On the order dimension of locally countable partial orderings

TL;DR

This paper determines the order dimension of certain infinite partial orders, showing it depends on the cardinality and cofinality, and explores implications for the Turing degrees' structure.

Contribution

It establishes exact order dimension values for the partial order of finite subsets and bounds for locally countable partial orders, revealing new structural insights.

Findings

Order dimension of finite subsets of 6 is 4(4(6))

Order dimension of locally countable partial orders of size 4^+ is at most 4

Turing degrees' dimension can be less than the continuum under ZFC

Abstract

We show that the order dimension of the partial order of all finite subsets of $\kappa$ under set inclusion is ${\log}_{2}({\log}_{2}(\kappa))$ whenever $\kappa$ is an infinite cardinal. We also show that the order dimension of any locally countable partial ordering $(P, <)$ of size $\kappa^+$, for any $\kappa$ of uncountable cofinality, is at most $\kappa$. In particular, this implies that it is consistent with ZFC that the dimension of the Turing degrees under partial ordering can be strictly less than the continuum.

Equations

$${\log }_2(\kappa )=\min \{\lambda :2^{\lambda }\ge \kappa \}$$

$${\left({\exp }_{r-1}(\rho )\right)}^{+}\to {\left({\rho }^{+}\right)}_{\rho }^r.$$

$$A_{\alpha }\cap \bigcup _{{\alpha }^{\prime }<\alpha }{A}_{{\alpha }^{\prime }}=A_{\alpha }\cap \bigcup _{{\alpha }^{\prime }\in B_{\alpha }}{A}_{{\alpha }^{\prime }}.$$

$$r^{\prime \prime }\left(\bigcup _{{\alpha }^{\prime }\in B_{\alpha }}{A}_{{\alpha }^{\prime }}\right)\subseteq \eta \cdot \gamma .$$

$$P=\{\{\alpha \}:\alpha <{\omega }_1\}\cup \{\{\gamma :\gamma <\beta \}\setminus \{\alpha \}:\alpha ,\beta <{\omega }_1\}.$$

Full text

We also show that the order dimension of any locally countable partial ordering $(P,<)$ of size $\kappa^{+}$, for any $\kappa$ of uncountable cofinality, is at most $\kappa$. In particular, this implies that it is consistent with ZFC that the dimension of the Turing degrees under partial ordering can be strictly less than the continuum.

The first author was partially supported by Grant-in-Aid for JSPS Fellows. The second author was partially supported by NSF grant DMS-160022; he also wishes to thank the National University of Singapore for its hospitality during his sabbatical year when most of this research was carried out. The third author was partially supported by the Singapore Ministry of Education’s research grant number MOE2017-T2-2-125. The fourth author was partially supported in part by the Singapore Ministry of Education Academic Research Fund Tier 2 grant MOE2016-T2-1-019 / R146-000-234-112.

1. Introduction

This paper arose from a question posed by the first to the third author at the Computability Theory and the Foundations of Mathematics conference at Tokyo in 2015 regarding a set-theoretic question about a computability-theoretic structure:

Question 1.1** (Higuchi).**

What is the order dimension of the Turing degrees regarded as a partial order?

Higuchi had already shown that this dimension must be uncountable and asked whether it is the continuum. This paper provides a partial answer: It is consistent with ZFC that the dimension is less than the continuum. But Higuchi’s question raised a number of related questions to which we give some answers in this paper, all about the order dimension of locally finite and locally countable partial orders.

We start with some definitions:

Definition 1.2** (Dushnik, Miller [4], Ore [7]).**

Given a partial order $\mathcal{P}=(P,\prec)$, the order dimension (or simply dimension) of $\mathcal{P}$ is the smallest cardinality of a collection of linearizations of $\prec$ which intersect to $\prec$.

So, for example, the dimension of a linear order is clearly 1, and the dimension of an antichain is easily seen to be 2. It is also easy to see that the dimension of an infinite partial order $\mathcal{P}$ can be at most $|P|$: For each pair $x,y\in P$ with $y\not\preceq x$, fix a linearization $<_{x,y}$ of $\prec$ with $x<_{x,y}y$.

Definition 1.3**.**

Call a partial order $\mathcal{P}=(P,\prec)$ locally finite (locally countable, respectively), if for each $x\in P$, the set $\{y\in P\mid y\prec x\}$ is finite or countable, respectively.

Partial orders which are locally finite (or locally countable, respectively) are also often said to have the finite predecessor property (or the countable predecessor property).

The order dimension of the Turing degrees can be thought of as a new cardinal invariant because it is between ${\aleph}_{1}$ and ${2}^{{\aleph}_{0}}$.

Definition 1.4**.**

Let $\left\langle\mathcal{D},{<}_{T}\right\rangle$ denote the class of Turing degrees equipped with the ordering of Turing reducibility. The cardinal ${\mathfrak{dim}}_{T}$ denotes the order dimension of $\left\langle\mathcal{D},{<}_{T}\right\rangle$.

Since $\mathcal{D}$ has cardinality ${2}^{{\aleph}_{0}}$, ${\mathfrak{dim}}_{T}\leq{2}^{{\aleph}_{0}}$, and by Higuchi’s Proposition 4.3, ${\aleph}_{1}\leq{\mathfrak{dim}}_{T}$. Thus the cardinal ${\mathfrak{dim}}_{T}$ sits between ${\aleph}_{1}$ and ${2}^{{\aleph}_{0}}$, like many of the standard cardinal invariants of the continuum such as $\mathfrak{b}$, ${\mathfrak{d}}$, ${\mathfrak{a}}$, etc. The reader is referred to Blass [2] for a general survey of combinatorial cardinal characteristics of the continuum. Of course, under $\mathrm{CH}$, ${\mathfrak{dim}}_{T}={\aleph}_{1}={2}^{{\aleph}_{0}}$. In this paper, we will show that ${\mathfrak{dim}}_{T}$ is smaller than ${2}^{{\aleph}_{0}}$ “most of the time”. More precisely, we will show that there are only three circumstances under which ${\mathfrak{dim}}_{T}={2}^{{\aleph}_{0}}$ is possibly consistent: ${2}^{{\aleph}_{0}}={\aleph}_{1}$, or ${2}^{{\aleph}_{0}}$ is a limit cardinal (either singular or weakly inaccessible), or ${2}^{{\aleph}_{0}}$ is the successor of a singular cardinal of countable cofinality.

We will now present some results on the dimension of locally finite partial orders in Section 2, and on the dimension of locally countable partial orders in Section 3, and close with some results on the dimension of degree structures from computability theory in Section 4.

2. The dimension of ${\left[\kappa\right]}^{<\omega}$

We determine the order dimension of $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$, which is universal among locally finite posets of cardinality $\kappa$, i.e., every locally finite poset $\mathcal{P}=(P,\prec)$ with $|P|=\kappa$ embeds into $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$ by assigning $a\in P$ to $\{f(b):b\in P,b\preceq a\}$, where $f:P\to\kappa$ is any injection. Thus our result in this section provides an upper bound for every locally finite poset.

Definition 2.1**.**

For any infinite cardinal $\kappa$,

[TABLE]

The main theorem of this section is the following.

Theorem 2.2**.**

Let $\kappa\geq\omega$ be any cardinal. Then the order dimension of $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$ is ${\log}_{2}({\log}_{2}(\kappa))$.

It follows from Definition 2.1 that ${\log}_{2}({\log}_{2}(\kappa))$ is the minimal $\lambda$ such that ${2}^{{2}^{\lambda}}\geq\kappa$. Theorem 2.2 says in particular that $\left\langle{\left[{\omega}_{1}\right]}^{<\omega},\subsetneq\right\rangle$ and indeed $\left\langle{\left[{2}^{{2}^{\omega}}\right]}^{<\omega},\subsetneq\right\rangle$ have countable dimension. But $\left\langle{\left[{\left({2}^{{2}^{\omega}}\right)}^{+}\right]}^{<\omega},\subsetneq\right\rangle$ has uncountable dimension. Theorem 2.2 will be proved via two lemmas that establish upper and lower bounds on the dimension of $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$. The lower bound is established using a result from partition calculus. Recall the following consequence of the well-known Erdős-Rado theorem.

Theorem 2.3** (Corollary 17.11 in [5]).**

For any cardinal $\rho$ and any nonzero $r<\omega$,

[TABLE]

The ${\exp}_{i}(\rho)$ operation is defined by recursion on $i<\omega$: ${\exp}_{0}(\rho)=\rho$; ${\exp}_{i+1}(\rho)={2}^{{\exp}_{i}(\rho)}$. Setting $r=3$ in Theorem 2.3 gives ${\left({2}^{{2}^{\rho}}\right)}^{+}\rightarrow{\left({\rho}^{+}\right)}^{3}_{\rho}$.

Lemma 2.4**.**

Let $\kappa\geq\omega$ be a cardinal. Then the order dimension of $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$ is at least ${\log}_{2}({\log}_{2}(\kappa))$.

Proof.

Let $\rho$ denote the order dimension of $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$. Suppose for a contradiction that $\rho<{\log}_{2}({\log}_{2}(\kappa))$. In other words, ${2}^{{2}^{\rho}}<\kappa$, and so $\theta=\max\{\omega,{\left({2}^{{2}^{\rho}}\right)}^{+}\}\leq\kappa$. Let $\left\langle{<}_{\iota}:\iota<\rho\right\rangle$ be a sequence of linear orders witnessing that $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$ has dimension $\rho$. Thus for all $X,Y\in{\left[\kappa\right]}^{<\omega}$, if $X\subsetneq Y$, then $\forall\iota<\rho\left[X\>{<}_{\iota}\>Y\right]$, while if $X\not\subseteq Y$, then $\exists\iota<\rho\left[Y\>{<}_{\iota}\>X\right]$. Now define a coloring $c:{\left[\theta\right]}^{3}\rightarrow\rho$ as follows. Given any $\{\alpha,\beta,\gamma\}\in{\left[\theta\right]}^{3}$ with $\alpha<\beta<\gamma$, there exists $\iota<\rho$ such that $\{\alpha,\gamma\}\>{<}_{\iota}\>\{\beta\}$ because $\{\alpha,\gamma\},\{\beta\}\in{\left[\kappa\right]}^{<\omega}$ and $\{\beta\}\not\subseteq\{\alpha,\gamma\}$. Define $c(\{\alpha,\beta,\gamma\})=\min\{\iota<\rho:\{\alpha,\gamma\}\>{<}_{\iota}\>\{\beta\}\}$. Using Ramsey’s Theorem when $\rho$ is finite and Theorem 2.3 otherwise, fix $\alpha<\beta<\gamma<\delta<\theta\leq\kappa$ and $\iota<\rho$ so that $\{\alpha,\gamma\}\>{<}_{\iota}\>\{\beta\}$ and $\{\beta,\delta\}\>{<}_{\iota}\>\{\gamma\}$. However, we now have $\{\gamma\}\>{<}_{\iota}\>\{\alpha,\gamma\}\>{<}_{\iota}\>\{\beta\}\>{<}_{\iota}\>\{\beta,\delta\}\>{<}_{\iota}\>\{\gamma\}$, which is impossible. This contradiction completes the proof of the lemma. ∎

Note that the proof of Lemma 2.4 shows that $\left\langle{\left[\kappa\right]}^{\leq 2},\subsetneq\right\rangle$ has order dimension at least ${\log}_{2}({\log}_{2}(\kappa))$. Also the proof only needed the partition relation ${\left({2}^{{2}^{\rho}}\right)}^{+}\rightarrow{\left(4\right)}^{3}_{\rho}$, and $\omega\rightarrow{\left(4\right)}^{3}_{\rho}$ when $\rho<\omega$ (actually something even weaker; just a monochromatic shift is needed).

Definition 2.5**.**

Let $\theta\geq\omega$ be a cardinal. For $X\subseteq\theta$, we use ${X}^{0}$ to denote $X$ and ${X}^{1}$ to denote $\theta\setminus X$. We say that a sequence $\left\langle{X}_{i}:i\in I\right\rangle$ of subsets of $\theta$ is independent if for each nonempty finite set $F\subseteq I$ and for each function $\sigma:F\rightarrow 2$, the set ${\bigcap}_{i\in F}{{X}^{\sigma(i)}_{i}}$ has cardinality $\theta$.

Theorem 2.6** (Hausdorff; see Exercise (A6) in Chapter VIII of [6]).**

For each cardinal $\theta\geq\omega$, there is an independent family $\left\langle{X}_{\alpha}:\alpha<{2}^{\theta}\right\rangle$ of subsets of $\theta$.

Lemma 2.7**.**

Let $\kappa\geq\omega$ be a cardinal. Then the order dimension of $\left\langle{\left[\kappa\right]}^{<\omega},\subsetneq\right\rangle$ is at most ${\log}_{2}({\log}_{2}(\kappa))$.

Proof.

Let $\theta={\log}_{2}({\log}_{2}(\kappa))$, and let $\lambda={2}^{\theta}$. Then ${2}^{\lambda}\geq\kappa$, and so it is possible to find a sequence $\left\langle{f}_{\alpha}:\alpha<\kappa\right\rangle$ of $\kappa$ many distinct functions from $\lambda$ to $2$. To avoid confusion, we use ${}^{\lambda}{2}$ to denote the collection of all functions from $\lambda$ to $2$. Next, use Theorem 2.6 to fix an independent family $\left\langle{X}_{\xi}:\xi<\lambda\right\rangle$ of subsets of $\theta$. For each $\iota<\theta$, define a function ${g}_{\iota}:\lambda\rightarrow 2$ by stipulating that ${g}_{\iota}(\xi)=0$ iff $\iota\in{X}_{\xi}$. Given $f,g\in{}^{\lambda}{2}$ with $f\neq g$, let $\Delta(f,g)=\min\{\xi<\lambda:f(\xi)\neq g(\xi)\}$. Unfixing $\iota<\theta$, we are going to define a linear order ${<}_{\iota}$ on ${\left[\kappa\right]}^{<\omega}$ for each $\iota<\theta$. First declare, for each $\iota<\theta$ and all $u,v\in{\left[\kappa\right]}^{<\omega}$, that $u\>{\prec}_{\iota}\>v$ if and only if either $u\subsetneq v$ or $f_{\alpha}$ coincides with $g_{\iota}$ on $\Delta(u,f_{\alpha})=\{\Delta(f_{\beta},f_{\alpha}):\beta\in u\}$ for some $\alpha\in v\setminus u$, i.e., $f_{\alpha}(\xi)=g_{\iota}(\xi)$ for each $\xi\in\Delta(u,f_{\alpha})$.

Claim 2.8**.**

For each $\iota<\theta$, ${\prec}_{\iota}$ is a partial order on ${\left[\kappa\right]}^{<\omega}$.

Proof.

Clearly, $u\>{\nprec}_{\iota}\>u$ because both the defining clauses of ${\prec}_{\iota}$ fail in this case. So it remains to check that ${\prec}_{\iota}$ is transitive. Consider $u,v,w\in{\left[\kappa\right]}^{<\omega}$ and suppose that $u\>{\prec}_{\iota}\>v$ and that $v\>{\prec}_{\iota}\>w$. There are four cases to consider.

Case 1: $u\subsetneq v$ and $v\subsetneq w$. Then $u\subsetneq w$, whence $u\>{\prec}_{\iota}\>w$.

Case 2: $u\subsetneq v$ and $f_{\alpha}$ coincides with $g_{\iota}$ on $\Delta(v,f_{\alpha})$ for some $\alpha\in w\setminus v$. Then, such $\alpha$ is an element of $w\setminus u$, and $f_{\alpha}$ coincides with $g_{\iota}$ on $\Delta(u,f_{\alpha})$ since $\Delta(u,f_{\alpha})\subset\Delta(v,f_{\alpha})$ by $u\subsetneq v$. So $u\>{\prec}_{\iota}\>w$.

Case 3: $f_{\alpha}$ coincides with $g_{\iota}$ on $\Delta(u,f_{\alpha})$ for some $\alpha\in u\setminus v$ and $v\subsetneq w$. Then, such $\alpha\in w\setminus u$, and $f_{\alpha}$ coincides with $g_{\iota}$ on $\Delta(u,f_{\alpha})$. So $u\>{\prec}_{\iota}\>w$.

Case 4: $f_{\beta}$ coincides with $g_{\iota}$ on $\Delta(u,f_{\beta})$ for some $\beta\in v\setminus u$, and $f_{\gamma}$ coincides with $g_{\iota}$ on $\Delta(v,f_{\gamma})$ for some $\gamma\in w\setminus v$ . Note that $f_{\beta}$ and $f_{\gamma}$ coincide with $g_{\iota}$ on $\Delta(u,f_{\beta})$ and $\Delta(v,f_{\gamma})$, respectively, and $f_{\beta}(\Delta(f_{\beta},f_{\gamma}))\neq f_{\gamma}(\Delta(f_{\beta},f_{\gamma}))=g_{\iota}(\Delta(f_{\beta},f_{\gamma}))$. Hence we have $\Delta(f_{\beta},f_{\gamma})\not\in\Delta(u,f_{\beta})$, which implies $\gamma\not\in u$ and so $\gamma\in w\setminus u$. Fixing $\alpha\in u$, we next show that $f_{\gamma}(\Delta(f_{\alpha},f_{\gamma}))=g_{\iota}(\Delta(f_{\alpha},f_{\gamma}))$. From the above conclusion, we know that $\Delta(f_{\alpha},f_{\beta})\neq\Delta(f_{\beta},f_{\gamma})$. There are two subcases to check.

Subcase 4.1: $\Delta(f_{\alpha},f_{\beta})<\Delta(f_{\beta},f_{\gamma})$. Then $\Delta(f_{\alpha},f_{\gamma})=\Delta(f_{\alpha},f_{\beta})$, and therefore $f_{\gamma}(\Delta(f_{\alpha},f_{\gamma}))=f_{\gamma}(\Delta(f_{\alpha},f_{\beta}))=f_{\beta}(\Delta(f_{\alpha},f_{\beta}))=g_{\iota}(\Delta(f_{\alpha},f_{\beta}))=g_{\iota}(\Delta(f_{\alpha},f_{\gamma}))$.

Subcase 4.2: $\Delta(f_{\alpha},f_{\beta})>\Delta(f_{\beta},f_{\gamma})$. Then $\Delta(f_{\alpha},f_{\gamma})=\Delta(f_{\beta},f_{\gamma})$, and therefore $f_{\gamma}(\Delta(f_{\alpha},f_{\gamma}))=f_{\gamma}(\Delta(f_{\beta},f_{\gamma}))=g_{\iota}(\Delta(f_{\beta},f_{\gamma}))=g_{\iota}(\Delta(f_{\alpha},f_{\gamma}))$.

We conclude that $\gamma\in w\setminus u$ and that ${f}_{\gamma}$ coincides with $g_{\iota}$ on $\Delta(u,f_{\gamma})$. So $u\>{\prec}_{\iota}\>w$.

This completes the proof of the claim. ∎

Now, for each $\iota<\theta$, let ${<}_{\iota}$ be any linear order on ${\left[\kappa\right]}^{<\omega}$ extending ${\prec}_{\iota}$. Note that for any $u,v\in{\left[\kappa\right]}^{<\omega}$, if $u\subsetneq v$, then $u\>{\prec}_{\iota}\>v$, and hence $u\>{<}_{\iota}\>v$. Thus $\left\langle{<}_{\iota}:\iota<\theta\right\rangle$ is a sequence of linear orders on ${\left[\kappa\right]}^{<\omega}$ compatible with $\subsetneq$. To see that this sequence of linear orders witnesses that the dimension of ${\left[\kappa\right]}^{<\omega}$ is at most $\theta$, fix any $u,v\in{\left[\kappa\right]}^{<\omega}$ and suppose that $v\not\subseteq u$. It suffices to find an $\iota<\theta$ such that $u\>{\prec}_{\iota}\>v$. Fix $\alpha\in v\setminus u$. If $u=\emptyset$, then $\emptyset\subsetneq v$, and so for every $\iota<\theta$, $u\>{\prec}_{\iota}\>v$. Therefore we may assume that $u\neq\emptyset$. Define $F=\Delta(u,f_{\alpha})$. $F$ is a nonempty finite subset of $\lambda$. Define $\sigma:F\rightarrow 2$ by setting $\sigma(\xi)={f}_{\alpha}(\xi)$, for all $\xi\in F$. Since $\left\langle{X}_{\xi}:\xi<\lambda\right\rangle$ is an independent family of subsets of $\theta$, ${\bigcap}_{\xi\in F}{{X}^{\sigma(\xi)}_{\xi}}$ is nonempty. Let $\iota=\min\left({\bigcap}_{\xi\in F}{{X}^{\sigma(\xi)}_{\xi}}\right)$. Then $\iota\in\theta$ and for each $\beta\in u$, ${g}_{\iota}(\Delta({f}_{\alpha},{f}_{\beta}))=\sigma(\Delta({f}_{\alpha},{f}_{\beta}))={f}_{\alpha}(\Delta({f}_{\alpha},{f}_{\beta}))$. In other words, $f_{\alpha}$ coincides with $g_{\iota}$ on $\Delta(u,f_{\alpha})$. So $u\>{\prec}_{\iota}\>v$ as claimed. ∎

3. The dimension of locally countable partial orderings

The setting of locally countable partial orders (for which the Turing degrees and many other degree structures from computability theory form natural examples) is quite a bit more complicated.

Even though the following theorem is not needed for the proof of our main result, it provides information about arbitrary locally countable partial orders which is likely to be useful in their analysis. It proves the existence of a ranking function on them.

Lemma 3.1**.**

Suppose $\mathcal{P}=(P,\prec)$ is any locally countable partial ordering. Then there is a function $r:P\to\eta\cdot\omega_{1}$ such that for all $x,y\in P$, $x\prec y$ implies $r(x)<r(y)$. (Here, $\eta\cdot\omega_{1}$ is the order product of these two order types under the antilexicographical ordering.)

Proof.

If $P$ is empty, then there is nothing to prove. So we assume that $P$ is non-empty. Let $\kappa=\left|P\right|$. From an enumeration $\langle y_{\beta}:\beta<\kappa\rangle$ of $P$, we construct a cofinal sequence $\langle x_{\alpha}:\alpha<\lambda\rangle$ of elements of $P$ by recursion as follows: Let $x_{0}=y_{0}$, and for $\alpha>0$, let $x_{\alpha}=y_{\beta}$ for the least $\beta$ such that $y_{\beta}\not\preceq x_{\alpha^{\prime}}$ for any $\alpha^{\prime}<\alpha$. (The recursion stops at the least ordinal $\lambda$ when there is no such $\beta$.)

Now, for each $\alpha<\lambda$, let $A_{\alpha}=\{x\in P:x\preceq x_{\alpha}\}$ and recall that by local countability, each $A_{\alpha}$ is countable. Now define $r$ by recursion on $\alpha<\lambda$: Let $r\upharpoonright A_{0}$ map $A_{0}$ into $\eta\cdot\{0\}$ using any linearization of $\prec\upharpoonright A_{0}^{2}$. For $\alpha$ with $0<\alpha<\lambda$, assume that $r\upharpoonright\left({\bigcup}_{\alpha^{\prime}<\alpha}{{A}_{\alpha^{\prime}}}\right)$ has been defined. Find a countable set $B_{\alpha}\subseteq\alpha$ such that

[TABLE]

Fix $\gamma<\omega_{1}$ such that

[TABLE]

Now extend the definition of $r$ to $A_{\alpha}\setminus\left(\bigcup_{\alpha^{\prime}\in B_{\alpha}}A_{\alpha^{\prime}}\right)$ by mapping this set into $\eta\cdot\{\gamma\}$ using any linearization of $\prec$ on this set. It is clear that such countable $\gamma$ must exist because $\bigcup_{\alpha^{\prime}\in B_{\alpha}}A_{\alpha^{\prime}}$ is countable, giving us the desired map $r$. ∎

Our main theorem on locally countable partial orders is the following:

Theorem 3.2**.**

Let $\kappa$ be a regular uncountable cardinal and $\mathcal{P}=(P,\prec)$ a locally countable partial order of size $\kappa^{+}$. Then $\mathcal{P}$ has dimension at most $\kappa$.

Our proof of this theorem uses two lemmas:

Lemma 3.3**.**

For $\kappa$ and $\mathcal{P}=(P,\prec)$ as in Theorem 3.2, there is a sequence $\langle f_{\alpha}:\alpha<\kappa\rangle$ of functions $f_{\alpha}:P\to 2$ such that for any countable subset $A\subseteq P$ and any $y\in P\setminus A$, there is $\alpha<\kappa$ such that $f_{\alpha}^{\prime\prime}A=\{0\}$ and $f_{\alpha}(y)=1$.

Proof.

The hypothesis is that $\left|P\right|={\kappa}^{+}$. Let $\left\langle{x}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ be a 1–1 enumeration of $P$. The proof will proceed via three claims.

Claim 3.4**.**

Suppose that there is a sequence $\left\langle{g}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ such that for each $\xi<{\kappa}^{+}$, ${g}_{\xi}:\kappa\rightarrow 2$, and that for every countable $B\subseteq{\kappa}^{+}$ and every $\zeta\in{\kappa}^{+}\setminus B$, there exists $\alpha<\kappa$ so that $\forall\xi\in B\left[{g}_{\xi}(\alpha)=0\right]$ and ${g}_{\zeta}(\alpha)=1$. Then there exists a sequence of functions $\left\langle{f}_{\alpha}:\alpha<\kappa\right\rangle$ as in the statement of the lemma.

Proof.

Let $\left\langle{g}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ be given. Fix $\alpha<\kappa$ and define ${f}_{\alpha}:P\rightarrow 2$ by stipulating that ${f}_{\alpha}({x}_{\xi})={g}_{\xi}(\alpha)$ for each $\xi<{\kappa}^{+}$. Suppose that $A\subseteq P$ is countable and that $y\in P\setminus A$. Let $B=\{\xi<{\kappa}^{+}:{x}_{\xi}\in A\}$. Then $B$ is a countable subset of ${\kappa}^{+}$ because we have a 1–1 enumeration of $P$. Let $\zeta<{\kappa}^{+}$ be so that $y={x}_{\zeta}$. Note $\zeta\in{\kappa}^{+}\setminus B$. So there is $\alpha<\kappa$ so that $\forall\xi\in B\left[{g}_{\xi}(\alpha)=0\right]$ and ${g}_{\zeta}(\alpha)=1$. If $x\in A$, then $x={x}_{\xi}$ for some $\xi\in B$, and so ${f}_{\alpha}(x)={f}_{\alpha}({x}_{\xi})={g}_{\xi}(\alpha)=0$. Therefore $\forall x\in A\left[{f}_{\alpha}(x)=0\right]$ and ${f}_{\alpha}(y)={f}_{\alpha}({x}_{\zeta})={g}_{\zeta}(\alpha)=1$, as required. ∎

Claim 3.5**.**

Suppose that there is a sequence $\left\langle{E}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ such that the following conditions hold:

1. (1)

$\forall\xi<{\kappa}^{+}\left[{E}_{\xi}\subseteq\kappa\right]$; 2. (2)

for any countable $B\subseteq{\kappa}^{+}$ and any $\zeta\in{\kappa}^{+}\setminus B$, there is a finite set $F\subseteq\kappa$ such that $\forall\xi\in B\left[{E}_{\xi}\cap F\neq{E}_{\zeta}\cap F\right]$.

Then there is a sequence of functions $\left\langle{g}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ as in Claim 3.4.

Proof.

Let $L=\{\left\langle s,H\right\rangle:s\in{\left[\kappa\right]}^{<\omega}\wedge H\subseteq\mathcal{P}(s)\}$. Clearly $\left|L\right|=\kappa$. So it suffices to find a sequence $\left\langle{h}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ of functions ${h}_{\xi}:L\rightarrow 2$ satisfying the properties of the sequence $\left\langle{g}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ from Claim 3.4. Fix $\xi<{\kappa}^{+}$ and define ${h}_{\xi}:L\rightarrow 2$ by stipulating, for all $\left\langle s,H\right\rangle\in L$, that ${h}_{\xi}(\left\langle s,H\right\rangle)=1$ if and only if ${E}_{\xi}\cap s\in H$. Let $B\subseteq{\kappa}^{+}$ be countable and suppose $\zeta\in{\kappa}^{+}\setminus B$. By (2), there is a finite set $F\subseteq\kappa$ such that $\forall\xi\in B\left[{E}_{\xi}\cap F\neq{E}_{\zeta}\cap F\right]$. Let $H=\{F\cap{E}_{\zeta}\}$. Then $F\in{\left[\kappa\right]}^{<\omega}$ and $H\subseteq\mathcal{P}(F)$. Therefore $\left\langle F,H\right\rangle\in L$. Since ${E}_{\zeta}\cap F\in H$, ${h}_{\zeta}(\left\langle F,H\right\rangle)=1$. On the other hand, for any $\xi\in B$, ${E}_{\xi}\cap F\notin H$ because ${E}_{\xi}\cap F\neq{E}_{\zeta}\cap F$. Hence ${h}_{\xi}(\left\langle F,H\right\rangle)=0$. So we have $\forall\xi\in B\left[{h}_{\xi}(\left\langle F,H\right\rangle)=0\right]$, as required. ∎

Claim 3.6**.**

There is a sequence $\left\langle{E}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ satisfying (1) and (2) of Claim 3.5.

Proof.

By well-known arguments (see, e.g., Kunen [6, Theorem 1.2]), there is a sequence $\left\langle{E}_{\xi}:\xi<{\kappa}^{+}\right\rangle$, which is an almost disjoint family of subsets of $\kappa$. In other words, $\forall\xi<{\kappa}^{+}\left[{E}_{\xi}\in{\left[\kappa\right]}^{\kappa}\right]$, and $\forall\xi,\zeta<{\kappa}^{+}\left[\xi\neq\zeta\implies\left|{E}_{\xi}\cap{E}_{\zeta}\right|<\kappa\right]$. The standard argument for the existence of such an almost disjoint family (like the proof of [6, Theorem 1.2]) uses the regularity of $\kappa$. Suppose $B\subseteq{\kappa}^{+}$ is countable and that $\zeta\in{\kappa}^{+}\setminus B$. Then $\left|{E}_{\zeta}\cap{E}_{\xi}\right|<\kappa$ for all $\xi\in B$, and since $\operatorname{cf}(\kappa)>\omega$, $\left|{\bigcup}_{\xi\in B}{\left({E}_{\zeta}\cap{E}_{\xi}\right)}\right|<\kappa$. So choose $\alpha\in{E}_{\zeta}\setminus\left({\bigcup}_{\xi\in B}{{E}_{\xi}}\right)$. Then $F=\{\alpha\}$ is a finite subset of $\kappa$ and for any $\xi\in B$, $F\cap{E}_{\xi}=\emptyset$, while $F\cap{E}_{\zeta}=\{\alpha\}$. So $F$ is as required. ∎

Now Claim 3.6 says that the hypotheses of Claim 3.5 are satisfied. So there is a sequence $\left\langle{g}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ satisfying the hypotheses of Claim 3.4. Therefore, there is a sequence $\left\langle{f}_{\alpha}:\alpha<\kappa\right\rangle$ as in the statement of the lemma. ∎

Lemma 3.7**.**

Fix $\kappa$ and $\mathcal{P}=(P,\prec)$ as in Theorem 3.2, $\{f_{\alpha}\}_{\alpha<\kappa}$ as in Lemma 3.3, and $\alpha<\kappa$. Then there is a linearization $<_{\alpha}$ of $\prec$ such that for any $x,y\in P$, if $f_{\alpha}(u)=1$ for some $u\in P$ with $u\preceq x$ and $f_{\alpha}(z)=0$ for all $z\in P$ with $z\preceq y$, then $y<_{\alpha}x$.

Proof.

We first define a relation $<^{\prime}_{\alpha}$ on $P\times P$ by setting $y<^{\prime}_{\alpha}x$ iff $f_{\alpha}(u)=1$ for some $u\in P$ with $u\preceq x$ and $f_{\alpha}(z)=0$ for all $z\in P$ with $z\preceq y$. Note that $<^{\prime}_{\alpha}$ is irreflexive. We now only need to show that the transitive closure of $\prec\cup<^{\prime}_{\alpha}$ on $P$ is irreflexive, since any linearization $<_{\alpha}$ of this transitive closure will be as desired.

So, for the sake of a contradiction, define $x<^{\prime\prime}y$ iff $x\prec y$ or $x<^{\prime}_{\alpha}y$, and assume that there is a finite sequence $x_{0}<^{\prime\prime}x_{1}<^{\prime\prime}\dots<^{\prime\prime}x_{n}=x_{0}$. Since $\prec$ is irreflexive, we may assume, by a cyclic permutation of the $x_{i}$, that $x_{0}<^{\prime}_{\alpha}x_{1}$. Furthermore, if $x_{i}<^{\prime}_{\alpha}x_{i+1}\prec x_{i+2}$, then $x_{i}<^{\prime}_{\alpha}x_{i+2}$ by the definition of $<^{\prime}_{\alpha}$, and we can delete $x_{i+1}$ from this chain. Since $<^{\prime}_{\alpha}$ is irreflexive, we have $x_{0}<^{\prime}_{\alpha}x_{1}<^{\prime}_{\alpha}x_{2}$. So we have that on the one hand, $f_{\alpha}(u)=1$ for some $u\in P$ with $u\preceq x_{1}$ by the relation $x_{0}<^{\prime}_{\alpha}x_{1}$, and on the other hand, $f_{\alpha}(u)=0$ for all $u\in P$ with $u\preceq x_{1}$ by the relation $x_{1}<^{\prime}_{\alpha}x_{2}$. This is a contradiction. ∎

Proof of Theorem 3.2.

Assume that $x\not\preceq y$. Let $A=\{z\in P:z\preceq y\}$, which is a countable set, so apply Lemma 3.3 to find a function $f_{\alpha}$ such that $f_{\alpha}^{\prime\prime}A=\{0\}$ and $f_{\alpha}(x)=1$, and then apply Lemma 3.7 with this function to obtain a linearization $<_{\alpha}$ of $\prec$ with $y<_{\alpha}x$.

Thus, the at most $\kappa$ many linearizations $<_{\alpha}$ witness that the dimension of $\mathcal{P}$ is at most $\kappa$. ∎

Note that the proof of Theorem 3.2 really shows that if $\lambda$ is some uncountable cardinal such that ${\mathfrak{a}}(\lambda)\geq{\kappa}^{+}$, then the dimension of $(P,<)$ is at most $\lambda$, where ${\mathfrak{a}}(\lambda)$ is the almost disjointness number at $\lambda$.

Next, we consider the case when $\kappa$ is possibly singular. We are able to prove Theorem 3.2 also in that case if $\operatorname{cf}(\kappa)>\omega$. The proof is almost the same.

Theorem 3.8**.**

Suppose $\kappa$ is any cardinal such that $\operatorname{cf}(\kappa)>\omega$ and $\mathcal{P}=(P,\prec)$ is any locally countable partial order of size ${\kappa}^{+}$. Then $\mathcal{P}$ has dimension at most $\kappa$.

Proof.

The only place in the proof of Theorem 3.2 where the regularity of $\kappa$ is used is in Claim 3.6. All of the other lemmas and claims needed for the proof of Theorem 3.2 go through for any uncountable cardinal $\kappa$. So we assume that $\kappa$ is a singular cardinal with $\operatorname{cf}(\kappa)>\omega$. To establish the analogue of Claim 3.6, we use a basic fact from PCF theory. Let $\mu=\operatorname{cf}(\kappa)$. By hypothesis, $\mu$ is an uncountable regular cardinal. By [1, Theorem 2.23], there exist sequences $\left\langle{\lambda}_{\alpha}:\alpha<\mu\right\rangle$ and $\left\langle{f}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ satisfying the following conditions:

1. (1)

for each $\alpha<\mu$, ${\lambda}_{\alpha}<\kappa$; 2. (2)

for each $\xi<{\kappa}^{+}$, ${f}_{\xi}$ is a function, $\operatorname{dom}({f}_{\xi})=\mu$, and $\forall\alpha<\mu\left[{f}_{\xi}(\alpha)\in{\lambda}_{\alpha}\right]$; 3. (3)

for all $\xi<\zeta<{\kappa}^{+}$, ${Z}_{\xi,\zeta}\{\alpha<\mu:{f}_{\xi}(\alpha)\geq{f}_{\zeta}(\alpha)\}$ is bounded in $\mu$.

Note that for each $\xi<{\kappa}^{+}$, ${f}_{\xi}\subseteq\mu\times\kappa$. Furthermore $\left|{f}_{\xi}\right|=\mu$, and if $\zeta\neq\xi$, then $\left|{f}_{\zeta}\cap{f}_{\xi}\right|\leq\left|\{\alpha<\mu:{f}_{\zeta}(\alpha)={f}_{\xi}(\alpha)\}\right|<\mu$. Therefore if $B\subseteq{\kappa}^{+}$ is countable and $\zeta\in{\kappa}^{+}\setminus B$, then $\left|{\bigcup}_{\xi\in B}{{f}_{\xi}\cap{f}_{\zeta}}\right|<\mu$ because $\mu>\omega$ is a regular cardinal. So we may find $\left\langle\alpha,\delta\right\rangle\in{f}_{\zeta}\setminus\left({\bigcup}_{\xi\in B}{{f}_{\xi}\cap{f}_{\zeta}}\right)$. As in the proof of Claim 3.6, let $F=\{\left\langle\alpha,\delta\right\rangle\}$. $F$ is a finite subset of $\mu\times\kappa$, and ${f}_{\zeta}\cap F=\{\left\langle\alpha,\delta\right\rangle\}$, while $\forall\xi\in B\left[{f}_{\xi}\cap F=\emptyset\right]$. Thus the sequence $\left\langle{f}_{\xi}:\xi<{\kappa}^{+}\right\rangle$ satisfies the analogue of conditions (1) and (2) of Claim 3.5. Since $\left|\mu\times\kappa\right|=\kappa$, we have shown that the hypotheses of Claim 3.5 are satisfied. Now the rest of the proof follows the proof of Theorem 3.2. ∎

The following corollary is immediate from Theorems 3.2 and 3.8.

Corollary 3.9**.**

If the order dimension of any locally countable partial order is ${2}^{{\aleph}_{0}}$, then either $\mathrm{CH}$ holds, or ${2}^{{\aleph}_{0}}$ is a limit cardinal, or ${2}^{{\aleph}_{0}}$ is the successor of a singular cardinal of countable cofinality.

Our conjecture is that the cases besides $\mathrm{CH}$ are also realized. In other words, we think that there are models where ${\mathfrak{dim}}_{T}={2}^{{\aleph}_{0}}={\aleph}_{{\omega}_{1}}$, ${\mathfrak{dim}}_{T}={2}^{{\aleph}_{0}}={\aleph}_{\omega+1}$, and ${\mathfrak{dim}}_{T}={2}^{{\aleph}_{0}}$ where ${2}^{{\aleph}_{0}}$ is weakly inaccessible.

4. The dimension of some degree structures

from computability theory

In this section, we state some results on the dimension of three degree structures from computability theory, the Turing degrees, the Medvedev degrees and the Muchnik degrees.

We start with the Turing degrees since the original motivation for our investigation was determining the dimension of the Turing degrees under the partial ordering, for which we obtain two partial results.

Definition 4.1**.**

For a partial order $\mathcal{P}=(P,\prec)$ and a subset $S$ of $P$, we say that $S$ is a strongly independent antichain if for any subset $T$ of $S$ with $|T|<|S|$ and for any $x\in S\setminus T$, there is an upper bound $y\in P$ of $T$ with $y\not\succeq x$.

Lemma 4.2**.**

Let $\mathcal{P}=(P,\prec)$ be a partial order with a strongly independent antichain $S$. Then the dimension of $\mathcal{P}$ is at least $|S|$.

Proof.

We provide a proof by contradiction. Suppose that the dimension $\rho$ of $\mathcal{P}$ is less than $|S|$. Let $\hat{S}\subset S$ satisfy that $|\hat{S}|>\rho$ is a successor cardinal and let $\kappa=|\hat{S}|$. Choose linear extensions $\{<_{\alpha}\}_{\alpha<\rho}$ of $\prec$ whose intersection is $\prec$.

Fix $x\in\hat{S}$. Let $\{T_{\alpha}\}_{\alpha<\kappa}$ be any increasing sequence of subsets of $\hat{S}$ of cardinality $<|\hat{S}|$ such that $\bigcup_{\alpha<\kappa}T_{\alpha}=\hat{S}\setminus\{x\}$. By the strong independence, we can find a sequence $\{y_{\alpha}\}_{\alpha<\kappa}$ of upper bounds of $T_{\alpha}$’s such that for each $\alpha<\kappa$, $x\not\preceq y_{\alpha}$, which means that there exists $\beta<\rho$ such that $y_{\alpha}<_{\beta}x$ by the choice of $\{<_{\beta}\}_{\beta<\rho}$. Since $\rho<\kappa$ and $\kappa$ is regular (or finite), there must exist a fixed $\beta<\rho$ such that $y_{\alpha}<_{\beta}x$ holds for unboundedly many $\alpha<\kappa$. By the choice of $\{T_{\alpha}\}_{\alpha<\kappa}$, every element of $\hat{S}$ distinct from $x$ is in almost all $\{T_{\alpha}\}_{\alpha<\kappa}$, and therefore, $y<_{\beta}x$ must hold for each $y\in\hat{S}\setminus\{x\}$. Hence we conclude that for any $x\in\hat{S}$, there exists $\beta<\rho$ such that $y<_{\beta}x$.

Since again $\rho<\kappa=|\hat{S}|$, there must exist a common $\beta<\rho$ and distinct $x_{0},x_{1}\in\hat{S}$ such that $y<_{\beta}x_{i}$ holds for each $y\in\hat{S}\setminus\{x_{i}\}$ and $i\in\{0,1\}$. But this gives us $x_{0}<_{\beta}x_{1}<_{\beta}x_{0}$, and hence $x_{0}<_{\beta}x_{0}$, a contradiction. ∎

We can now state two partial results about the dimension of the Turing degrees as a partial order:

Proposition 4.3** (Higuchi).**

The dimension of the Turing degrees is uncountable.

Proof.

By Sacks [9], every locally countable partial order of cardinality $\aleph_{1}$ is embeddable into the Turing degrees. Thus it is enough to find such a partial order whose dimension is at least $\aleph_{1}$. Let us consider the suborder $(P,\subsetneq)$ of $\mathcal{P}({\aleph}_{1})$ under set inclusion whose underlying set is

[TABLE]

It is easy to see that $\{\{\alpha\}:\alpha<\omega_{1}\}$ is a strongly independent antichain of cardinality $\aleph_{1}$ in the partial order $(P,\subsetneq)$, and therefore, the dimension of $(P,\subsetneq)$ is at least $\aleph_{1}$ by Lemma 4.2. ∎

Theorem 4.4**.**

It is consistent with ZFC that the dimension of the Turing degrees is strictly less than the continuum. More precisely, the dimension of the Turing degrees can be continuum only if either $\mathrm{CH}$ holds, or ${2}^{{\aleph}_{0}}$ is a limit cardinal, or ${2}^{{\aleph}_{0}}$ is the successor of a singular cardinal of countable cofinality.

Proof.

The first part is a direct corollary of Theorems 3.2: Work in a model in which $2^{\aleph_{0}}=\aleph_{2}$ and apply the theorem with $\kappa=\aleph_{1}$. The second part combines Theorems 3.2 and 3.8. ∎

We now turn our attention to the Medvedev and the Muchnik degrees. By a cardinality argument, the dimension of both can be at most $2^{2^{\aleph_{0}}}$.

Theorem 4.5** (Pouzet 1969).**

Let $\mathcal{P}=(P,\prec)$ be a partial order. Then the dimension of $(\operatorname{IniSeg}(\mathcal{P}),\subset))$ is the chain covering number of $\mathcal{P}$, where $\operatorname{IniSeg}(\mathcal{P})$ is the set of initial segments of $\mathcal{P}$ and the chain covering number of $\mathcal{P}$ is the least cardinal $\kappa$ such that there exists a set $\mathcal{C}$ of chains of $\mathcal{P}$ with $|C|=\kappa$ and $\bigcup\mathcal{C}=P$.

It is known that the Muchnik degrees are isomorphic to the set of all final segments of the Turing degrees ordered by $\supset$. Note that the dimension of a partial order does not change if we reverse the order. Thus the dimension of the Muchnik degrees is the chain covering number of the Turing degrees, which is $2^{\aleph_{0}}$ since there are at most $2^{\aleph_{0}}$ many Turing degrees and the Turing degrees contain an antichain of size $2^{\aleph_{0}}$. Since the Muchnik degrees can be seen a suborder of the Medvedev degrees, the dimension of the latter is at least $2^{\aleph_{0}}$, and by a cardinality argument at most $2^{2^{\aleph_{0}}}$.

We can thus determine the dimension of the Muchnik degrees in ZFC but leave open the following questions:

Question 4.6**.**

1. (1)

Does ZFC determine the dimension of the Turing degrees? 2. (2)

Does ZFC determine the dimension of the Medvedev degrees?

5. Some examples

One of the goals of our paper was to find out whether the order dimension of the structure of Turing degrees is $\aleph_{1}$ in all models of ZFC. More generally, we can pose the following

Question 5.1**.**

Is there a partially ordered set $(F,<)$ such that, in all models of ZFC, the following holds?

1. (1)

The order $<$ is locally countable; 2. (2)

for every at most countable subset $G\subseteq F$ there is an upper bound $x$ of $G$, i.e., $y\leq x$ for all $y\in F$; 3. (3)

the cardinality of $F$ is $2^{\aleph_{0}}$; and 4. (4)

the order dimension of $(F,<)$ is $\aleph_{1}$.

Each of the examples below will satisfy three of the properties but the fourth at most partially. Note that additional set-theoretical assumptions like $2^{\aleph_{0}}\in\{\aleph_{1},\aleph_{2}\}$ might make the examples have all the desired properties.

Example 5.2**.**

Let $F$ be the set of all hereditarily countable sets and let $<$ be the transitive closure of the element-relation. As every set bounds only countably many other sets in a hereditary way, the partial order is locally at most countable, and it is well-founded. Furthermore, every at most countable ordinal $\alpha$ can be identified with the at most countable set $\{\beta:\beta<\alpha\}$, and these sets are hereditarily countable. The set $\{\{\{\alpha\}\}:\alpha$ is an at most countable ordinal$\}$ is an antichain of size $\aleph_{1}$ consisting of hereditarily countable sets; as each of its at most countable subsets is hereditarily countable, these sets witness that the antichain is indeed a strong antichain and that therefore the order dimension is at least $\aleph_{1}$. It is known that the set of hereditarily countable sets has cardinality $2^{\aleph_{0}}$.

Example 5.3**.**

Let $F$ be the set of all functions $f$ from an ordinal $\alpha<\omega_{1}$ into $\omega_{1}$; order $F$ by letting $f<g$ iff there are ordinals $\alpha,\beta\in\operatorname{dom}(g)$ such that for all $\gamma\in\operatorname{dom}(f)$, $f(\gamma)=g(\alpha+1+\gamma)$, $g(\alpha)=g(\beta)$, $f(\gamma)<g(\alpha)$ and $\alpha+1+\operatorname{dom}(f)=\beta$. It is easy to see that $<$ is transitive and locally countable. Furthermore, one can easily see that for at most countably many functions $f_{0},f_{1},\ldots$, there is a common upper bound $g$ by choosing an ordinal $\alpha$ strictly larger than all ordinals occurring in the $f_{k}$, considering an $\omega$-power $\omega^{\gamma}$ larger than the domains of all $f_{k}$, letting the domain of $h$ be $\omega^{\gamma+1}$, and setting $h(\omega^{\gamma}\cdot k+1+\gamma)=f_{k}(\gamma)$ for all $\gamma\in\operatorname{dom}(f_{k})$ and $h(\delta)=\alpha$ for all $\delta$ in the domain of $h$ where $h$ is not yet defined. So every at most countable set $G$ of members of $F$ has a common upper bound, which strictly bounds from above exactly the members of $G$ and those members of $F$ which are below a member of $G$. The cardinality of $F$ is $2^{\aleph_{0}}$. Furthermore, the set of all functions with domain $\{0\}$ forms a strong antichain and therefore, the order dimension is at least $\aleph_{1}$.

Example 5.4**.**

Let $F$ be the set of all subsets of ${\mathbb{R}}\times{\omega_{1}}$ such that $F$ is the union of finitely many sets of the form $A_{x,y,z}=\{x\}\times\{u\in M:y\leq u\leq z\}$ and $B_{x,y,z}=\{x\}\times\{u\in M:y\leq u<z\}$, and order this set by set inclusion. The set $(F,<)$ satisfies all four conditions except for the second, which is weakened to the existence of common upper bounds of finitely many elements.

One can see from the definition that every set of the form $A_{x,y,z}$ or $B_{x,y,z}$ has at most countably many subsets of this form in $F$; furthermore each member of $F$ is countable. Thus this is a locally countable partially ordered set.

Furthermore, as the finite union of any members of $F$ is again a member of $F$, one has also that finitely many subsets have an upper bound. However, this does not extend to all countable subsets of $F$.

The cardinality of $F$ is $2^{\aleph_{0}}$. The lower bound is seen by looking at all sets $A_{x,0,0}$ with $x\in\mathbb{R}$. The upper bound stems from the fact that there are $2^{\aleph_{0}}$ many countable subsets of $\mathbb{R}$.

Furthermore, let $p,q$ be rational numbers with $p<q$ and $y\in{\omega_{1}}$. For each $p,q,y$, one defines a linear order $<_{p,q,y}$ as a linear extension of the partial order defined by $A<_{p,q,y}B$ iff $B$ has more elements than $A$ of the form $(x,y)$ with $p\leq x\leq q$; note that each set has only finitely many such elements. Now if $B\not\subseteq A$ then there is an $(x,y)\in B\setminus A$ and there are rationals $p,q$ such that $x$ is the unique real number $z$ with $p\leq z\leq q$ and $(z,y)\in A\cup B$. It follows that $A<_{p,q,y}B$. So $<$ is the intersection of all $<_{p,q,y}$ with $p,q\in\mathbb{Q}$ and $y\in\omega_{1}$ and $p<q$. It follows that the order dimension of $(F,<)$ is at most $\aleph_{1}$.

Now consider the set $C$ of all $A_{0,y,y}$, which are all singletons. The set $C$ forms a strong antichain, as for every $A_{0,x,x}$ and every at most countable set $D$ of sets of the form $A_{0,y,y}$ with $y\neq x$, there is an upper bound $z$ of all these $y$. For this upper bound $z$, now consider the set $E=B_{0,0,x}\cup A_{0,x+1,z}$, which is a superset of all $A_{0,y,y}\in D$ but not a superset of $A_{0,x,x}$. Now by Lemma 4.2, $(F,<)$ has order dimension at least $\aleph_{1}$, as this is the cardinality of $C$.

Example 5.5**.**

Given $F$ as in Example 5.4, the subset $G=\{A\in F:A\subseteq\{0\}\times{\omega_{1}}\}$ satisfies the first, second and last property, but differs from the third in all models of set theory where $\aleph_{1}\neq 2^{\aleph_{0}}$.

Example 5.6**.**

The set $F$ of all countable subsets of $\omega_{1}$ with the order of inclusion satisfies the property that every countable subset of $F$ has an upper bound in $F$ and the cardinality is $2^{\aleph_{0}}$. Furthermore, $(F,<)$ has order-dimension $\aleph_{1}$.