the unique 1-to-1 monotone map.

Each ${(H^{\lambda}_{(m)},{\cal K}^{\lambda}_{(m)},{\cal U}^{\lambda}_{(m)})}$ is a convenient triple. To show (1) and (2) note that each $D_{h(\lambda)}\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\lambda}_{(m)})}$ by Lemma 35 and ${\bf h}_{P}^{{\cal K}^{<\lambda}_{(m)}}({D_{h(\lambda)}})={\bf h}_{P}^{{\cal K}^{<h(\lambda)}}({D_{h(\lambda)}})$ for $P\in\{\,{\langle{S^{m}}\rangle},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}\,\}$ by Lemma 36 and ${\cal K}^{<\lambda}_{(m)}\subseteq{\cal K}^{<h(\lambda)}$. ∎

The next definition and lemma introduce the concept of stability as well as a way to pass to a stable subset in some cases.

Definition 19**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ and $\xi\subseteq\omega_{1}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$. Let $P\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ be closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. Call $P$ $({\cal C},\xi)$-stable (or simply stable if $({\cal C},\xi)$ is clear from the context) in ${(G,{\cal K},{\cal U})}$ if for any infinite $P^{\prime}\subseteq P$ and any fpse-chain ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma\leq\gamma^{1}\,\}}$ of finite type above ${(G,{\cal K},{\cal U})}$ relative to $({\cal C},\xi)$ ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{\gamma^{1}}}({P^{\prime}})={\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}}({P})$ provided $P^{\prime}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\gamma^{1}})$.

Lemma 39**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ and $\xi\subseteq\omega_{1}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$. Let $P\subseteq\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ be closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. Then there exists an infinite $P^{\prime}\subseteq P$ and an fpse-chain ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma<\gamma^{1}\,\}}$ above ${(G,{\cal K},{\cal U})}$ close to ${\langle{S^{m}}\rangle}$ such that $P^{\prime}$ is $({\cal C},\xi)$-stable in ${(H^{\gamma^{1}},{\cal K}^{\gamma^{1}},{\cal U}^{\gamma^{1}})}$.

Proof.

Let $\delta<\omega_{1}$ be the smallest ordinal such that there exists an infinite $P^{\prime}\subseteq P$ and an fpse-chain ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{0}\leq\gamma<\gamma^{1}\,\}}$ above ${(G,{\cal K},{\cal U})}$ close to ${\langle{S^{m}}\rangle}$ such that $P^{\prime}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\gamma^{1}})$ and ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{\gamma^{1}}}({P^{\prime}})=\delta$.

If $P^{\prime\prime}\subseteq P^{\prime}$ is not stable in ${(H^{\gamma^{1}},{\cal K}^{\gamma^{1}},{\cal U}^{\gamma^{1}})}$ for some infinite $P^{\prime\prime}$, there exists an fpse-chain $C={\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma^{1}\leq\gamma\leq\gamma^{2}\,\}}$ of finite type above ${(H^{\gamma^{1}},{\cal K}^{\gamma^{1}},{\cal U}^{\gamma^{1}})}$ relative to $({\cal C},\xi)$ such that ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{\gamma^{2}}}({P^{\prime\prime\prime}})<{\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{\gamma^{1}}}({P^{\prime}})=\delta$ for some infinite $P^{\prime\prime\prime}\subseteq P^{\prime\prime}$ closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\gamma^{2}})$.

Using the $m$-reduction $C^{(m)}$ of $C$ instead of $C$ and Lemmas 36 and 38 leads to a contradiction with the minimality of $\delta$. ∎

Call the $\delta$ in the proof above the $({\cal C},\xi)$-stable ${\cal K}$-height of $P$ above ${\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}$.

6 Graded boolean group extensions

This section introduces an algebraic mechanism for keeping track and altering of the sequential order of points.

Definition 20**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple, ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ and $\xi={\{\,\sigma^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$. Let $P\subseteq[{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}]_{\Sigma}^{\mathop{k_{\omega}}({\cal K})}$ for some $m\in\omega$ and $\Sigma<\sigma^{m}-1$.

Call $W={\{\,(H^{\alpha},{\cal K}^{\alpha},{\cal U}^{\alpha},S_{\alpha},s_{\alpha}):\alpha^{0}\leq\alpha\leq\Omega\,\}}$ a well-graded group stack above $(P,G,{\cal K},{\cal U},\Sigma)$ or wggs for short if it satisfies the following properties:

1. (1)

${\{\,{(H^{\alpha},{\cal K}^{\alpha},{\cal U}^{\alpha})}:\alpha^{0}\leq\alpha\leq\Omega\,\}}$* is an fpse-chain of finite type above ${(G,{\cal K},{\cal U})}$ along ${\{\,S_{\alpha}:\alpha\leq\Omega\,\}}$ relative to ${\cal C}$ close to ${\langle{S^{m}}\rangle}$;* 2. (2)

$S_{\alpha}\to s_{\alpha}\in H^{\alpha}\setminus H^{<\alpha}$* in $\mathop{k_{\omega}}({\cal K}^{\alpha})$;* 3. (3)

if $\alpha$ is a limit ordinal there exist increasing $\alpha_{n}<\alpha$ such that $\alpha_{n}\to\alpha$, $\mathop{\mathfrak{so}}(s_{\alpha_{n}},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\alpha_{n}}))$ is increasing, and $S_{\alpha}={\{\,s_{\alpha_{n}}:n\in\omega\,\}}$; 4. (4)

if $\alpha$ is a successor ordinal there exist $p_{n}\in P$ such that $S_{\alpha}={\{\,p_{n}:n\in\omega\,\}}$; 5. (5)

if $g\in H^{\alpha}\setminus H^{<\alpha}$, $\alpha\leq\Omega$ then $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\alpha}))\geq\mathop{\mathfrak{so}}(s_{\alpha},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\alpha}))$; 6. (6)

$\top(W)=\mathop{\mathfrak{so}}(s_{\Omega},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\Omega}))>\mathop{\mathfrak{so}}(s_{\alpha},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\alpha}))$* for any $\alpha<\Omega$;*

Lemma 40**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ and $\xi={\{\,\sigma^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$. Let ${\{\,{(H^{\gamma},{\cal K}^{\gamma},{\cal U}^{\gamma})}:\gamma<\alpha\,\}}$ be a an fpse-chain above ${(G,{\cal K},{\cal U})}$ relative to $({\cal C},\xi)$ where $\alpha$ is a limit ordinal.

Let $m\in\omega$ and $S={\{\,s^{n}:n\in\omega\,\}}\subseteq\overline{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})}$ be such that $s_{n}\in H^{\gamma_{n}}\setminus H^{<\gamma_{n}}$ where $\gamma_{n}\to\alpha$ is strictly increasing and cofinal in $\alpha$. Suppose $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma_{n}}))\geq\sigma_{n}<\sigma^{m}$ for any $g\in H^{\gamma_{n}}\setminus H^{<\gamma_{n}}$. Then $S$ is independent, ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{<\alpha}}({S})\geq\sigma=\sup\sigma_{n}$ and ${\langle{S}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{<\alpha})$. In particular, if $\mathop{\mathfrak{so}}(s^{n},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma_{n}}))=\sigma_{n}$ then ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{<\alpha}}({S})=\sigma$.

Proof.

Let $K\in{\cal K}^{<\alpha}$ and $\beta<\sigma$. Then $K\in{\cal K}^{\gamma}$ for some $\gamma<\alpha$. Let $\gamma_{i}>\gamma$ be such that $\mathop{\mathfrak{so}}(s_{i},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma_{i}}))=\sigma_{i}>\beta$. If $a+g=s_{i}$ for some $a\in K$ and $g\in H^{<\alpha}$ then $g\in H^{\gamma_{i}}\setminus H^{<\gamma_{i}}$. Then either $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))\geq\sigma^{m}>\sigma_{i}>\beta$ or $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\alpha}))=\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma_{i}}))\geq\mathop{\mathfrak{so}}(s_{i},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma_{i}}))=\sigma_{i}>\beta$ by the assumption and Lemma 32.

Thus for any such $K$ the set $S\setminus(K+[{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}]_{\beta}^{\mathop{k_{\omega}}({\cal K}^{<\alpha})})\not=\varnothing$ for every $\beta<\sigma$ so ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{<\alpha}}({S})\geq\sigma$.

If $s\in{\langle{S}\rangle}$ then $s=\sum_{n\in I}s_{n}$ for some finite $I\subseteq\omega$ and $s\in H^{\gamma^{\prime}}\setminus H^{<\gamma^{\prime}}$ where $\gamma^{\prime}=\max_{n\in I}\gamma_{n}$ since $\gamma_{n}$ are strictly increasing. Thus if $s\not\in{\langle{F}\rangle}$ where $F={\{\,s_{n}:\gamma_{n}\leq\gamma\,\}}$ then $s\not\in K$. Therefore ${\langle{S}\rangle}\cap K$ is finite and ${\langle{S}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{<\alpha})$. ∎

Lemma 41**.**

Let ${(G,{\cal K},{\cal U})}$ be a convenient triple. Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ and $\xi={\{\,\sigma^{i}:i\in\omega\,\}}$ be such that $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$. Let $m\in\omega$ and $P={\{\,p^{n}:n\in\omega\,\}}\subseteq[{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}]_{\Sigma}^{\mathop{k_{\omega}}({\cal K})}$ be an independent subset of $G$ such that ${\langle{P}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$, ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}}({P})=\Sigma<\sigma^{m}-1$, and $P$ is $({\cal C},\xi)$-stable in ${(G,{\cal K},{\cal U})}$. Let $D={\{\,d^{n}:n\in\omega\,\}}\subseteq G$ and $d\in[D]_{\mu}^{\mathop{k_{\omega}}({\cal K})}$ where $\mu\geq 1$ is a successor ordinal, $\Sigma+\mu\leq\sigma\leq\sigma^{m}-1$ for some successor ordinal $\sigma$.

Then there exists a wggs $W={\{\,(H^{\alpha},{\cal K}^{\alpha},{\cal U}^{\alpha},S_{\alpha},s_{\alpha}):\alpha\leq\Omega\,\}}$ above $(P,G,{\cal K},{\cal U},\Sigma)$ such that $\top(W)=\sigma$ and the following property holds.

1. (1)

$d+s_{\Omega}\in[{\{\,d^{n}+p^{n}:n\in\omega\,\}}]_{\eta}^{\mathop{k_{\omega}}({\cal K}^{\Omega})}$* where*

[TABLE]

In addition, if $P^{\prime}\subseteq G$ is such that $P^{\prime}\cap P=\varnothing$, $P\cup P^{\prime}$ is independent, and $\langle P^{\prime}\cup P\rangle$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$ then one may assume the following property holds.

1. (1)

${\langle{P^{\prime}}\rangle}$* is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\Omega})$;*

Proof.

Let $\sigma=\Sigma+1<\sigma^{m}$. Then $\mu=1$ so by passing to subsets and reindexing we may assume that $d^{n}\to d$ in ${\mathop{k_{\omega}}({\cal K})}$. Use Lemma 10 to find an $S_{0}\subseteq P$, and ${\cal U}^{0}\supseteq{\cal U}$ such that $\overline{{\langle{S^{\prime}}\rangle}}^{\tau({\cal U}^{0})}\cap G={\langle{S^{\prime}}\rangle}$ for every infinite $S^{\prime}\subseteq S_{0}$ and $\overline{{\langle{2{S_{0}}}\rangle}}^{\tau({\cal U}^{0})}\cap\overline{{{\langle{S_{0}}\rangle}\setminus{\langle{2{S_{0}}}\rangle}}}^{\tau({\cal U}^{0})}=\varnothing$. By passing to subsets and applying Lemma 6 we may assume that $S_{0}\to s_{0}$ in $\tau({\cal U}^{0})$ and $S_{0}$ satisfies ${\mathbb{M}}({\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},{\cal K})$ and ${\mathbb{M}}({\langle{S^{i}}\rangle},{\cal K})$ for every $i\in\omega$. Let ${(H^{0},{\cal K}^{0},{\cal U}^{0})}$ be the fpse of ${(G,{\cal K},{\cal U})}$ over $S_{0}$. Put ${(H^{<0},{\cal K}^{<0},{\cal U}^{<0})}={(G,{\cal K},{\cal U})}$ and $W={\{\,(H^{i},{\cal K}^{i},{\cal U}^{i},S_{i},s_{i}):i<1\,\}}$. Properties (1)–(4) are immediate, (6) holds vacuously, and (1) holds by the construction.

Let $g\in H_{0}\setminus G$ and let $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{0}))=\sigma^{\prime}$. If $\sigma^{\prime}\leq\Sigma$ then $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},{\mathop{k_{\omega}}({\cal K})})=\sigma^{\prime}$ by Lemma 11, in particular $g\in G$. Thus $\sigma^{\prime}>\Sigma$ so $\sigma^{\prime}\geq\Sigma+1\geq\mathop{\mathfrak{so}}(s_{0},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{0}))$ and (5) holds.

Suppose $K\in{\cal K}$ is such that $(K+L)\cap{\langle{P^{\prime}}\rangle}$ is infinite where $L=\overline{{\langle{S_{0}}\rangle}}^{\tau({\cal U}^{0})}$. Let $a_{n}+d_{n}=p_{n}\in{\langle{P^{\prime}}\rangle}$ be distinct points in ${\langle{P^{\prime}}\rangle}$ such that $a_{n}\in K$ and $d_{n}\in L$. Then $d_{n}\in G$ so by (2) $d_{n}\in{\langle{S_{0}}\rangle}\subseteq{\langle{P}\rangle}$. If the set ${\{\,d_{n}:n\in\omega\,\}}$ is infinite then $K\cap{\langle{P^{\prime}\cup P}\rangle}$ is infinite since $P^{\prime}$ and $P$ are disjoint and $P^{\prime}\cup P$ is independent, contradicting ${\langle{P^{\prime}\cup P}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$. Thus ${\{\,d_{n}:n\in\omega\,\}}$ is finite so $K^{\prime}\cap{\langle{P^{\prime}}\rangle}$ is infinite for some $K^{\prime}\in{\cal K}$ contradicting ${\langle{P^{\prime}\cup P}\rangle}$ is closed and discrete in ${\mathop{k_{\omega}}({\cal K})}$, so (1) holds.

Suppose $W$ that satisfy (1) and (1) can be constructed for every $\sigma^{\prime}<\sigma$ where $\Sigma+1<\sigma\leq\sigma^{m}-1$.

If $\mu>1$ let $\mu_{n}\to\mu-1$ be increasing such that $d_{n}^{\prime}\to d$ in $\mathop{k_{\omega}}({\cal K})$ where $\mathop{\mathfrak{so}}(d_{n}^{\prime},D,{\mathop{k_{\omega}}({\cal K})})=\mu_{n}$. Using regularity, construct disjoint $I_{n}\subseteq\omega$ such that $\mathop{\mathfrak{so}}(d_{n}^{\prime},D_{n},{\mathop{k_{\omega}}({\cal K})})=\mu_{n}$ where $D_{n}={\{\,d^{i}:i\in I_{n}\,\}}$.

If $\mu=1$ assume, by possibly thinning out and reindexing that $d^{n}\to d$, put $\mu_{n}=1$ and let $I_{n}\subseteq\omega$ be arbitrary infinite disjoint subsets.

Let $\sigma_{i}\to\sigma-1>\Sigma$ be successor ordinals such that $\sigma_{i}\geq\Sigma+\mu_{i}$. Let $P_{n}={\{\,p^{i}:i\in I_{n}\,\}}$. Note that every subset of $P$ (including $P_{n}$, and their arbitrary unions) is $({\cal C},\xi)$-stable in ${(G,{\cal K},{\cal U})}$ and ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}}({P_{n}})=\Sigma$.

Build, by induction, a sequence of $W_{i}={\{\,(H^{\alpha},{\cal K}^{\alpha},{\cal U}^{\alpha},S_{\alpha},s_{\alpha}):\gamma_{i-1}<\alpha\leq\gamma_{i}\,\}}$ such that $W_{0}$ is a wggs above $(P_{0},G,{\cal K},{\cal U},\Sigma)$, $W_{i+1}$ is a wggs above $(P_{i+1},H^{\gamma_{i}},{\cal K}^{\gamma_{i}},{\cal U}^{\gamma_{i}},\Sigma)$ for $i\in\omega$, and the following properties hold.

1. (1)

the subgroup ${\langle{P^{\prime}\cup\cup_{j>i}P_{j}}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\gamma_{i}})$; 2. (2)

for each $i\in\omega$ $\mathop{\mathfrak{so}}(s_{\gamma_{i}},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma_{i}}))=\sigma_{i}$; 3. (3)

$d_{n}^{\prime}+s_{\gamma_{n}}\in[{\{\,d^{i}+p^{i}:i\in I_{n}\,\}}]_{\eta_{n}}^{\mathop{k_{\omega}}({\cal K}^{\gamma_{i}})}$ where

[TABLE]

At the $n$-th step the set $\cup_{j>n}P_{j}$ is closed, discrete, and $({\cal C},\xi)$-stable in ${(H^{\gamma_{n}},{\cal K}^{\gamma_{n}},{\cal U}^{\gamma_{n}})}$, ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{\gamma_{n}}}({\cup_{j>n}P_{j}})=\Sigma$, and $\sigma_{i}<\sigma$ so $W_{n+1}$ exists by the hypothesis.

Let $\Omega=\lim\gamma_{i}$ and consider the conventient triple ${(H^{<\Omega},{\cal K}^{<\Omega},{\cal U}^{<\Omega})}$. Put $S_{\Omega}^{\prime}={\{\,s_{\gamma_{i}}:i\in\omega\,\}}$ and let $g\in H^{<\Omega}$ be such that $g\in H^{\alpha}\setminus H^{<\alpha}$ and $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\Omega}))=\sigma^{\prime}$ for some $\sigma^{\prime}<\sigma^{m}$ and $\alpha<\Omega$. Then $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\alpha}))=\sigma^{\prime}$ by (1), Lemma 32, and the construction of $W_{i}$, in particular, $\mathop{\mathfrak{so}}(s_{\gamma_{i}},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\Omega}))=\sigma_{i}$.

The sequence of ordinals $\gamma_{i}$ is cofinal in $\Omega$ and $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\gamma_{i}}))\geq\sigma_{i}$ for every $g\in H^{\gamma_{i}}\setminus H^{<\gamma_{i}}$ by (5) and the hypothesis so ${\bf h}_{{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle}}^{{\cal K}^{<\Omega}}({S_{\Omega}})=\sigma-1$ for any infinite subset $S_{\Omega}\subseteq S_{\Omega}^{\prime}$ and ${\langle{S_{\Omega}^{\prime}}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}^{<\Omega})$ by Lemma 40.

Construct an fpse ${(H^{\Omega},{\cal K}^{\Omega},{\cal U}^{\Omega})}$ of ${(H^{<\Omega},{\cal K}^{<\Omega},{\cal U}^{<\Omega})}$ over some infinite $S_{\Omega}\subseteq S_{\Omega}^{\prime}$ such that $S_{\Omega}\to s_{\Omega}$ in $\mathop{k_{\omega}}({\cal K}^{\Omega})$ using Lemma 29 and let $L=\overline{{\langle{S_{\Omega}}\rangle}}^{\tau({\cal U}^{\Omega})}$.

Suppose $\mathop{\mathfrak{so}}(g^{\prime},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\Omega}))<\sigma$ for some $g^{\prime}\in H^{\Omega}$. Then $\mathop{\mathfrak{so}}(g^{\prime},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\Omega}))\leq_{L}\sigma-1$ so by Lemma 11 $\mathop{\mathfrak{so}}(g^{\prime},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{<\Omega}))\leq_{L}\sigma-1<\omega_{1}$ and $g^{\prime}\in H^{<\Omega}$. Thus $\mathop{\mathfrak{so}}(g,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\Omega}))\geq\sigma$ for any $g\in H^{\Omega}\setminus H^{<\Omega}$, in particular, $\mathop{\mathfrak{so}}(s_{\Omega},{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},\mathop{k_{\omega}}({\cal K}^{\Omega}))=\sigma$.

Now (1)–(6) hold by construction, (1) holds by the choice of $d_{n}^{\prime}$.

Let $K\in{\cal K}^{<\Omega}$ be such that $K+L\cap{\langle{P^{\prime}}\rangle}$ is infinite. Suppose $p_{n}=a_{n}+s_{n}$ are distinct points in ${\langle{P^{\prime}}\rangle}$ such that $a_{n}\in K$ and $s_{n}\in L$. Then $s_{n}\in H^{<\Omega}$ so $s_{n}\in{\langle{S_{\Omega}}\rangle}$ by (2). Let $K\in{\cal K}^{\beta}$ where $\beta<\Omega$. If the set ${\{\,s_{n}:n\in\omega\,\}}$ is infinite there is an $n\in\omega$ such that $s_{n}\in H^{\alpha}\setminus H^{<\alpha}$ for some $\alpha\geq\beta$ so $p_{n}=a_{n}+s_{n}\not\in G$ contradicting $p_{n}\in P^{\prime}$. Thus ${\{\,s_{n}:n\in\omega\,\}}$ is finite and $p_{n}\in K^{\prime}\in{\cal K}^{\gamma_{i}}$ for some $\gamma_{i}<\Omega$. Then ${\langle{P^{\prime}}\rangle}$ is not closed and discrete in $\mathop{k_{\omega}}({\cal K}^{\gamma_{i}})$ contradicting the properties of $W_{i}$. Hence (1) holds. ∎

Lemma 42**.**

Let ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ be discrete separated in some $\mathop{k_{\omega}}$-pair $(G,{\cal K})$. Let $g\in\overline{P}^{\mathop{k_{\omega}}({\cal K})}$ for some $g\in G$ and $P={\{\,p_{j}:j\in\omega\,\}}\subseteq G$. Let $D={\{\,d_{j}:j\in\omega\,\}}$ where $d_{j}=\sum_{i\in I}d_{j}^{i}$ for some finite $I\subseteq\omega$, and $d_{j}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$. Suppose $p_{j}-d_{j}\in K$ for some $K\in{\cal K}$ and every $j\in\omega$.

Then there are $d^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$, $i\in I$ and $a\in K$ such that

1. (1)

$g=a+\sum_{i\in I}d^{i}$; 2. (2)

for any $U$, $U_{K}$, $U(i)$, $i\in I$ open in ${\mathop{k_{\omega}}({\cal K})}$ such that $g\in U$, $a\in U_{K}$, and $d^{i}\in U(i)$ there is an $n\in\omega$ such that $p_{n}\in U$, $p_{n}-d_{n}\in U_{K}$, and $d_{n}^{i}\in U(i)$;

Proof.

Suppose the Lemma holds for all $g^{\prime}\in G$ such that $\mathop{\mathfrak{so}}(g^{\prime},P,{\mathop{k_{\omega}}({\cal K})})=\sigma^{\prime}<\sigma$ and let $\mathop{\mathfrak{so}}(g,P,{\mathop{k_{\omega}}({\cal K})})=\sigma$. If $\sigma=0$ the proof is immediate so assume $\sigma>0$. Then there are $g_{n}\in[P]_{\sigma_{n}}$ where $\sigma_{n}<\sigma$ and $g_{n}\to g$. By the hypothesis there are $a_{n}\in K$ and $d^{i,n}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$ for $i\in I$ such that $g_{n}=a_{n}+\sum_{i\in I}d^{i,n}$ and for any $m\in\omega$ and any open in ${\mathop{k_{\omega}}({\cal K})}$ subsets $U\ni g_{m}$, $U_{K}\ni a_{m}$, $U(i)\ni d^{i,m}$ where $i\in I$ there is an $n\in\omega$ such that $p_{n}\in U$, $d_{n}^{i}\in U(i)$, $p_{n}-d_{n}\in U_{K}$.

Since ${\{\,\sum_{i\in I}d^{i,n}:n\in\omega\,\}}\subseteq K^{\prime}$ for some $K^{\prime}\in{\cal K}$ and ${\cal C}$ is discrete separated by ${\cal K}$ in $G$, there is an $i\in I$ such that the set ${\{\,d^{i,n}:n\in\omega\,\}}$ is not closed and discrete. By thinning out and reindexing we then may assume that $d^{i,n}\to d^{i}$ for some $d^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K})}$. Replacing $K^{\prime}$ with some $K^{\prime\prime}\supseteq K^{\prime}+{\{\,d^{i,n}:n\in\omega\,\}}$ and repeating this argument we may find $d^{i}$ such that $d^{i,n}\to d^{i}$ for each $i\in I$. By thinning out and reindexing we may assume that $a_{n}\to a$ for some $a\in K$. Then $g=a+\sum_{i\in I}d^{i}$.

Let $U$, $U_{K}$, $U(i)$, $i\in I$ be open in ${\mathop{k_{\omega}}({\cal K})}$ such that $g\in U$, $a\in U_{K}$, and $d^{i}\in U(i)$ for every $i\in I$. Pick an $m\in\omega$ such that $g_{m}\in U$, $a_{m}\in U_{K}$, and $d^{i,m}\in U(i)$ for every $i\in I$. By the hypothesis there is an $n\in\omega$ such that $p_{n}\in U$, $p_{n}-d_{n}\in U_{K}$, and $d_{n}^{i}\in U(i)$, $i\in I$. ∎

Lemma 43**.**

Let $\xi={\{\,\sigma^{m}:m\in\omega\,\}}\subseteq\omega_{1}\setminus 2$, $m\in\omega$ be a family of successor ordinals. Then there exists a convenient triple $(G,{\cal K},{\cal U})$ and a family ${\cal C}={\{\,S^{i}:i\in\omega\,\}}$ of discrete separated subsets of $G$ such that $\mathop{\mathfrak{so}}(0,{\langle{S^{m}}\rangle}\setminus{\langle{2S^{m}}\rangle},{\mathop{k_{\omega}}({\cal K})})=\sigma^{m}$ for every $m\in\omega$ and $({\cal C},\xi)$ is a sequential scale in ${(G,{\cal K},{\cal U})}$.

Proof.

For each $m\in\omega$ let ${(G(m),{\cal K}(m),{\cal U}(m))}$ be a convenient triple such that $\mathop{k_{\omega}}({\cal K}(m))$ is neither discrete nor compact. Let ${\cal C}_{m}={\{\,S(n):n\in\omega\,\}}$ be a family of countable disjoint subsets of $G(m)\setminus\{0\}$ such that $\cup_{n\in\omega}S(n)$ is an independent set and ${\langle{\cup_{n\in\omega}S(n)}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}(m))$. Then $({\cal C}_{m},\xi)$ is a trivial sequential scale in ${(G(m),{\cal K}(m),{\cal U}(m))}$.

Let $\sigma^{m}_{n}\to\sigma^{m}-1$. Let $S(m)=\cup_{n\in\omega}S(m,n)$ where $S(m,n)$ are infinite and disjoint. Each $S(m,n)$ is trivially $({\cal C}_{m},\xi)$-stable in ${(G(m),{\cal K}(m),{\cal U}(m))}$ (its ${\cal K}(m)$-height is [math]). Put ${(G^{-1},{\cal K}^{-1},{\cal U}^{-1})}={(G(m),{\cal K}(m),{\cal U}(m))}$, $\gamma_{-1}(m)=-1$, and use Lemma 41 to construct wggs

[TABLE]

above $(S(m,n),G^{\gamma_{n-1}(m)},{\cal K}^{\gamma_{n-1}(m)},{\cal U}^{\gamma_{n-1}(m)},0)$ relative to $({\cal C}_{m},\xi)$ such that $\top(W_{n}(m))=\sigma^{m}_{n}$ for every $n\in\omega$. Let $\alpha(m)=\lim\gamma_{n}(m)$.

Just as in the proof of Lemma 41, the set $S={\{\,s_{\gamma(n)}:n\in\omega\,\}}$ is closed and discrete and ${\bf h}_{{{\langle{S(m)}\rangle}\setminus{\langle{2{S(m)}}\rangle}}}^{{\cal K}^{<\alpha(m)}}({S})=\sigma^{m}-1$. Use Lemma 29 to find an fpse ${(G^{\alpha}(m),{\cal K}^{\alpha}(m),{\cal U}^{\alpha}(m))}$ of ${(G^{<\alpha(m)},{\cal K}^{<\alpha(m)},{\cal U}^{<\alpha(m)})}$ such that $S\to s\not\in G^{<\alpha(m)}$ in $\mathop{k_{\omega}}({\cal K}^{\alpha(m)})$. Then $\mathop{\mathfrak{so}}(s,{{\langle{S(m)}\rangle}\setminus{\langle{2{S(m)}}\rangle}},\mathop{k_{\omega}}({\cal K}^{\alpha(m)}))=\sigma^{m}-1$, the set $s+S(m)$ is independent, ${\langle{s+S(m)}\rangle}\setminus{\langle{2{(s+S(m))}}\rangle}=s+{{\langle{S(m)}\rangle}\setminus{\langle{2{S(m)}}\rangle}}$ so ${\mathbb{S}}({\cal K}^{\alpha(m)},s+S(m),\sigma^{m})$ holds. Put $S^{m}=s+S(m)$, ${(G_{m},{\cal K}_{m},{\cal U}_{m})}={(G^{\alpha(m)},{\cal K}^{\alpha(m)},{\cal U}^{\alpha(m)})}$.

Let $G$ be the direct sum of $G_{m}$, ${\cal U}$ be the appropriate basis of the topology inherited from the product of $(G_{m},\tau({\cal U}_{m}))$, and ${\cal K}$ be the closure of $\cup_{m\in\omega}{\cal K}_{m}$ under finite sums. Put ${\cal C}={\{\,S_{m}:m\in\omega\,\}}$. Then the desired properties hold by the construction.

Note that in the proof above, the full sequential scale ${\cal C}_{m}$ was only required to formally satisfy the conditions of Lemma 41. ∎

Let $\sigma^{m}$ be an increasing sequence of successor ordinals in $\omega_{1}$. Let $\sigma^{i}_{k}\to\sigma^{i}-1$ be an increasing sequence of successor ordinals for every $i\in\omega$ such that $\sup\sigma^{i}_{k}=\sigma^{i}-1$.

Since all groups $G_{\alpha}$ in the construction have cardinality $2^{\omega}$ we will assume that every $G_{\alpha}$ is a subgroup (algebraically) of $2^{\omega}$. Let ${\{\,C_{\alpha}:\alpha<\omega_{1}\,\}}\subseteq[2^{\omega}]^{\omega}$ and ${\{\,P_{\alpha}:\alpha<\omega_{1}\hbox{ is a limit ordinal}\,\}}\subseteq[2^{\omega}]^{\omega}$.

Let ${(G_{0},{\cal K}_{0},{\cal U}_{0})}={(G,{\cal K},{\cal U})}$ where the convenient triple ${(G,{\cal K},{\cal U})}$ and the sequential scale $({\cal C},\xi)$ have been constructed in Lemma 43. Below we use the notation ${\bf c}{C_{\alpha}}$ for the closure of $C_{{\bf m}(\alpha)}$ in $\mathop{k_{\omega}}({\cal K}_{<{\bf m}(\alpha)})$.

Lemma 44**.**

There exists an fpse-chain ${\{\,{(G_{\alpha},{\cal K}_{\alpha},{\cal U}_{\alpha})}:\alpha<\omega_{1}\,\}}$ relative to $({\cal C},\xi)$ and an increasing continuous ${\bf m}:\omega_{1}\to\omega_{1}$ such that

1. (1)

if $\alpha$ is a limit ordinal and $P_{\alpha}\subseteq G_{{\bf m}(\alpha+1)}$ is an infinite subset that is closed and discrete in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})$ then there exists an infinite $S_{\alpha}\subseteq P_{\alpha}$ such that $S_{\alpha}\to s_{\alpha}$ in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+2)})$; 2. (2)

if $\alpha$ is a limit ordinal and $0\in\overline{P_{\alpha}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}$ then $\mathop{\mathfrak{so}}(0,P_{\alpha},\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+2)}))\leq\sigma^{m}$ for some $m\in\omega$; 3. (3)

if $\alpha$ is a limit ordinal then one of the following conditions holds

1. a.

if ${\bf h}_{{\cal C}}^{{\cal K}_{<{\bf m}(\alpha)}}({{\bf c}{C_{\alpha}}\cap U})=\omega$ for every $U\in{\cal U}_{<{\bf m}(\alpha)}$ then $S\to 0$ in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha)})$ for some $S\subseteq{\bf c}{C_{\alpha}}$; 2. b.

suppose ${\bf c}{C_{\alpha}}\subseteq\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<{\bf m}(\alpha)})}$ for some finite $I\subset\omega$ (so (3.a) does not hold), and ${\bf c}{C_{\alpha}}\cap\sum_{i\in I^{\prime}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<{\bf m}(\alpha)})}=\varnothing$ for any $I^{\prime}\subset I$, $I^{\prime}\not=I$; if there exists an fpse-chain $C={\{\,{(G_{\gamma},{\cal K}_{\gamma},{\cal U}_{\gamma})}:{\bf m}(\alpha)\leq\gamma\leq\gamma^{0}\,\}}$ such that $\overline{{\bf c}{C_{\alpha}}}^{\mathop{k_{\omega}}({\cal K}_{\gamma^{0}})}\cap\sum_{i\in I^{\prime}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma^{0}})}\not=0$ for some $I^{\prime}\subset I$, $I^{\prime}\not=I$; then $\overline{{\bf c}{C_{\alpha}}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}\cap\sum_{i\in I^{\prime\prime}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}\not=0$ for some $I^{\prime\prime}\subset I$, $I^{\prime\prime}\not=I$; 3. c.

suppose neither (3.a) nor (3.b) holds, ${\bf c}{C_{\alpha}}\subseteq\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<{\bf m}(\alpha)})}$ for some finite $I\subset\omega$, and ${\bf c}{C_{\alpha}}\cap\sum_{i\in I^{\prime}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<{\bf m}(\alpha)})}=\varnothing$ for any $I^{\prime}\subset I$, $I^{\prime}\not=I$; suppose there exists an fpse-chain

[TABLE]

and for some $\gamma_{n}$, $n\in\omega$, strictly increasing and cofinal in $\gamma^{0}$ the following properties hold:

there are $d(n)\in\overline{{\bf c}{C_{\alpha}}}^{\mathop{k_{\omega}}({\cal K}_{\gamma^{0}})}$ such that $d(n)=\sum_{i\in I}d(i,n)$ for every $n\in\omega$ where $d(i,n)\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma^{0}})}\cap\bigcap{\{\,\overline{U}^{\tau({\cal U}_{\gamma^{0}})}:U\in{\cal U}_{<{\bf m}(\alpha)}\,\}}$, and

$\mathop{\mathfrak{so}}(g,{\langle{S^{i(n)}}\rangle},\mathop{k_{\omega}}({\cal K}_{\gamma^{0}}))\geq\sigma^{i(n)}_{n}$* for any $g\in G_{\gamma_{n}}\setminus G_{<\gamma_{n}}$ where $d({i(n)},n)\in G_{\gamma_{n}}\setminus G_{<\gamma_{n}}$ for some choice of $i(n)\in I$ and $d(i,n)\not\in G_{<{\bf m}(\alpha)}$ for $i\in I$; then the condition above holds if $\gamma^{0}$ is replaced by ${\bf m}(\alpha+1)$.* 4. (4)

if $\alpha$ is a limit ordinal then for every finite $I\subset\omega$

[TABLE]

where $H(I)=\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+3)})}$ and for any $U\in{\cal U}_{{\bf m}(\alpha+3)}$ there is a $U^{\prime}\subseteq U$ such that $U^{\prime}\in{\cal U}_{{\bf m}(\alpha+3)}$ and

[TABLE]

Proof.

Put ${\bf m}(0)=0$. If $\alpha$ is a successor such that $\alpha=\beta+n$ where $n\in\omega$ and $\beta$ is a limit and $n>3$ or $\beta=0$ and $n>0$ let ${\bf m}(\alpha)={\bf m}(\alpha-1)+1$ and let ${(G_{{\bf m}(\alpha)},{\cal K}_{{\bf m}(\alpha)},{\cal U}_{{\bf m}(\alpha)})}$ be an arbitrary fpse of ${(G_{{\bf m}(\alpha-1)},{\cal K}_{{\bf m}(\alpha-1)},{\cal U}_{{\bf m}(\alpha-1)})}$.

Let $\alpha$ be a limit ordinal and $G_{\alpha^{\prime}}$ etc., ${\bf m}(\alpha^{\prime})$ have been constructed for all $\alpha^{\prime}<\alpha$. Then ${\bf m}(\alpha)$ is uniquely defined by continuity.

Let ${\cal K}_{<{\bf m}(\alpha)}={\{\,K_{i}:i\in\omega\,\}}$ and ${\cal U}_{<{\bf m}(\alpha)}={\{\,U_{i}:i\in\omega\,\}}$.

Suppose ${\bf h}_{{\cal C}}^{{\cal K}_{<{\bf m}(\alpha)}}({{\bf c}{C_{\alpha}}\cap U})=\omega$ for every $U\in{\cal U}_{{\bf m}(\alpha)}$. Find an infinite $S\subseteq{\bf c}{C_{\alpha}}$ such that $S\to 0$ in $\tau({\cal U}_{<{\bf m}(\alpha)})$ and $S$ satisfies ${\mathbb{M}}({\cal C},{\cal K}_{<{\bf m}(\alpha)})$ using Lemma 6. Use Lemma 31 to find an fpse ${(G_{{\bf m}(\alpha)},{\cal K}_{{\bf m}(\alpha)},{\cal U}_{{\bf m}(\alpha)})}$ of ${(G_{<{\bf m}(\alpha)},{\cal K}_{<{\bf m}(\alpha)},{\cal U}_{<{\bf m}(\alpha)})}$ over $S$ such that $S\to 0$ in ${\cal K}_{{\bf m}(\alpha)}$. Let ${(G_{{\bf m}(\alpha+1)},{\cal K}_{{\bf m}(\alpha+1)},{\cal U}_{{\bf m}(\alpha+1)})}$ be an arbitrary fpse of ${(G_{{\bf m}(\alpha)},{\cal K}_{{\bf m}(\alpha)},{\cal U}_{{\bf m}(\alpha)})}$ and put ${\bf m}(\alpha+1)={\bf m}(\alpha)+1$ to satisfy (3.a).

If the wggs from the statement of (3.b) or (3.c) exists put ${\bf m}(\alpha+1)=\gamma^{0}$.

Suppose $P_{\alpha}$ is closed and discrete in $G_{{\bf m}(\alpha+1)}$.

If ${\bf h}_{{\cal C}}^{{\cal K}_{{\bf m}(\alpha+1)}}({P_{\alpha}})=\omega$ use Lemma 29 to find an fpse ${(G_{{\bf m}(\alpha+2)},{\cal K}_{{\bf m}(\alpha+2)},{\cal U}_{{\bf m}(\alpha+2)})}$ of ${(G_{{\bf m}(\alpha+1)},{\cal K}_{{\bf m}(\alpha+1)},{\cal U}_{{\bf m}(\alpha+1)})}$ over $S_{\alpha}\subseteq P_{\alpha}$ such that $S_{\alpha}\to s\in G_{{\bf m}(\alpha+2)}$ in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+2)})$ where ${\bf m}(\alpha+2)={\bf m}(\alpha+1)+1$.

Otherwise $P_{\alpha}\subseteq\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}+K$ for some $K\in{\cal K}_{{\bf m}(\alpha+1)}$. Pick an infinite $D={\{\,d_{n}=\sum_{i\leq m}d^{i}_{n}:n\in\omega\,\}}$ where $d^{i}_{n}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}$ and $p_{n}=d_{n}+a_{n}$ for some distinct $p_{n}\in P_{\alpha}$ and $a_{n}\in K$. By passing to subsequences and reindexing we may assume that $a_{n}\to a$ for some $a\in K$ and either $d^{i}_{n}\to d^{i}$ or $D^{i}={\{\,d^{i}_{n}:n\in\omega\,\}}$ is closed and discrete in ${\cal K}_{<{\bf m}(\alpha)}$. Repeatedly using Lemma 29 find a finite fpse-chain

[TABLE]

for some $k\leq m$ such that ${\{\,d^{i}_{n}:n\in J\,\}}\to d^{i}$ in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+2)})$ for some infinite $J\subseteq\omega$.

Let $0\in\overline{P_{\alpha}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}$ and suppose ${{}^{\omega}{\bf h}_{{\cal C}}^{{\cal K}_{{\bf m}(\alpha+1)}}({P_{\alpha}})}<\omega$. Let $P_{\alpha}={\{\,p_{j}:j\in\omega\,\}}$ and $p_{j}-d_{j}\in K$ for some $K\in{\cal K}_{{\bf m}(\alpha+1)}$ where $d_{j}=\sum_{i\in I}d_{j}^{i}$ for some finite $I\subset\omega$, and $d_{j}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}$.

Using Lemma 42 find $d^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}$ and $a\in K$ such that $0=a+\sum_{i\in I}d^{i}$ and for any open in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})$ subsets $U\ni 0$, $U_{K}\ni a$, $U(i)\ni d^{i}$, $i\in I$ there is an $n\in\omega$ such that $p_{n}\in U$, $d_{n}^{i}\in U(i)$, and $p_{n}-d_{n}\in U_{K}$, $i\in I$.

Replacing each $d_{n}^{i}$ with $d_{n}^{i}+d^{i}$, each $d_{n}$ with $d_{n}+a$ and picking $K\in{\cal K}_{{\bf m}(\alpha+1)}$ so that $d^{i}\in K$, $i\in I$, we may assume that the following properties hold.

1. (1)

$d_{j}=\sum_{i\in I}d_{j}^{i}$, $d_{j}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}$, and $p_{j}-d_{j}\in K$ for every $j\in\omega$; 2. (2)

for any $U\ni 0$ open in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})$ there exists an $n\in\omega$ such that $p_{n},d_{n}^{i},p_{n}-d_{n}\in U$ for every $i\in I$;

Indeed, let $U$ be open in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})$. Find a $U^{\prime}\ni 0$, open in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})$ such that $U^{\prime}+\cdots+U^{\prime}\subseteq U$ where the sum has $|I|$ terms. Put $U_{K}=U^{\prime}+a$, $U(i)=U^{\prime}+d^{i}$ and let $n\in\omega$ be such that $p_{n}\in U^{\prime}$, $d_{n}^{i}\in U(i)$, and $p_{n}-d_{n}\in U_{K}$. Then $d_{n}^{i}+d^{i}\in U^{\prime}$ and $p_{n}-(d_{n}+a)=p_{n}-d_{n}+a\in U^{\prime}$.

Let ${(G(0),{\cal K}(0),{\cal U}(0))}={(G_{{\bf m}(\alpha+1)},{\cal K}_{{\bf m}(\alpha+1)},{\cal U}_{{\bf m}(\alpha+1)})}$. By induction on $k\in\omega$ build convenient triples ${(G(k),{\cal K}(k),{\cal U}(k))}$ and points $d(i,k)\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}(k))}$ where $i\in I$, such that the following properties hold for $k>0$:

1. (1)

${(G(k),{\cal K}(k),{\cal U}(k))}={(G_{\gamma(k)},{\cal K}_{\gamma(k)},{\cal U}_{\gamma(k)})}$ for some $\gamma(k)<\omega_{1}$ where

[TABLE]

is an fpse-chain above ${(G(k-1),{\cal K}(k-1),{\cal U}(k-1))}$ relative to $({\cal C},\xi)$; 2. (2)

either $d(i,k)=0$ or $d(i,k)\in G_{\gamma(i,k)}\setminus G_{<\gamma(i,k)}$ where $\gamma(k-1)<\gamma(i,k)\leq\gamma(k)$; 3. (3)

if $g\in G_{\gamma(i,k)}\setminus G_{<\gamma(i,k)}$ then $\mathop{\mathfrak{so}}(g,{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},{\cal K}_{\gamma(i,k)})\geq\sigma^{i}_{k}$; 4. (4)

$d(i,k)\in\overline{U}^{\tau({\cal U}(k))}$ and $p(k)-(\sum_{i\in I}d(i,k))\in K\cap\overline{U}^{\tau({\cal U}(k))}$ for every $U\in{\cal U}(k-1)$ where $p(k)\in[P_{\alpha}]_{\eta(k)}^{\mathop{k_{\omega}}({\cal K}(k))}$, and $\eta(k)\leq\max_{i\in I}\sigma^{i}-1$;

Suppose ${(G(k^{\prime}),{\cal K}(k^{\prime}),{\cal U}(k^{\prime}))}$ etc. that satisfy (1)–(4) have been built for all $k^{\prime}<k$. Let ${\cal K}(k-1)={\{\,K_{n}:n\in\omega\,\}}$ and ${\cal U}(k-1)={\{\,U_{n}:n\in\omega\,\}}$. Pick $n(j)\in\omega$ by induction using (1) and (2) so that

[TABLE]

In the rest of the construction below an infinite set $J\subseteq{\{\,n(j):j\in\omega\,\}}$ will be selected so that $D(J,i)={\{\,d_{j}^{i}:j\in J\,\}}$ are independent subsets that satisfy some additional properties. Note that $D(J^{\prime},i)\to 0$ in $\tau({\cal U}(k-1))$ for every $i\in I$ and any infinite $J^{\prime}\subseteq J$.

Let $i\in I$. If $D(J,i)$ is not closed and discrete in $\mathop{k_{\omega}}({\cal K}(k-1))$ then by (2) there exists an infinite $J_{i}\subseteq J$ such that $D(J_{i},i)\to 0$ in $\mathop{k_{\omega}}({\cal K}(k-1))$. Otherwise one can find an infinite $J_{i}\subseteq J$ such that for any infinite $J^{\prime}\subseteq J_{i}$ ${\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}(k-1)}({D(J_{i},i)})={\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}(k-1)}({D(J^{\prime},i)})$.

Repeating this argument for every $i\in I$ in the natural order one can build $J_{i}$ such that $J_{i^{\prime}}\subseteq J_{i}$ if $i^{\prime}\leq i$ and one of the two alternatives above holds. After replacing $J$ with $J=\cap_{i\in I}J_{i}$ we may assume that either $D(J,i)\to 0$ in $\mathop{k_{\omega}}({\cal K}(k-1))$ or $D(J,i)$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}(k-1))$ and ${\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}(k-1)}({D(J,i)})={\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}(k-1)}({D(J^{\prime},i)})$ for any infinite $J^{\prime}\subseteq J$.

Let $I=I_{u}\cup I_{z}\cup I_{h}$ where $D(J,i)\to 0$ in $\mathop{k_{\omega}}({\cal K}(k-1))$ for every $i\in I_{u}$, $D(J,i)$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}(k-1))$ for $i\in I_{z}\cup I_{h}$, ${\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}(k-1)}({D(J^{\prime},i)})\geq\sigma^{i}-1$ for each infinite $J^{\prime}\subseteq J$ and $i\in I_{z}$, and ${\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}(k-1)}({D(J,i)})<\sigma^{i}-1$ for every $i\in I_{h}$.

Passing to a subset of $J$ if necessary we may assume that ${\langle{D(J,i)}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}(k-1))$ for every $i\in I_{z}\cup I_{h}$. Since $({\cal C},\xi)$ is a sequential scale the set $\cup_{i\in I_{z}\cup I_{h}}D(J,i)$ is independent and the group ${\langle{\cup_{i\in I_{z}\cup I_{h}}D(J,i)}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}(k-1))$. If $i\in I_{z}$ use $D(J,i)\to 0$ in $\tau({\cal U}(k-1))$ and Lemma 31 to find, after thinning out $J$ if necessary, an fpse ${(G_{\gamma(k-1)+1},{\cal K}_{\gamma(k-1)+1},{\cal U}_{\gamma(k-1)+1})}$ of ${(G(k-1),{\cal K}(k-1),{\cal U}(k-1))}$ such that $D(J,i)\to 0$ in $\mathop{k_{\omega}}({\cal K}_{\gamma(k-1)+1})$, $D(J,i^{\prime})\to 0$ in $\tau({\cal U}_{\gamma(k-1)+1})$ for every $i^{\prime}\in I_{z}\setminus\{i\}$ and $D(J,i^{\prime\prime})$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}_{\gamma(k-1)+1})$ for every $i^{\prime\prime}\in I_{h}\cup I_{z}\setminus\{i\}$. Note that while formally Lemma 31 does not guarantee that each $D(J,i^{\prime\prime})$ remains closed and discrete in $\mathop{k_{\omega}}({\cal K}_{\gamma(k-1)+1})$ for $i^{\prime\prime}\in I_{h}\cup I_{z}\setminus\{i\}$, it follows from $D(J,i^{\prime\prime})\to 0$ in $\tau({\cal U}_{\gamma(k-1)+1})$ that if $D(J,i^{\prime\prime})$ is not closed and discrete in $\mathop{k_{\omega}}({\cal K}_{\gamma(k-1)+1})$, after passing to a smaller $J$ if necessary we may assume that $D(J,i^{\prime\prime})\to 0$ in $\mathop{k_{\omega}}({\cal K}_{\gamma(k-1)+1})$. We may thus move such $i^{\prime\prime}$ to $I_{u}$ and proceed with the argument. Alternatively, one may note that the fpse constructed above is of finite type and use Lemma 37.

Repeating this argument for every $i\in I_{z}$ in the natural order and possibly passing to a subset of $J$ if necessary one can build a finite fpse-chain ${\{\,{(G_{\gamma},{\cal K}_{\gamma},{\cal U}_{\gamma})}:\gamma(k-1)<\gamma\leq\gamma_{z}=\gamma(k-1)+|I_{z}|\,\}}$ above ${(G(k-1),{\cal K}(k-1),{\cal U}(k-1))}$ such that $D(J,i)\to 0$ in $\mathop{k_{\omega}}({\cal K}_{\gamma_{z}})$ for every $i\in I_{u}\cup I_{z}$ and $D(J,i)$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}_{\gamma_{z}})$ for every $i\in I_{h}$.

Let $i\in I_{h}$. Use Lemma 39 to find an fpse-chain ${\{\,{(G_{\gamma},{\cal K}_{\gamma},{\cal U}_{\gamma})}:\gamma_{z}<\gamma<\gamma_{s}(i)\,\}}$ above ${(G_{\gamma_{z}},{\cal K}_{\gamma_{z}},{\cal U}_{\gamma_{z}})}$ close to ${\langle{S^{i}}\rangle}$ and an infinite $J_{i}\subseteq J$ such that $D(J_{i},i)$ is closed, discrete, and stable in ${(G_{\gamma_{s}(i)},{\cal K}_{\gamma_{s}(i)},{\cal U}_{\gamma_{s}(i)})}$. Note that $D(J,i^{\prime})$ remains closed discrete in ${(G_{\gamma_{s}(i)},{\cal K}_{\gamma_{s}(i)},{\cal U}_{\gamma_{s}(i)})}$ for every $i^{\prime}\in I_{h}$ by Lemma 37.

Repeating this argument for every $i\in I_{h}$ in the natural order and possibly passing to a subset of $J$ if necessary one may build an fpse-chain ${\{\,{(G_{\gamma},{\cal K}_{\gamma},{\cal U}_{\gamma})}:\gamma_{z}<\gamma<\gamma_{s}\,\}}$ of finite type over ${(G_{\gamma_{z}},{\cal K}_{\gamma_{z}},{\cal U}_{\gamma_{z}})}$ such that each $D(J,i)$, $i\in I_{h}$ is closed, discrete, and stable in ${(G_{<\gamma_{s}},{\cal K}_{<\gamma_{s}},{\cal U}_{<\gamma_{s}})}$. Put $\delta_{i}={\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}_{<\gamma_{s}}}({D(J,i)})$ and note that $\delta_{i}<\sigma^{i}-1$ for every $i\in I_{h}$.

For every $i\in I_{h}$ pick a $K(i)\in{\cal K}_{<\gamma_{s}}$, a $D^{+}(J,i)={\{\,d_{j}(i):j\in J\,\}}\subseteq[{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}]_{\delta_{i}}^{\mathop{k_{\omega}}({\cal K}_{<\gamma_{s}})}$ and $d^{i}_{j}=d_{j}(i)+a_{j}(i)$ for every $j\in J$ where $a_{j}(i)\in K(i)$. By thinning out $J$ we may assume that $a_{j}(i)\to a(i)$ for some $a(i)\in K(i)$.

Let $\nu=|I_{h}|$ and $I_{h}=\{\,i_{1},\ldots,i_{\nu}\,\}$ where the order of $i_{j}$ is chosen so that the following assumption holds. Let successor $\eta^{i}(k)<\omega_{1}$ for $i\in I_{h}$ be chosen so that $\sigma^{i}-1\geq\delta_{i}+\eta^{i}(k)\geq\sigma^{i}_{k}$ and assume that $\eta^{i_{j}}(k)$ are increasing for each $k\in\omega$. Put $d_{n}^{\prime}=\sum_{i\in I_{u}\cup I_{z}}d_{n}(i)$. Then $d_{n}^{\prime}\to 0$ in $\mathop{k_{\omega}}({\cal K}_{<\gamma_{s}})$. Use Lemma 41 to construct a wggs

[TABLE]

above

[TABLE]

so that

[TABLE]

and

[TABLE]

Recursively applying Lemma 41 and using $\delta_{i_{j}}+\eta^{i_{j-1}}\leq\delta_{i_{j}}+\eta^{i_{j}}$ for $j>1$ construct wggs

[TABLE]

above

[TABLE]

for every $j\leq\nu$ so that

[TABLE]

and

[TABLE]

Put $\gamma(k)=\gamma(i_{\nu},k)$, ${(G(k),{\cal K}(k),{\cal U}(k))}={(G_{\gamma(k)},{\cal K}_{\gamma(k)},{\cal U}_{\gamma(k)})}$ and $d(i,k)=a(i)+s_{\gamma(i,k)}$ if $i\in I_{h}$, otherwise put $d(i,k)=0$.

If $U\in{\cal U}(k-1)$ then $D(J,i)\subseteq^{*}U$ and ${\{\,a_{j}(i):j\in J\,\}}\subseteq^{*}a(i)+U$ so $D^{+}(J,i)\subseteq^{*}a(i)+U$. Now $d(i,k)\in\overline{a(i)+D^{+}(J,i)}^{\mathop{k_{\omega}}({\cal K}(k))}\subseteq\overline{U}^{\tau({\cal U}(k))}$.

By the construction, $\sum_{i\in I_{h}}s_{\gamma(i,k)}\in[{\{\,\sum_{i\in I}d_{n}(i):n\in J\,\}}]_{\eta^{i_{\nu}}(k)}^{\mathop{k_{\omega}}({\cal K}(k))}$ and $\sum_{i\in I}a_{n}(i)\to\sum_{i\in I}a(i)$, so $\sum_{i\in I}d(i,k)\in[{\{\,\sum_{i\in I}d^{i}_{n}:n\in J\,\}}]_{\eta(k)}^{\mathop{k_{\omega}}({\cal K}(k))}$ by Lemma 1, where $\eta(k)=\eta^{i_{\nu}}(k)\leq\max{\{\,\sigma^{i}-1:i\in I\,\}}$.

Since $p_{n}-\sum_{i\in I}d^{i}_{n}\in K$ for all but finitely many $n\in J$ by the choice of $p_{n(j)}$, there exists a $p(k)\in[P_{\alpha}]_{\eta(k)}^{\mathop{k_{\omega}}({\cal K}(k))}$ such that $p(k)-\sum_{i\in I}d(i,k)\in K$ by Lemma 2. Since $d_{n}^{i},p_{n},p_{n}-\sum_{i\in I}d^{i}_{n}\in\overline{U}^{\tau({\cal U}(k))}$ for all but finitely many $n\in J$, $p(k)\in\overline{U}^{\tau({\cal U}(k))}$ so (4) holds.

Let $g\in G_{\gamma(i,k)}\setminus G_{<\gamma(i,k)}$ and $\mathop{\mathfrak{so}}(g,{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}(k)))<\sigma^{i}_{k}\leq\sigma^{i}-1$. Then $i=i_{j}\in I_{h}$ for some $j\leq\nu$ by (1), Lemma 37 and the construction of $W_{j}$. Then by Lemma 32 $\mathop{\mathfrak{so}}(g,{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}_{\gamma(i,k)}))<\sigma^{i}_{k}$ contradicting $\mathop{\mathfrak{so}}(g,{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}_{\gamma(i,k)}))\geq\mathop{\mathfrak{so}}(s_{\gamma(i,k)},{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}_{\gamma(i,k)}))\geq\sigma^{i}_{k}$ by (5) and the construction of $W_{j}$. Thus (3) holds.

Let $\gamma(\omega)=\lim_{k\in\omega}\gamma(k)$ and ${(G(\omega),{\cal K}(\omega),{\cal U}(\omega))}={(G_{<\gamma(\omega)},{\cal K}_{<\gamma(\omega)},{\cal U}_{<\gamma(\omega)})}$.

If $U^{\prime}\in{\cal U}(\omega)$ then $U^{\prime}=\overline{U}^{\tau({\cal U}(\omega))}$ where $U\in{\cal U}(k-1)$ for some $k\in\omega$. By (4) $d(i,n),p(n),p(n)-\sum_{i\in I}d(i,n)\in\overline{U}^{\tau({\cal U}(n))}\subseteq U^{\prime}$ for all $n\geq k$. Since $b(n)=p(n)-\sum_{i\in I}d(i,n)\in K$ by (4), it follows that $b(n)\to 0$ in $\mathop{k_{\omega}}({\cal K}(\omega))$.

By thinning out and reindexing we may assume that $d(i,n)\not=0$ if $i\in I_{0}$ and $d(i,n)=0$ if $i\not\in I_{0}$ for some $I_{0}\subseteq I$ and every $n\in\omega$. If $g\in G_{\gamma(i,n)}\setminus G_{<\gamma(i,n)}$ then $\mathop{\mathfrak{so}}(g,{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}_{\gamma(i,n)}))\geq\sigma^{i}_{n}$ by (3) and $\gamma(i,n)$ is cofinal in $\lim_{k\in\omega}\gamma(k)$. Thus by (2) and Lemma 40 ${\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}(\omega)}({D^{i}})=\sigma^{i}-1$ where $D^{i}={\{\,d(i,n):n\in\omega\,\}}$ for every $i\in I_{0}$ and ${\langle{\cup_{i\in I_{0}}D^{i}}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}(\omega))$.

Using Lemma 31 we can build a finite fpse-chain

[TABLE]

above ${(G(\omega),{\cal K}(\omega),{\cal U}(\omega))}$ relative to $({\cal C},\xi)$ such that $D^{i}_{0}\to 0$ in $\mathop{k_{\omega}}({\cal K}_{\gamma_{t}})$ for each $i\in I_{0}$ where $D^{i}_{0}={\{\,d(i,n):n\in J_{0}\,\}}$ for some infinite $J_{0}\subseteq\omega$. Then ${\{\,p(n):n\in J_{0}\,\}}\to 0$ and $p(n)\in[P_{\alpha}]_{\eta^{\prime}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{t}})}$ where $\eta^{\prime}=\sup_{k\in\omega}\eta(k)\leq\max{\{\,\sigma^{i}-1:i\in I\,\}}$ so $0\in[P_{\alpha}]_{\eta}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{t}})}$ where $\eta\leq\max{\{\,\sigma^{i}:i\in I\,\}}$. Put ${\bf m}(\alpha+2)=\gamma_{t}$.

Put ${\bf m}(\alpha+3)={\bf m}(\alpha+2)+1$. Let ${(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime})}$ be an arbitrary fpse of ${(G_{{\bf m}(\alpha+2)},{\cal K}_{{\bf m}(\alpha+2)},{\cal U}_{{\bf m}(\alpha+2)})}$. Using Lemmas 22, 9, and 30 extend ${\cal U}^{\prime}$ to a countable family of open (in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+2)})$) subgroups of $G_{{\bf m}(\alpha+2)}$ of finite index ${\cal U}^{\prime\prime}\supseteq{\cal U}^{\prime}$ so that (4) holds after replacing ${\cal U}_{{\bf m}(\alpha+3)}$ with ${\cal U}^{\prime\prime}$ and let ${(G_{{\bf m}(\alpha+3)},{\cal K}_{{\bf m}(\alpha+3)},{\cal U}_{{\bf m}(\alpha+3)})}=(G^{\prime},{\cal K}^{\prime},{\cal U}^{\prime\prime})$. ∎

Suppose $\diamondsuit$ holds and let ${\{\,C_{\alpha}:\alpha<\omega_{1}\,\}}$ be a $\diamondsuit$-sequence. Identifying $\omega_{1}$ and $2^{\omega}$ we may assume that each $C_{\alpha}\subseteq 2^{\omega}$. Let ${\{\,P_{\alpha}:\alpha<\omega_{1}\hbox{ is a limit ordinal}\,\}}$ list all infinite countable subsets of $2^{\omega}$ so that each $P_{\alpha}$ is listed $\omega_{1}$ times.

Lemma 45**.**

Let ${(G_{\omega_{1}},{\cal K}_{\omega_{1}},{\cal U}_{\omega_{1}})}={(G_{<\omega_{1}},{\cal K}_{<\omega_{1}},{\cal U}_{<\omega_{1}})}$ where ${(G_{\alpha},{\cal K}_{\alpha},{\cal U}_{\alpha})}$, $\alpha<\omega_{1}$ have been constructed in Lemma 44. Then $\mathop{k_{\omega}}({\cal K}_{\omega_{1}})=\tau({\cal U}_{\omega_{1}})$, $G_{\omega_{1}}$ is countably compact, and $\mathop{\mathfrak{so}}(\mathop{k_{\omega}}({\cal K}_{\omega_{1}}))=\sup{\{\,\sigma^{m}:m\in\omega\,\}}$.

Proof.

Suppose $A\subseteq G_{\omega_{1}}$ is such that $0\not\in A$, $A\cap K$ is closed for every $K\in{\cal K}_{\omega_{1}}$ and $A\cap U\not=\varnothing$ for every $U\in{\cal U}_{\omega_{1}}$.

Let $\theta$ be a large enough cardinal. Consider the sets of the form $A\cap M$ where $M$ is a countable elementary submodel of $H(\theta)$ and $X\in M$ is a countable set containing the details of the construction of $G_{\omega_{1}}$. The set

[TABLE]

is a club in $\omega_{1}$. Thus $C_{\gamma}=A\cap M$ for some $\gamma<\omega_{1}$ where $M\cap\omega_{1}=\gamma$. Note that $\gamma={\bf m}(\alpha)$ for some limit $\alpha<\omega_{1}$ and ${\bf c}{C_{\alpha}}=A\cap G_{<{\bf m}(\alpha)}$.

Let ${\cal K}_{<\gamma}={\{\,K_{i}:i\in\omega\,\}}$, ${\cal U}_{\omega_{1}}\cap M={\{\,U_{i}:i\in\omega\,\}}$. Note that for any $U\in{\cal U}_{<\gamma}$ there is a $U_{i}$ such that $U_{i}\cap G_{<\omega}=U$ (since $\overline{U}^{{\cal K}_{\omega_{1}}}\in{\cal U}_{\omega_{1}}$). Pick points $d_{n}\in A\cap(\cap_{i\leq n}U_{i})\cap M$ so that

[TABLE]

If the recursion does not terminate by (3.a) there exists an $S\subseteq D={\{\,d_{n}:n\in\omega\,\}}\subseteq A$ such that $S\to 0$ in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})$ contradicting the choice of $A$.

If the recursion stops at some $n\in\omega$ let $K^{\prime}\in{\cal K}_{<\gamma}\cap M$ be such that $\sum_{i<n}K_{i}+{\langle{{\{\,d_{i}:i<n\,\}}}\rangle}\subseteq K^{\prime}$. Then for every $d\in A\cap(\cap_{i\leq n}U_{i})\cap M$ there is an $s\in\sum_{m<n}\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<\gamma})}\subseteq\sum_{m<n}\overline{{\langle{S^{m}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})}$ such that $d\in s+K^{\prime}$.

By elementarity $A\cap U\subseteq\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})}+K$ for some $m\in\omega$, $U\in{\cal U}_{<\omega_{1}}$, and $K\in{\cal K}_{\omega_{1}}$. We may assume that $A=A\cap U$ by picking a subset if necessary.

Pick points $d_{n}\in A\cap(\cap_{i\leq n}U_{i})\cap M$ so that $d_{n}=a_{n}+\sum_{i\leq m}d_{n}^{i}$ where $d_{n}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<\gamma})}$ and $a_{n}\in K$. By passing to a subsequence and reindexing if necessary, assume $a_{n}\to a$. Now $d_{n}\to 0$ so $\sum_{i\leq m}d_{n}^{i}\to a$ in $\tau({\cal U}_{<\gamma})$. Let $K\in{\cal K}_{\beta}$ for some $\beta<\gamma$.

If $a\not\in\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\beta^{\prime}})}$ for any $\beta^{\prime}<\gamma$ let $\alpha^{\prime}\in\omega_{1}$ be a limit such that $\beta<{\bf m}(\alpha^{\prime}+3)<\gamma$. Such $\alpha^{\prime}$ exists since $\beta,{\bf m}\in M$. Then $a\not\in\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha^{\prime}+3)})}$ and by (4) there is a clopen $U\in{\cal U}_{{\bf m}(\alpha^{\prime}+3)}$ such that $a\not\in U$ and $\sum_{i\leq m}{{\langle{S^{i}}\rangle}}\subseteq U$. Note that $U=U_{i}\cap G_{{\bf m}(\alpha^{\prime}+3)}$ for some $i\in\omega$ so $a\not\in U_{i}$ and $\sum_{i\leq m}d_{n}^{i}\in U_{i}$ for every $n\in\omega$ contradicting $\sum_{i\leq m}d_{n}^{i}\to a$ in $\tau({\cal U}_{<\gamma})$.

Thus $a\in\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<\gamma})}$. By (1) there exists a $\gamma_{0}<\omega_{1}$ such that $\gamma_{0}>\gamma$ and $D^{i}\to d^{i}$ in $\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})$ for each $i\leq m$ where $D^{i}\subseteq{\{\,d_{n}^{i}:n\in\omega\,\}}$ is infinite. By thinning out and reindexing, we may assume that $d_{n}^{i}\to d^{i}$ in $\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})$. Then $d_{n}\to d=a+\sum_{i\leq m}d^{i}$ in $\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})$ so $d\in A\cap\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}$ by Lemma 1.

Let $U\in{\cal U}_{\omega_{1}}\cap M$ then $d_{n}\in U$ for all but finitely many $n\in\omega$ so $d\in U$. Thus for every $U\in{\cal U}_{\omega_{1}}\cap M$ there is a $\gamma^{\prime}<\omega_{1}$ and a $d\in A\cap(\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma^{\prime}})})\cap\overline{U}^{\tau({\cal U}_{\gamma^{\prime}})}$. Then the set $A^{\prime}=A\cap(\sum_{i\leq m}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})})$ is closed in $\mathop{k_{\omega}}({\cal K}_{\omega_{1}})$ and by elementarity $0\in\overline{A^{\prime}}^{\tau({\cal U}_{\omega_{1}})}$ so assume that $A\subseteq\sum_{i\leq m}{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})}$.

Let $I\subseteq m$ be such that $0\in\overline{A^{\prime}}^{\tau({\cal U}_{\omega_{1}})}$ and $0\not\in\overline{A^{\prime\prime}}^{\tau({\cal U}_{\omega_{1}})}$ for any $A^{\prime\prime}=A\cap(\sum_{i\in I^{\prime}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})})$ where $A^{\prime}=A\cap(\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})})$, $I^{\prime}\subset I$, $I^{\prime}\not=I$. Using this property of $I$ and picking a closed (in $\mathop{k_{\omega}}({\cal K}_{\omega_{1}})$) subset of $A^{\prime}$ if necessary we may assume that every $d\in A$ has the property that $d=\sum_{i\in I}d^{i}\in A$ where $d^{i}\not=0$ and $d^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})}$.

Pick points $d_{n}=\sum_{i\in I}d_{n}^{i}\in A\cap(\cap_{k\leq n}U_{k})\cap M$ so that $d_{n}^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{<\gamma})}\cap(\cap_{k\leq n}U_{k})$ for every $i\in I$.

If the recursion terminates at some $n\in\omega$ then there exists a $U\in{\cal U}_{\omega_{1}}\cap M$ such that for every $d\in A\cap M$ $d\not\in\sum_{i\in I}(\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma^{\prime}})}\cap U)$ for any $\gamma^{\prime}\in M$. Let $\gamma^{\prime}<\gamma$ be such that $U\cap G_{\gamma^{\prime}}\in{\cal U}_{\gamma^{\prime}}$. Then $U\cap G_{\gamma^{\prime\prime}}\in{\cal U}_{\gamma^{\prime\prime}}$ for any $\gamma^{\prime\prime}\geq\gamma^{\prime}$. Let $\alpha^{\prime}<\gamma$ be a limit such that $\gamma^{\prime}<{\bf m}(\alpha^{\prime}+3)<\gamma$. By (4) there exists a $U^{\prime}\in{\cal U}_{\omega_{1}}\cap M$ such that $U^{\prime}\subseteq U$, $U^{\prime}\cap G_{{\bf m}(\alpha^{\prime}+3)}\in{\cal U}_{{\bf m}(\alpha^{\prime}+3)}$, and $U^{\prime}\cap\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha^{\prime}+3)})}=\sum_{i\in I}(\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha^{\prime}+3)})}\cap U^{\prime})$. Let $\gamma_{0}\geq{\bf m}(\alpha^{\prime}+3)$ by such that $\gamma_{0}<\gamma$ and $d=\sum_{i\in I}d^{i}\in A\cap U^{\prime}\cap M$ for some $d^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}$. Such $d$ and $\gamma_{0}$ exist by elementarity since $A,U^{\prime},{\cal K}_{\omega_{1}},{\cal C}\in M$. Note that for every $i\in I$ the set $\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}\cap U^{\prime}$ is closed in $\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})$ so by Lemma 22 the set $\sum_{i\in I}(\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}\cap U^{\prime})$ is closed in $\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})$. Since $U^{\prime}\cap\sum_{i\in I}{\langle{S^{i}}\rangle}$ is dense in $U^{\prime}\cap\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}$ this shows that $U^{\prime}\cap\sum_{i\in I}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}=\sum_{i\in I}(\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}\cap U^{\prime})$. Then $d^{i}\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}\cap U^{\prime}\subseteq\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{0}})}\cap U$ (note that $d$ uniquely determines $d^{i}$ by (2)) contradicting the choice of $U$.

Thus $D^{i}={\{\,d_{n}^{i}:n\in\omega\,\}}\to 0$ in $\tau({\cal U}_{<\gamma})$ for each $i\in I$. If $D^{i}$ is not closed and discrete in $\mathop{k_{\omega}}({\cal K}_{<\gamma})$ for some $i\in I$ then for some infinite $J\subseteq\omega$ ${\{\,d_{n}^{i}:i\in J\,\}}\to 0$ in $\mathop{k_{\omega}}({\cal K}_{\omega_{1}})$. Using (1) and possibly thinning out $J$ we may assume that ${\{\,d_{n}^{j}:n\in J\,\}}\to d(j)$ for every $j\in I$ in $\mathop{k_{\omega}}({\cal K}_{\omega_{1}})$. Then $d=\sum_{i^{\prime}\in I}d(i^{\prime})\in\sum_{i^{\prime}\in I\setminus\{i\}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\omega_{1}})}$ and $d\in A$ contradicting the minimality of $I$.

Thus each $D^{i}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}_{<\gamma})$. If ${\bf h}_{{\langle{S^{i^{\prime}}}\rangle}}^{{\cal K}_{<\gamma}}({D^{i^{\prime}}})\geq\sigma^{i^{\prime}}-1$ for some $i^{\prime}\in I$ then by Lemma 31 there exists an fpse ${(G_{\gamma},{\cal K}_{\gamma},{\cal U}_{\gamma})}$ of ${(G_{<\gamma},{\cal K}_{<\gamma},{\cal U}_{<\gamma})}$ such that ${\{\,d_{n}^{i^{\prime}}:n\in J\,\}}\to 0$ in $\mathop{k_{\omega}}({\cal K}_{\gamma})$ and a finite fpse-chain ${\{\,{(G_{\lambda},{\cal K}_{\lambda},{\cal U}_{\lambda})}:\gamma<\lambda<\gamma+k\,\}}$ for some $k\in\omega$ such that ${\{\,d_{n}^{i}:n\in J\,\}}\to d(i)$ in $\mathop{k_{\omega}}({\cal K}_{<\gamma+k})$ for some $d(i)\in G_{<\gamma+k}$ and some infinite $J\subseteq\omega$. As above $d=\sum_{i\in I}d(i)\in\sum_{i\in I\setminus\{i^{\prime}\}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{\gamma})}$ and $d\in\overline{A}^{{\cal K}_{\gamma}}$. Thus by (3.b) $A\cap\sum_{i\in I^{\prime}}\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}\not=\varnothing$ for some $I^{\prime}\subset I$, $I^{\prime}\not=I$ contradicting the minimality of $I$.

Now ${\bf h}_{{\langle{S^{i}}\rangle}}^{{\cal K}_{<\gamma}}({D^{i}})<\sigma^{i}-1$ for every $i\in I$. Pick an $i^{\prime}\in I$ and use Lemma 10 to build an fpse-chain ${\{\,{(G_{\gamma},{\cal K}_{\gamma},{\cal U}_{\gamma})}:{\bf m}(\alpha)\leq\gamma\leq{\bf m}(\alpha)+k\,\}}$ of finite type above ${(G_{<{\bf m}(\alpha)},{\cal K}_{<{\bf m}(\alpha)},{\cal U}_{<{\bf m}(\alpha)})}$ away from ${\langle{S^{i^{\prime}}}\rangle}$ for some $k\in\omega$ so that for some infinite $J\subseteq\omega$ ${\{\,d_{n}^{i}:n\in J\,\}}\to d^{i}\not\in G_{<{\bf m}(\alpha)}$ in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha)+k})$ for $i\in I\setminus\{i^{\prime}\}$. Then $P={\{\,d_{n}^{i^{\prime}}:n\in J\,\}}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha)+k})$ by Lemma 37.

Use Lemma 19 to construct an fpse-chain ${\{\,{(G_{\gamma},{\cal K}_{\gamma},{\cal U}_{\gamma})}:{\bf m}(\alpha)+k<\gamma\leq\gamma_{0}\,\}}$ such that, after possibly thinning out $J$, $P$ is stable in ${(G_{\gamma_{0}},{\cal K}_{\gamma_{0}},{\cal U}_{\gamma_{0}})}$ and ${\langle{P}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}{{\cal K}_{\gamma_{0}}}$.

Let $J=\cup_{j\in\omega}J_{j}$ be such that $J_{j}$ are infinite and disjoint and $P^{j}={\{\,d_{n}^{i^{\prime}}:n\in J_{j}\,\}}$.

Repeatedly apply Lemma 41 to build wggs

[TABLE]

over $(D_{j},G_{\gamma_{j}},{\cal K}_{\gamma_{j}},{\cal U}_{\gamma_{j}},\delta)$ where $\delta={\bf h}_{{{\langle{S^{i^{\prime}}}\rangle}\setminus{\langle{2{S^{i^{\prime}}}}\rangle}}}^{{\cal K}_{\gamma_{j}}}({P^{j}})={\bf h}_{{{\langle{S^{i^{\prime}}}\rangle}\setminus{\langle{2{S^{i^{\prime}}}}\rangle}}}^{{\cal K}_{\gamma_{0}}}({P})$ and $D_{j}\subseteq[{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}]_{\delta}^{\mathop{k_{\omega}}({\cal K}_{<\gamma_{j}})}$ is such that $P^{j}\subseteq D_{j}+K(j)$ and $D_{j}\subseteq P^{j}+K(j)$ for some $K(j)\in{\cal K}_{\gamma_{j}}$ (due to the stability of $P$ we may assume that $K(j)=K\in{\cal K}_{\gamma_{0}}$ but this stronger property is not needed) such that

[TABLE]

and ${\langle{\cup_{k>j}P^{j}}\rangle}$ is closed and discrete in $\mathop{k_{\omega}}({\cal K}_{\gamma_{j+1}})$. Let $\gamma^{0}=\lim\gamma_{j}$.

Use Corollary 1 to find a $d(i^{\prime},n)\in\overline{P^{n}}^{\mathop{k_{\omega}}({\cal K}_{\gamma_{n+1}})}$ such that $d(i^{\prime},n)=a(n)+s_{\gamma_{n+1}}$ for some $a(n)\in K(n)$. Note that $d(i^{\prime},n)\in G_{\gamma_{n+1}}\setminus G_{<\gamma_{n+1}}$ since $a(n)\in K(n)\in{\cal K}_{<\gamma_{n+1}}$ and by (5)

[TABLE]

for any $g\in G_{\gamma_{n+1}}\setminus G_{<\gamma_{n+1}}$. Let $d(i,n)=d^{i}$ for $i\in I\setminus\{i^{\prime}\}$. Put $d(n)=\sum_{i\in I}d(i,n)$. Then $d(n)\in\overline{A}^{\mathop{k_{\omega}}({\cal K}_{\gamma^{0}})}$ by Corollary 1. Since $D^{i}\to 0$ in $\tau({\cal U}_{<\gamma})$, $d(i,n)\in\cap{\{\,\overline{U}^{\tau({\cal U}_{<\gamma^{0}})}:U\in{\cal U}_{<\gamma}\,\}}$.

Now by (3.c) for every $n\in\omega$ there exists a $d(n)\in G_{{\bf m}(\alpha+1)}\setminus G_{<\gamma}$ such that $d(n)\in\overline{A}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}$ and $d(n)=\sum_{i\in I}d(i,n)$ where

[TABLE]

and $\mathop{\mathfrak{so}}(d(i^{\prime},n),{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)}))\geq\sigma^{i^{\prime}}_{n}$ for some $i^{\prime}\in I$. Note that $d(i,n)\not\in M$.

Thus for any $U\in{\cal U}_{\omega_{1}}\cap M$, any $\gamma^{\prime}\in\omega_{1}\cap M$, and any $n\in\omega$ there exists a $d(n)=\sum_{i\in I}d(i,n)\in A$ where $d(i,n)\in\overline{{\langle{S^{i}}\rangle}}^{\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)})}\cap U$, $d(i,n)\not\in G_{\gamma^{\prime}}$, and $\mathop{\mathfrak{so}}(d(i^{\prime},n),{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle},\mathop{k_{\omega}}({\cal K}_{{\bf m}(\alpha+1)}))\geq\sigma^{i^{\prime}}_{n}$ for some $i^{\prime}\in I$, $\alpha<\omega_{1}$.

By elementarity such $d(n)$ and $d(i,n)$ exist in every $(\cap_{j\leq n}U_{j}\setminus G_{<\gamma_{n}})\cap M$ for some $\gamma_{n}$ cofinal in $\gamma$. Now ${\bf h}_{{\langle{S^{i}}\rangle}\setminus{\langle{2S^{i}}\rangle}}^{{\cal K}_{<\gamma}}({{\{\,d({i^{\prime}},n):n\in\omega\,\}}})\geq\sigma^{i^{\prime}}-1$ by Lemma 40 and $d(i^{\prime},n)\to 0$ in ${\cal U}_{<{\bf m}(\alpha)}$ so by an argument similar to the one above using (3.b) one obtains a contradiction with the minimality of $I$.

The upper bound estimate for $\mathop{\mathfrak{so}}(G_{\omega_{1}})$ follows from (2), while the lower bound estimate follows from Lemma 32. Countable compactness follows from (1). ∎

Theorem 1** ($\diamondsuit$).**

Let $\sigma\leq\omega_{1}$. There exists a countably compact sequential group $G$ such that $\mathop{\mathfrak{so}}(G)=\sigma$.

Proof.

The existence of such $G$ for $\sigma<\omega_{1}$ follows from Lemma 45. For $\sigma=\omega_{1}$ consider a $\Sigma$ product of $\omega_{1}$ countably compact sequential groups $G_{\alpha}$ such that $\mathop{\mathfrak{so}}(G_{\alpha})=\alpha$.

Note that the group of sequential order $\omega_{1}$ constructed above is not separable unlike the examples of smaller sequential order. A modified construction similar to that of Lemma 45 may be used to construct a separable example of sequential order $\omega_{1}$, as well, although we omit the details. ∎

Corollary 3**.**

The existence of a countably compact sequential non Fréchet group is independent of the axioms of ZFC.

Proof.

See Theorem 1 and [17] Theorem 2. ∎

We conclude by listing some open questions.

Question 1**.**

Does Theorem 1 follow from CH alone?

Question 2**.**

Is the existence of a countably compact sequential group $G$ such that $K\subseteq G$ for some compact subspace $K$, $\mathop{\mathfrak{so}}(K)\geq 2$ consistent with ZFC?

Question 3**.**

Can the construction in Theorem 1 be made Cohen-indestructible (see, for example, [6] for the relevant definitions)?

The last question requires some clarification. Since the group $G$ will contain compact subspaces homeomorphic to the Cantor cube $2^{\omega}$, the addition of any reals will destroy the sequentiality of $G$. Each such subspace, as well as $G$ itself, will inherit a precompact topology from the ground model, however, so the group $\overline{G}\supseteq G$ in the extension may be defined by taking the Raikov completion of each compact subspace in ${\cal K}_{\omega_{1}}$ and letting the topology of $\overline{G}$ be determined by the new family of compact subspaces.

7 Acknowledgements.

The research in this paper was started when one of the authors (Alexander Shibakov) was visiting the Department of Mathematics at Ehime University in the Summer of 2017. He would like to thank the Department and Prof. Shakhmatov for the support that made this visit possible and productive.