The Frame of Nuclei of an Alexandroff Space

TL;DR

This paper characterizes when the frame of nuclei of the open set frame of an Alexandroff space is spatial, linking it to the embeddability of an infinite binary tree into the space's specialization preorder.

Contribution

It establishes a precise condition involving the embedding of an infinite binary tree for the spatiality of the frame of nuclei in Alexandroff spaces.

Findings

The frame of nuclei is spatial iff the infinite binary tree does not embed into the space.

Provides a characterization connecting topological and order-theoretic properties.

Bridges the structure of Alexandroff spaces with combinatorial tree properties.

Abstract

Let $\mathcal{O}S$ be the frame of open sets of a topological space $S$, and let $N(\mathcal{O}S)$ be the frame of nuclei of $\mathcal{O}S$. For an Alexandroff space $S$, we prove that $N(\mathcal{O}S)$ is spatial iff the infinite binary tree $\mathscr T_2$ does not embed isomorphically into $(S, \le)$, where $\le$ is the specialization preorder of $S$.

Equations

$$\eta (a)=\{x\in X_L\mid a\in x\}.$$

$$\{\eta (a)\mid a\in L\}\text{mbox}and\{\eta (a)\setminus \eta (b)\mid a,b\in L\}$$

$$\uparrow A=\{x\in X_L\mid a\le x\text{mbox}forsomea\in A\},$$

$$\downarrow A=\{x\in X_L\mid x\le a\text{mbox}forsomea\in A\}.$$

$$\epsilon (s)=\{U\in \mathcal{O}S\mid s\in U\}.$$

$$x\in \downarrow \epsilon (s)$$

$$x\in X_L\setminus \eta \left(S\setminus \overline{\{s\}}\right)⟺S\setminus \overline{\{s\}}\notin x.$$

$$x\in \downarrow \epsilon (s)\text{mbox}iffx\in X_L\setminus \eta \left(S\setminus \overline{\{s\}}\right),$$

Full text

1. Introduction

Nuclei play an important role in pointfree topology as they characterize homomorphic images of frames (or dually sublocales of locales). For a frame $L$, let $N(L)$ be the frame of nuclei of $L$, also known as the assembly of $L$. The frame $N(L)$ has been investigated by many authors; see, e.g., [7, 10, 16, 2, 17, 13, 12, 14, 11, 20, 15, 6, 3, 19, 4] (which are listed in chronological order). For example, Beazer and Macnab [2] gave a necessary and sufficient condition for $N(L)$ to be boolean; Niefield and Rosenthal [14] gave necessary and sufficient conditions for $N(L)$ to be spatial, and derived that if $N(L)$ is spatial, then so is $L$; Simmons [17] proved that if $L$ is the frame of opens of a $T_{0}$-space $S$, then $N(L)$ is boolean iff $S$ is scattered; and Isbell [11] proved that if $L$ is the frame of opens of a sober space $S$, then $N(L)$ is spatial iff $S$ is weakly scattered (see Section 2 for definitions).

In [6] the study of $N(L)$ using the spectrum of $L$ was initiated. We utilized this approach in [1] to generalize the results mentioned above (and also to give alternate proofs of these results). One of the main results of [1] gives a necessary and sufficient condition for $N(L)$ to be spatial in terms of the spectrum of $L$, from which it is derived that if $L=\mathcal{O}S$ is the frame of opens of a topological space $S$, then $N(L)$ is spatial iff the soberification of $S$ is weakly scattered.

In the present paper we restrict our attention to Alexandroff spaces (in which each point has a least neighborhood). It is well known that Alexandroff spaces correspond to preordered sets, and Alexandroff $T_{0}$-spaces to partially ordered sets. Thus, the frame of opens of an Alexandroff space $S$ is isomorphic to the frame of upward closed sets of a preordered set. We prove that for an Alexandroff $T_{0}$-space $S$, the frame $N(\mathcal{O}S)$ is spatial iff the infinite binary tree $\mathscr{T}_{2}$ is not isomorphic to a subposet of $S$. From this we derive that for an arbitrary Alexandroff space $S$, the frame $N(\mathcal{O}S)$ is spatial iff the infinite binary tree $\mathscr{T}_{2}$ does not embed isomorphically into $S$.

We point out that if $S$ is a poset, then Simmons’s characterization of when $N(\mathcal{O}S)$ is boolean takes on the following form: $N(\mathcal{O}S)$ is boolean iff $S$ is noetherian (has no infinite ascending chains). Since $S$ being noetherian is equivalent to $S$ being sober, Isbell’s characterization of when $N(\mathcal{O}S)$ is spatial for sober $S$ does not yield any examples of posets $S$ such that $N(\mathcal{O}S)$ is spatial but not boolean. Our main result yields many such examples. Indeed, it implies that if $S$ is a poset with no infinite antichains, then $N(\mathcal{O}S)$ is spatial. In particular, if $S$ is totally ordered, then $N(\mathcal{O}S)$ is spatial. Thus, each totally ordered set (or more generally a poset with no infinite antichains) that is not noetherian yields an example of a spatial $N(\mathcal{O}S)$ which is not boolean.

2. Preliminaries

Definition 2.1**.**

For a frame $L$, let $X_{L}$ be the set of prime filters of $L$. We will refer to $X_{L}$ as the spectrum of $L$.

If $\leq_{L}$ is the inclusion order, then $(X_{L},\leq_{L})$ is a poset (partially ordered set). For $a\in L$, let

[TABLE]

There are several topologies on $X_{L}$, two of which play an important role in our considerations. Define $\tau_{L}$ and $\pi_{L}$ on $X_{L}$ by letting

[TABLE]

be the bases for $\tau_{L}$ and $\pi_{L}$, respectively. It is well known that $\tau_{L}$ is a spectral topology (sober and coherent) and $\pi_{L}$ is the patch topology of $\tau_{L}$, hence $\pi_{L}$ is a Stone topology (compact, Hausdorff, zero-dimensional).

The ordered space $(X_{L},\pi_{L},\leq_{L})$ is a Priestley space; that is, a compact ordered space satisfying the Priestley separation axiom: $x\not\leq_{L}y$ implies there is a clopen upset containing $x$ and missing $y$. When there is no danger of confusion, we will abbreviate $(X_{L},\pi_{L},\leq_{L})$ by $X_{L}$. Since $L$ is a Heyting algebra, $X_{L}$ is in fact an Esakia space (the downset of clopen is clopen). In addition, since $L$ is complete, the closure of each open upset is a clopen upset. Such spaces are often referred to as extremally order-disconnected Esakia spaces (see, e.g., [1, Sec. 3] and the references therein).

For $A\subseteq X_{L}$ we recall that the upset ${\uparrow}A$ and the downset ${\downarrow}A$ are defined by

[TABLE]

It is well known that if $A$ is closed, then both ${\uparrow}A$ and ${\downarrow}A$ are closed.

The next definition originates in [6]. The current terminology was given in [1, Def. 4.1].

Definition 2.2**.**

Let $L$ be a frame and $X_{L}$ its spectrum.

1. (1)

We call a closed subset $F$ of $X_{L}$ nuclear provided ${\downarrow}(F\cap U)$ is clopen for each clopen $U$ of $X_{L}$. 2. (2)

Let $N(X_{L})$ be the set of nuclear subsets of $X_{L}$. 3. (3)

If $F=\{x\}$ is nuclear, then we call $x$ a nuclear point. 4. (4)

Let $Y_{L}$ be the subset of $X_{L}$ consisting of nuclear points of $X_{L}$.

Theorem 2.3**.**

[6*, Thm. 30]** Let $L$ be a frame and $X_{L}$ its spectrum. Then $N(L)$ is dually isomorphic to $N(X_{L})$. *

We denote the restrictions of $\tau_{L}$ and $\pi_{L}$ to $Y_{L}$ by $\tau$ and $\pi$, respectively. Let $\mathcal{O}_{\tau}(Y_{L})$ be the frame of opens of $(Y_{L},\tau)$ and $\mathcal{O}_{\pi}(Y_{L})$ the frame of opens of $(Y_{L},\pi)$.

Theorem 2.4**.**

[1, Thm. 5.9]** For a frame $L$, the following are equivalent.

1. (1)

The frame $N(L)$ is spatial. 2. (2)

If $N\in N(X_{L})$ is nonempty, then $N\cap Y_{L}\neq\varnothing$. 3. (3)

$N(L)$* is isomorphic to $\mathcal{O}_{\pi}(Y_{L})$.*

For $F$ a closed subset of $X_{L}$, let $\max F$ be the set of maximal points and $\min F$ the set of minimal points of $F$. It is well known that for each $x\in F$ there are $m\in\min F$ and $M\in\max F$ with $m\leq_{L}x\leq_{L}M$. Therefore, if $F\neq\varnothing$, then $\max F,\min F\neq\varnothing$. The following is a useful corollary of Theorem 2.4.

Corollary 2.5**.**

Let $L$ be a frame. Then $N(L)$ is spatial iff $\max U\cap Y_{L}\neq\varnothing$ for each nonempty clopen downset $U$ of $X_{L}$.

Proof.

First suppose that $N(L)$ is spatial. Let $U$ be a nonempty clopen downset of $X_{L}$. Then $U\in N(X_{L})$, so $\max U\in N(X_{L})$ by [1, Cor. 4.5]. Since $U\neq\varnothing$, we have $\max U\neq\varnothing$. Thus, $\max U\cap Y_{L}\neq\varnothing$ by Theorem 2.4.

Conversely, suppose that $\max U\cap Y_{L}\neq\varnothing$ for each clopen downset $U$ of $X_{L}$. Let $N\in N(L)$ be nonempty. Then $\max N$ is nonempty and $\max N\in N(L)$. If $U={\downarrow}N$, then $U$ is a clopen downset and $\max U=\max N$. Therefore, $\max U\in N(L)$ is nonempty. By our assumption, $\max U\cap Y_{L}\neq\varnothing$. This implies that $\max N\cap Y_{L}\neq\varnothing$. Thus, $N(L)$ is spatial by Theorem 2.4. ∎

Let $S$ be a topological space and $T$ a subspace of $S$. We recall that $x\in T$ is an isolated point of $T$ if $\{x\}=U\cap T$ for some open subset $U$ of $S$, and that $x$ is a weakly isolated point of $T$ if $x\in U\cap T\subseteq\overline{\{x\}}$ for some open subset $U$ of $S$. Then $X$ is scattered if each nonempty closed subspace of $X$ has an isolated point, and $X$ is weakly scattered if each nonempty closed subspace of $X$ has a weakly isolated point. For a spatial frame $L$, to the conditions of Theorem 2.4 and Corollary 2.5, we could add that $(Y_{L},\tau)$ is weakly scattered.

Theorem 2.6**.**

[1, Thm. 7.3]** Let $L$ be a spatial frame. Then $N(L)$ is spatial iff $(Y_{L},\tau)$ is weakly scattered.

Remark 2.7**.**

If in Theorem 2.6 we do not assume that $L$ is spatial, then to $(Y_{L},\tau)$ being weakly scattered we need to add the condition that $Y_{L}$ is dense in $(X_{L},\pi_{L})$ [1, Thm. 5.5].

Since $N(L)$ spatial implies that $L$ is spatial, from now on we will assume that $L$ is a spatial frame, so $L=\mathcal{O}S$ for some topological space $S$. There is a natural map $\varepsilon:S\to X_{L}$ given by

[TABLE]

For $U\in\mathcal{O}S$ we have that $\varepsilon^{-1}\eta(U)=U$. Therefore, $\varepsilon$ is a continuous map from $S$ to $(X_{L},\tau_{L})$, and it is an embedding iff $S$ is a $T_{0}$-space.

Theorem 2.8**.**

[1, Prop. 7.1]** $(Y_{L},\tau)$ is homeomorphic to the soberification of $S$.

We can view $\varepsilon$ as the soberification map from $S$ to $Y_{L}$.

Remark 2.9**.**

For the reader’s convenience, we give an elementary argument for why $\varepsilon[S]\subseteq Y_{L}$. To see this we must show that ${\downarrow}\varepsilon(s)$ is clopen in $(X_{L},\pi_{L})$. It is sufficient to observe that ${\downarrow}\varepsilon(s)=X_{L}\setminus\eta\left(S\setminus\overline{\{s\}}\right)$ for each $s\in S$. We have

[TABLE]

On the other hand,

[TABLE]

Since $S\setminus\overline{\{s\}}$ is the largest open set missing $s$, we conclude that

[TABLE]

yielding the desired equality.

One of the key techniques of Simmons in the study of $N(\mathcal{O}S)$ is the notion of the front topology on $S$. We recall that the front topology on $S$ is the topology $\tau_{F}$ generated by $\{U\setminus V\mid U,V\in\mathcal{O}S\}$.

Theorem 2.10**.**

[1, Lem. 7.9]** The map $\varepsilon:(S,\tau_{F})\to(X_{L},\pi_{L})$ is a compactification of $(S,\tau_{F})$.

Since $(X_{L},\pi_{L})$ is a Stone space, it follows that $(Y_{L},\pi)$ is a zero-dimensional Hausdorff space. But in general, $(Y_{L},\pi)$ is not compact.

3. Main Theorem

We recall that $S$ is an Alexandroff space if the intersection of an arbitrary family of open sets is open. Equivalently $S$ is Alexandroff iff each point of $S$ has a least open neighborhood. It is well known that Alexandroff spaces are in 1-1 correspondence with preordered sets. Indeed, the specialization preorder on $S$, defined by $s\leq t$ iff $s\in\overline{\{t\}}$, is reflexive and transitive, and $U$ is open in $S$ iff $U$ is an upset (that is, $s\in U$ and $s\leq t$ imply $t\in U$; equivalently ${\uparrow}U=U$). Moreover, $S$ is $T_{0}$ iff the specialization order is a partial order. From now on we will think of Alexandroff spaces as preorders $(S,\leq)$ and of the frame $\mathcal{O}S$ as the frame of upsets of $(S,\leq)$. Then closed sets are downsets ($s\leq t$ and $t\in F$ imply $s\in F$ or equivalently ${\downarrow}F=F$) and the closure of $A\subseteq S$ is ${\downarrow}A$.

For a preorder $(S,\leq)$ define an equivalence relation $\sim$ on $S$ by $x\sim y$ iff $x\leq y$ and $y\leq x$. Then $(S_{0},\leq_{0})$ is a partial order, known as the skeleton of $(S,\leq)$, where $S_{0}={S/{\sim}}$ and $[x]\leq_{0}[y]$ iff $x\leq y$. Topologically, the skeleton $S_{0}$ is the $T_{0}$-reflection of $S$. Since $\mathcal{O}S$ is isomorphic to $\mathcal{O}S_{0}$, we may restrict our attention to posets.

Let $S$ be a poset and let $L=\mathcal{O}S$. The spectrum $X_{L}$ of $L$ was described in [5, Sec. 3] as the Nachbin compactification of $S$. We recall that an ordered topological space $(X,\tau,\leq)$ is a Nachbin space if $X$ is compact (Hausdorff) and $\leq$ is closed in the product topology, and that an order-compactification of an ordered topological space $(X,\tau,\leq)$ is a Nachbin space $(Y,\pi,\leq)$ such that there is a topological and order embedding $e:X\to Y$ with $e[X]$ topologically dense in $Y$. A Nachbin compactification of $(X,\tau,\leq)$ is then the largest order-compactification of $(X,\tau,\leq)$. It is an order-topological analogue of the Stone-Čech compactification. In particular, every order-preserving continuous map from $X$ to a Nachbin space has a unique extension to the Nachbin compactification of $X$. Viewing a poset $S$ as an ordered topological space with the discrete topology, we have the following:

Theorem 3.1**.**

[5, Prop. 3.4]** Let $S$ be a poset and $L=\mathcal{O}S$ the frame of upsets of $S$. The Nachbin compactification of $S$ is order-homeomorphic to $(X_{L},\pi_{L},\leq_{L})$.

Convention 3.2**.**

To simplify notation, from now on we will drop the subscript from $(X_{L},\pi_{L},\leq_{L})$ and simply write $(X,\pi,\leq)$. We will also abbreviate $\tau_{L}$ by $\tau$. Similarly, we will write $Y$ instead of $Y_{L}$, so $(Y,\tau)$ is a subspace of $(X,\tau)$ and $(Y,\pi)$ is a subspace of $(X,\pi)$. We will write ${\sf cl}$ for the closure in $(X,\pi)$. Since $\varepsilon:S\to(Y,\tau)$ is the soberification of $S$, we identify $S$ with its image $\varepsilon[S]$ in $Y$, and view $S$ a subspace of $(Y,\tau)$.

Lemma 3.3**.**

Let $T$ be a subset of $S$.

1. (1)

If $A$ is an upset of $T$, then there is a clopen upset $U$ of $X$ with $A=U\cap T$. 2. (2)

If $B$ is a downset of $T$, then there is a clopen downset $V$ of $X$ with $B=V\cap T$.

Proof.

(1) Since $A$ is an upset of $T$, there is an upset $A^{\prime}$ of $S$ with $A=A^{\prime}\cap T$. Because $S$ is a subspace of $(Y,\tau)$, there is an open subset $U^{\prime}$ of $(Y,\tau)$ with $A^{\prime}=U^{\prime}\cap S$. By [1, Lem. 5.3(1)], there is a clopen upset $U$ of $X$ with $U^{\prime}=U\cap Y$. Therefore, $A=U\cap T$.

(2) The proof is similar to (1), but uses [1, Lem. 5.3(2)]. ∎

Since the topology on $S$ is discrete, the next lemma can be thought of as an order-theoretic analogue of [8, Cor. 3.6.4].

Lemma 3.4**.**

Let $D$ be a downset of $S$. If $A,B\subseteq D$ with $A\cap B=\varnothing$ and $B$ an upset of $D$, then ${\sf cl}(A)\cap{\sf cl}(B)=\varnothing$.

Proof.

First observe that $({\downarrow}A\cap D)\cap B=\varnothing$. Otherwise there is $b\in({\downarrow}A\cap D)\cap B$, so $b\leq a$ for some $a\in A$. Since $B$ is an upset of $D$, we have $a\in B$. Therefore, $a\in A\cap B$, which contradicts the assumption that $A,B$ are disjoint. Next we show that $({\downarrow}A\cap S)\cap({\uparrow}B\cap S)=\varnothing$. If not, then there are $b\in B$, $s\in S$, and $a\in A$ such that $b\leq s\leq a$. Since $a\in A\subseteq D$ and $B$ is an upset of $D$, we get $a\in B$, which is false as $A\cap B=\varnothing$. Now, define $f:S\to[0,1]$ by $f(s)=0$ if $s\in{\downarrow}A\cap S$ and $f(s)=1$ otherwise. Clearly $f$ is order-preserving. Thus, since $X$ is the Nachbin compactification of $S$ (see Theorem 3.1), there is a continuous order-preserving map $g:X\to[0,1]$ with $g|_{S}=f$. As $A\subseteq f^{-1}(0)$ and $B\subseteq f^{-1}(1)$, we then conclude that ${\sf cl}(A)\subseteq g^{-1}(0)$ and ${\sf cl}(B)\subseteq g^{-1}(1)$, and hence ${\sf cl}(A)\cap{\sf cl}(B)=\varnothing$. ∎

In what follows we will make heavy use of the technique of nets and net convergence (see, e.g., [8, Sec. 1.6]). We recall that a net in $X$ is a map $\mathfrak{n}$ from a directed set $\Gamma$ to $X$. We call a net $\mathfrak{n}:\Gamma\to X$ increasing if $\gamma\leq\delta$ implies $\mathfrak{n}(\gamma)\leq\mathfrak{n}(\delta)$.

We call a subset $A$ of $X$ up-directed if $A$ is a directed set with the induced order coming from $X$. If $A$ is up-directed, then the inclusion function $A\to X$ is an increasing net in $X$. Conversely, if $\mathfrak{n}:\Gamma\to X$ is an increasing net, then the image $\mathfrak{n}(\Gamma)$ is an up-directed subset of $X$.

Lemma 3.5**.**

Let $\mathfrak{n}$ be an increasing net in $Y$ converging to $x\in X$. Then $\mathfrak{n}(\Gamma)\subseteq{\downarrow}x$ and $x\in Y$.

Proof.

Let $A=\mathfrak{n}(\Gamma)$. We first show that $A\subseteq{\downarrow}x$. If not, then there is $a\in A$ with $a\not\leq x$. By the Priestley separation axiom, there is a clopen upset $U$ of $X$ with $a\in U$ and $x\notin U$. Since $X\setminus U$ is an open neighborhood of $x$ and $\mathfrak{n}$ is a net converging to $x$, there is $\gamma\in\Gamma$ such that for all $\delta\geq\gamma$, we have $\mathfrak{n}(\delta)\in X\setminus U$. Because $\mathfrak{n}$ is increasing, there is $\delta$ with $a,\mathfrak{n}(\gamma)\leq\mathfrak{n}(\delta)$. This implies $\mathfrak{n}(\delta)\in X\setminus U$, which is impossible since $\mathfrak{n}(\delta)\in U$ as $U$ is an upset and $a\in U$. The obtained contradiction proves that $A\subseteq{\downarrow}x$.

We next show that $x\in Y$. Let $V=X\setminus{\downarrow}x$, an open upset of $X$. Since $X$ is an extremally order-disconnected Esakia space, ${\sf cl}(V)$ is a clopen upset. Let $a\in A$. Then $a\leq x$, so $a\notin V$, and so ${\downarrow}a\cap V=\varnothing$ because $V$ is an upset. Since $a\in Y$, we have ${\downarrow}a$ is clopen, so ${\downarrow}a\cap{\sf cl}(V)=\varnothing$. This implies $A\cap{\sf cl}(V)=\varnothing$, and so ${\sf cl}(A)\cap{\sf cl}(V)=\varnothing$ as ${\sf cl}(V)$ is clopen. Since $x\in{\sf cl}(A)$, we conclude that $x\notin{\sf cl}(V)$, and hence ${\downarrow}x\cap{\sf cl}(V)=\varnothing$. This implies that $V={\sf cl}(V)$, so $V$ is clopen. Therefore, ${\downarrow}x$ is clopen. Thus, $x\in Y$. ∎

Lemma 3.6**.**

Let $A$ be an up-directed subset of $Y$. Viewing $A$ as a net, $A$ converges to a point $y\in Y$ with $A\subseteq{\downarrow}y$.

Proof.

Let $\mathfrak{n}:\Gamma\to X$ be an increasing net in $X$ with $\mathfrak{n}(\Gamma)=A$. Since $X$ is compact, $\mathfrak{n}$ has a convergent subnet $\mathfrak{n}\circ\varphi$ for some order preserving map $\varphi:\Lambda\to\Gamma$ whose image is cofinal in $\Gamma$. Set $B=\mathfrak{n}(\varphi(\Lambda))$. Let $y$ be the limit of $B$. By Lemma 3.5, $B\subseteq{\downarrow}y$ and $y\in Y$. We show that $y$ is the supremum of $B$ in $X$. Suppose $x$ is an upper bound of $B$. If $y\not\leq x$, then the Priestley separation axiom yields a clopen downset $V$ of $X$ containing $x$ but not $y$. Since $B\subseteq{\downarrow}x$, we have $B\subseteq V$, so $B\cap(X\setminus V)=\varnothing$, which is impossible because $X\setminus V$ is a neighborhood of $y$ and $y$ is the limit of $B$. Thus, $y\leq x$, and so $y$ is the supremum of $B$. Let $a\in A$. Since $B$ is cofinal in $A$, there is $b\in B$ with $a\leq b$. Consequently, $A\subseteq{\downarrow}y$, and so $y$ is also the supremum of $A$.

We show that $y$ is the limit of $A$. Suppose that $W$ is an open neighborhood of $y$. Then there are clopen upsets $U,V$ with $y\in U\setminus V\subseteq W$. Since $X\setminus V$ is a clopen downset and $y\in X\setminus V$, it follows that $A\subseteq{\downarrow}y\subseteq X\setminus V$. Since $U$ is an open neighborhood of $y$, there is $\lambda\in\Lambda$ such that if $\delta\geq\lambda$, then $\mathfrak{n}(\varphi(\delta))\in U$. Suppose that $\gamma\in\Gamma$ with $\gamma\geq\varphi(\lambda)$. Since $\mathfrak{n}$ is increasing, $\mathfrak{n}(\gamma)\geq\mathfrak{n}(\varphi(\lambda))$. We have $\mathfrak{n}(\varphi(\lambda))\in U$ and $U$ is an upset, so $\mathfrak{n}(\gamma)\in U$. Therefore, $\mathfrak{n}(\gamma)\in W$ for each $\gamma\geq\varphi(\lambda)$. Thus, $\mathfrak{n}$ converges to $y$. ∎

Lemma 3.7**.**

Let $E$ be a clopen downset of $X$ such that $\max E\cap Y=\varnothing$. If $A$ is a nonempty upset of $E\cap S$, then $A$ is not up-directed.

Proof.

Let $D=E\cap S$ and let $A$ be a nonempty upset of $D$. By Lemma 3.3(1), $A=U\cap D$ for some clopen upset $U$ of $X$. Suppose that $A$ is up-directed. Then the inclusion map $A\to X$ is an increasing net $\mathfrak{n}$. Lemma 3.6 implies that $\mathfrak{n}$ converges to a point $y\in Y$ such that $A\subseteq{\downarrow}y$. As $\max E\cap Y=\varnothing$, there is $x\in E$ with $y<x$. Because $U$ is a clopen upset containing $A$ and $A\subseteq{\downarrow}y$, we see that $y\in U$, so $x\in U$. Consequently, $x\in U\cap E$. By Theorem 2.10, $S$ is dense in $X$. Therefore, ${\sf cl}(D)={\sf cl}(E\cap S)=E\cap{\sf cl}(S)=E$ as $E$ is clopen. Thus, $D$ is dense in $E$, and hence $A$ is dense in $U\cap E$. From $A\subseteq{\downarrow}y$ it follows that ${\sf cl}(A)\subseteq{\downarrow}y$. Therefore, $x\in U\cap E={\sf cl}(A)\subseteq{\downarrow}y$. This is impossible since $y<x$. The obtained contradiction proves that $A$ is not up-directed. ∎

Lemma 3.8**.**

Suppose that $D$ is a downset of $S$ and there is $x\in X$ with $x\in{\sf cl}({\downarrow}x\cap D)$. Then there is an increasing net in $D$ which converges to $x$.

Proof.

Let $x\in{\sf cl}({\downarrow}x\cap D)$. Then there is a net $\mathfrak{n}:\Gamma\to{\downarrow}x\cap D$ converging to $x$. We build an increasing net in $D$ converging to $x$. Let $A={\downarrow}x\cap D$. Then all the terms of the net are in $A$. We show that $A$ is up-directed. Let $a,b\in A$. Since ${\uparrow}a\cap D$ and ${\uparrow}b\cap D$ are upsets of $D$, by Lemma 3.3(1) there are clopen upsets $U,V$ of $X$ with $U\cap D={\uparrow}a\cap D$ and $V\cap D={\uparrow}b\cap D$. Because $a,b\leq x$, we see that $x\in U\cap V$. Therefore, $U\cap V$ is an open neighborhood of $x$. Thus, $\mathfrak{n}(\gamma)\in U\cap V$ for some $\gamma$. This implies that $a,b\leq\mathfrak{n}(\gamma)$. Since $\mathfrak{n}(\gamma)\in A$, this shows that $A$ is up-directed. We may then view $A$ as an increasing net.

We show that $A$ converges to $x$. Let $W$ be an open neighborhood of $x$. Then there are clopen upsets $U,V$ of $X$ with $x\in U\setminus V\subseteq W$. As $X\setminus V$ is an open downset containing $x$ and $A\subseteq{\downarrow}x$, we have $A\subseteq X\setminus V$. There is $\delta\in\Gamma$ such that if $\gamma\geq\delta$, then $\mathfrak{n}(\gamma)\in U$. If $a\in A$ with $\mathfrak{n}(\delta)\leq a$, then $a\in U$ since $U$ is an upset. Consequently, for each $a\in A$ with $\mathfrak{n}(\delta)\leq a$, we have $a\in W$. This shows that the net $A$ converges to $x$. We have thus produced an increasing net in $D$ converging to $x$. ∎

Lemma 3.9**.**

Let $A,B\subseteq Y$ and $A\subseteq{\downarrow}B$. If $x$ is a limit point of $A$, then there is a limit point $y$ of $B$ with $x\leq y$.

Proof.

Since $x$ is a limit point of $A$ there is a net $\mathfrak{n}:\Gamma\to A$ converging to $x$. For each $\gamma$ choose $b_{\gamma}\in B$ with $\mathfrak{n}(\gamma)\leq b_{\gamma}$. Define a net $\mathfrak{m}:\Gamma\to Y$ by $\mathfrak{m}(\gamma)=b_{\gamma}$. Since $X$ is compact, there is a subnet $\mathfrak{m}\circ\varphi$ of $\mathfrak{m}$ converging to some $y\in X$, where $\varphi:\Lambda\to\Gamma$ is order preserving and its image is cofinal in $\Gamma$. Then $y$ is a limit point of $B$. Because $\mathfrak{n}$ converges to $x$, the subnet $\mathfrak{n}\circ\varphi$ also converges to $x$ (see, e.g., [8, Prop. 1.6.1]).

Suppose $x\not\leq y$. By the Priestley separation axiom, there is a clopen upset $U$ containing $x$ and missing $y$. Since $x\in U$ there is $\lambda\in\Lambda$ such that for each $\delta\geq\lambda$, we have $\mathfrak{n}(\varphi(\delta))\in U$. As $U$ is an upset, $\mathfrak{m}(\varphi(\delta))\in U$. Because $\mathfrak{m}\circ\varphi$ converges to $y$ and $y\in X\setminus U$, there is $\lambda^{\prime}$ such that for each $\delta\geq\lambda^{\prime}$ we have $\mathfrak{m}(\varphi(\delta))\in X\setminus U$. Then, for any $\delta\geq\lambda,\lambda^{\prime}$, we have $\mathfrak{m}(\varphi(\delta))\in U\cap(X\setminus U)$, which is impossible. Thus, $x\leq y$. ∎

Let $\mathscr{T}_{2}$ be the infinite binary tree shown below.

We think of $\mathscr{T}_{2}$ as built from combs where a comb is depicted below.

Namely, we start with the root of $\mathscr{T}_{2}$ and build a comb with the “round” and “square” points drawn below. The round points form the “spine” of the comb and the square points the “teeth” of the comb. Then for each square point we build a comb with the point as the root. Continuing this process yields $\mathscr{T}_{2}$.

Theorem 3.10**.**

$N(\mathcal{O}\mathscr{T}_{2})$* is not spatial.*

Proof.

Following Convention 3.2, we write $X$ for $X_{\mathcal{O}\mathscr{T}_{2}}$ and $Y$ for $Y_{\mathcal{O}\mathscr{T}_{2}}$. By Corollary 2.5, it is sufficient to show that there is a clopen downset $E$ of $X$ such that $\max E\cap Y=\varnothing$; and we show that $\max X\cap Y=\varnothing$. Since $\mathscr{T}_{2}$ is dense in $X$ by Theorem 2.10, we have that $X={\sf cl}(\mathscr{T}_{2})$. Suppose $y\in\max X\cap Y$. Then there is a net in $\mathscr{T}_{2}$ converging to $y$. Since $y\in Y$, we have ${\downarrow}y$ is clopen. Therefore, ${\downarrow}y={\sf cl}({\downarrow}y\cap\mathscr{T}_{2})$ (see the proof of Lemma 3.7). Consequently, by Lemma 3.8, there is an increasing net $\mathfrak{n}:\Gamma\to\mathscr{T}_{2}$ converging to $y$. Let $A=\mathfrak{n}(\Gamma)$. Then $A$ is an up-directed subset of $\mathscr{T}_{2}$, so $A$ is a chain in $\mathscr{T}_{2}$. Consider the comb that has $A$ as the spine. Let $B$ be the upset generated by the teeth of the comb. Then $A\cap B=\varnothing$ and $A\subseteq{\downarrow}B$. Therefore, by Lemma 3.9, there is a limit point $x$ of $B$ with $y\leq x$. By Lemma 3.4, ${\sf cl}(A)\cap{\sf cl}(B)=\varnothing$. Since $y\in{\sf cl}(A)$ and $x\in{\sf cl}(B)$, we conclude that $y\neq x$. Therefore, $y<x$, which is a contradiction to $y\in\max X$. Thus, $\max X\cap Y=\varnothing$, which shows that $N(\mathcal{O}\mathscr{T}_{2})$ is not spatial by Corollary 2.5. ∎

Lemma 3.11**.**

Let $S$ be a poset and $T$ a subposet of $S$. If $N(\mathcal{O}T)$ is not spatial, then neither is $N(\mathcal{O}S)$.

Proof.

Since $T$ is a subposet of $S$, we see that $\mathcal{O}T$ is a quotient of $\mathcal{O}S$. It follows from the proof of [18, Lem. 3.4] that $N(\mathcal{O}T)$ is isomorphic to an interval in $N(\mathcal{O}S)$. Thus, spatiality of $N(\mathcal{O}S)$ implies spatiality of $N(\mathcal{O}T)$. ∎

We are ready to prove the main result of this paper.

Theorem 3.12**.**

Let $S$ be a poset. Then $N(\mathcal{O}S)$ is not spatial iff $\mathscr{T}_{2}$ is isomorphic to a subposet of $S$.

Proof.

First suppose that $\mathscr{T}_{2}$ is isomorphic to a subposet of $S$. Then Theorem 3.10 and Lemma 3.11 yield that $N(\mathcal{O}S)$ is not spatial. Conversely, suppose that $N(\mathcal{O}S)$ is not spatial. Then Corollary 2.5 gives a nonempty clopen downset $E$ of $X$ with $\max E\cap Y=\varnothing$. Let $D=E\cap S$. By Lemma 3.7, each nonempty upset of $D$ is not up-directed. In particular, for each $x\in D$ the upset ${\uparrow}x\cap D$ of $D$ is not up-directed. Therefore, there are $y,z\in D$ with $x\leq y,z$ but no $w\in D$ with $y,z\leq w$. We build a copy of $\mathscr{T}_{2}$ inside $D$ by first building a comb inside $D$.

Let $x_{0}\in D$. Then there are $x_{1},y_{0}\in D$ with $x_{0}\leq x_{1},y_{0}$ such that nothing in $D$ is above both $x_{1},y_{0}$. Repeating this construction, for each $n$ we produce $x_{n}\in D$ and $x_{n+1},y_{n}\in D$ with $x_{n}\leq x_{n+1},y_{n}$ such that nothing in $D$ is above both $x_{n+1}$ and $y_{n}$. We claim that $C=\{x_{n},y_{n}\mid n\in\mathbb{N}\}$ is a comb inside $D$. By construction, $x_{0}<x_{1}<\cdots$ is a chain in $D$, and $x_{i}\leq y_{i}$ for each $i$. We need to show that $\{y_{n}\mid n\in\mathbb{N}\}$ is an antichain. Assume that there are $i\neq j$ with $y_{i}\leq y_{j}$. First suppose that $i<j$. The element $y_{j}$ is above both $y_{i}$ and $x_{j}$. Since $i<j$ and $\{x_{n}\}$ is an increasing chain, $y_{j}$ is above both $y_{i}$ and $x_{i+1}$. This is impossible by construction. Next, suppose that $j<i$. Then $y_{j}\geq y_{i}\geq x_{i}\geq x_{j+1}$. This is false by construction. Thus, $C$ is indeed a comb in $D$. By repeating this construction, we can build a comb in $D$ rooted at each $y_{n}$.

To see that the resulting poset is $\mathscr{T}_{2}$, if $i<j$, then we show that the combs rooted at $y_{i}$ and $y_{j}$ are disjoint. Suppose otherwise. Then there is $a\in D$ above both $y_{i}$ and $y_{j}$. Therefore, $a$ is above both $y_{i}$ and $x_{j}$. Since $i<j$, we have $x_{i+1}\leq x_{j}\leq a$. Thus, $a$ is above both $x_{i+1}$ and $y_{i}$, a contradiction. Hence, the combs above $y_{i}$ and $y_{j}$ are disjoint. The resulting subposet of $D$ is then isomorphic to $\mathscr{T}_{2}$, completing the proof. ∎

4. Consequences of the Main Theorem

We conclude the paper by deriving some consequences of Theorem 3.12. First we derive a characterization of when $N(\mathcal{O}S)$ is spatial for an arbitrary Alexandroff space. Let $S$ be an Alexandroff space, which we will view as a preordered set. Let $S_{0}$ be the skeleton ($T_{0}$-reflection) of $S$ and let $\rho:S\to S_{0}$ be the corresponding map sending $x\in S$ to $[x]\in S_{0}$. Then $\rho^{-1}:\mathcal{O}S_{0}\to\mathcal{O}S$ is an isomorphism of frames (see the beginning of Section 3).

Corollary 4.1**.**

For a preorder $S$ the following are equivalent.

1. (1)

$N(\mathcal{O}S)$* is not spatial.* 2. (2)

$\mathscr{T}_{2}$* is isomorphic to a subposet of $S_{0}$.* 3. (3)

$\mathscr{T}_{2}$* embeds isomorphically into $S$.*

Proof.

(1)$\Leftrightarrow$(2). Since $N(\mathcal{O}S)$ is isomorphic to $N(\mathcal{O}S_{0})$, we have that $N(\mathcal{O}S)$ is not spatial iff $N(\mathcal{O}S_{0})$ is not spatial. Now apply Theorem 3.12.

(2)$\Leftrightarrow$(3). Suppose that $\mathscr{T}_{2}$ is isomorphic to a subposet of $S_{0}$. We may identify $\mathscr{T}_{2}$ with its image in $S_{0}$. For each $t\in\mathscr{T}_{2}$ choose $s_{t}\in\rho^{-1}(t)$. Then sending $t$ to $s_{t}$ is the desired embedding of $\mathscr{T}_{2}$ into $S$. Conversely, suppose $\mathscr{T}_{2}$ embeds isomorphically into $S$. We may identify $\mathscr{T}_{2}$ with its image in $S$. Then $\rho(\mathscr{T}_{2})$ is a subposet of $S_{0}$ isomorphic to $\mathscr{T}_{2}$. ∎

We next recall that a poset $S$ is noetherian if $S$ has no infinite ascending chains. If $S$ is noetherian, it is clear that $\mathscr{T}_{2}$ does not embed in $S$. Therefore, Theorem 3.12 yields that $N(\mathcal{O}S)$ is spatial. In fact, $S$ is a noetherian poset iff $S$, viewed as an Alexandroff space, is scattered. Therefore, Simmons’s well-known theorem [17, Thm. 4.5] implies that $N(\mathcal{O}S)$ is moreover boolean.

It is natural to ask whether there exist posets $S$ such that $N(\mathcal{O}S)$ is spatial, but not boolean. Isbell’s theorem [11], that for a sober space $S$ the frame $N(\mathcal{O}S)$ is spatial iff $S$ is weakly scattered, does not resolve this question since for a poset $S$, the concepts of sober, weakly scattered, and scattered are all equivalent to $S$ being noetherian. We show that Theorem 3.12 resolves this question in the positive by providing many such examples. We recall that a poset $S$ is totally ordered if it is a chain; that is, $s\leq t$ or $t\leq s$ for all $s,t\in S$.

Corollary 4.2**.**

Let $S$ be a poset.

1. (1)

If $S$ has no infinite antichains, then $N(\mathcal{O}S)$ is spatial. 2. (2)

If $S$ is totally ordered, then $N(\mathcal{O}S)$ is spatial.

Proof.

(1) Suppose that $S$ has no infinite antichains. Since $\mathscr{T}_{2}$ has infinite antichains, $\mathscr{T}_{2}$ cannot be isomorphic to a subposet of $S$. Thus, $N(\mathcal{O}S)$ is spatial by Theorem 3.12.

(2) If $S$ is totally ordered, then $S$ has no infinite antichains. Now apply (1). ∎

Consequently, for each totally ordered set $S$ (or more generally for each poset $S$ with no infinite antichains), if $S$ is not noetherian, then $N(\mathcal{O}S)$ is spatial, but not boolean.

Remark 4.3**.**

The converse of Corollary 4.2(1) is clearly false. For example, if $S$ is an infinite antichain, then $N\mathcal{O}S$ is spatial by Theorem 3.12 since $\mathscr{T}_{2}$ is not isomorphic to a subposet of $S$.

By Theorem 2.4 and Corollary 4.2(2), for a totally ordered set $S$, the frame $N(\mathcal{O}S)$ is isomorphic to the frame of opens of $(Y,\pi)$. As we pointed out at the end of Section 2, $(Y,\pi)$ is a zero-dimensional Haudorff space. As our final result, we determine when $(Y,\pi)$ is compact, and hence a Stone space. Recall that a poset $S$ is artinian if there are no infinite descending chains in $S$; equivalently, if every nonempty subset of $S$ has a minimum.

Theorem 4.4**.**

For a totally ordered set $S$, the following are equivalent.

1. (1)

$Y=X$. 2. (2)

$S$* is artinian.* 3. (3)

$(Y,\pi)$* is compact.*

Proof.

(1) $\Rightarrow$ (2). Let $A$ be an infinite descending chain $a_{0}>a_{1}>\cdots$ in $S$. Then the closure of $A$ in $X$ is also a chain ([9, Thm. III.2.9]). Therefore, it has a unique minimum $x\in X$, and $x\notin A$ since $A$ is infinite. Thus, ${\downarrow}x\cap A=\varnothing$. This implies that ${\downarrow}x$ is not clopen, so $x\notin Y$. This is impossible since $Y=X$. Thus, $S$ is artinian.

(2) $\Rightarrow$ (3). Since $X$ is compact, it is sufficient to show that $X=Y$. Because $S$ is a chain and $S$ is dense in $X$, we have that $X$ is a chain (see [9, Thm. III.2.9]). Let $x\in X$. If $x$ is the maximum of $X$, then ${\downarrow}x=X$, so ${\downarrow}x$ is clopen, and hence $x\in Y$. Suppose not. Since $x$ is not the maximum of $X$, the set ${\uparrow}x\setminus\{x\}$ is nonempty. Because $X$ is a chain, ${\uparrow}x\setminus\{x\}=X\setminus{\downarrow}x$. Therefore, ${\uparrow}x\setminus\{x\}$ is a nonempty open subset of $X$. Thus, $({\uparrow}x\setminus\{x\})\cap S\neq\varnothing$. Since $S$ is artinian, $({\uparrow}x\setminus\{x\})\cap S$ has a minimum $s$. Because $X$ is a chain, $(x,s)=X\setminus({\downarrow}x\cup{\uparrow}s)$. Therefore, the interval $(x,s)$ is open. It misses $S$, so $(x,s)=\varnothing$ since $S$ is dense in $X$. Therefore, ${\downarrow}x=X\setminus{\uparrow}s$, so ${\downarrow}x$ is open. Thus, ${\downarrow}x$ is clopen, yielding $x\in Y$.

(3) $\Rightarrow$ (1). If $Y$ is compact, then $Y=X$ since $Y$ is a closed dense subset of $X$. ∎

Remark 4.5**.**

If $S$ is an artinian totally ordered set, then $X$ is in fact a compact ordinal. To see this, as we pointed out in the proof of Theorem 4.4, $X$ is a chain. In addition, the topology on $X$ is the interval topology (see, [9, Thm. III.2.17]). Thus, it is sufficient to show that $X$ is artinian. If not, then there is an infinite descending chain $x_{0}>x_{1}>\cdots$ in $X$. By Theorem 4.4, $X=Y$, so each ${\downarrow}x_{i}$ is clopen. Therefore, ${\downarrow}x_{i}\setminus{\downarrow}x_{i+1}$ is a nonempty clopen set in $X$. Since $S$ is dense in $X$, there is $s_{i}\in S$ with $s_{i}\in{\downarrow}x_{i}\setminus{\downarrow}x_{i+1}$. Because $X$ is a chain, $x_{i}\geq s_{i}>x_{i+1}\geq s_{i+1}$ for each $i$. This implies that $s_{0}>s_{1}>\cdots$ is an infinite descending chain in $S$, which is impossible since $S$ is artinian. Thus, $X$ is artinian, and hence is a compact ordinal.

Remark 4.6**.**

Theorem 4.4 is not true in general. Let $S$ be an infinite antichain. Then $S$ is artinian. On the other hand, $X$ is homeomorphic to the Stone-Čech compactification of $S$, and hence $Y=S\neq X$.