A new small Dowker space

Abstract

It is proved that a strong instance of the guessing principle $\clubsuit _{{{\,\textrm{AD}\,}}}$ on the first uncountable cardinal follows from either the principle , or the principle $\diamondsuit (\mathfrak b)$, or the existence of a Luzin set. In particular, any of the above hypotheses entails the existence of a Dowker space of size $\aleph _1$.

Appendix: Many Dowker spaces

In this section, $\kappa$ denotes a regular uncountable cardinal. By [12, §3], if $\clubsuit _{{{\,\textrm{AD}\,}}}({\mathcal {S}},1,2)$ holds for a partition ${\mathcal {S}}$ of some nonreflecting stationary subset of $\kappa$ into infinitely many stationary sets, then there exists a Dowker space of size $\kappa$. Here, we demonstrate the advantage of ${\mathcal {S}}$ being large.

Theorem A.1

Suppose that $\clubsuit _{{{\,\textrm{AD}\,}}}({\mathcal {S}},1,2)$ holds, where ${\mathcal {S}}$ is a partition of a nonreflecting stationary subset of $\kappa$ into infinitely many stationary sets. Denote $\mu :=|{\mathcal {S}}|$. Then there are $2^\mu$ many pairwise nonhomeomorphic Dowker spaces of size $\kappa$.

Proof

Fix an injective enumeration $\langle S^\zeta _{n}\mid \zeta<\mu , n<\omega \rangle$ of the elements of ${\mathcal {S}}$. As $\clubsuit _{{{\,\textrm{AD}\,}}}({\mathcal {S}},1,2)$ holds, we may fix a sequence $\langle A_{\alpha }\mid \alpha \in \bigcup {\mathcal {S}}\rangle$ such that

• (i) for every $\alpha \in \bigcup {\mathcal {S}}$, $A_\alpha$ is a subset of $\alpha$, and for every $\alpha '\in \alpha \cap \bigcup {\mathcal {S}}$, $\sup (A_{\alpha '}\cap A_\alpha )<\alpha '$;

• (ii) for all $B_0,B_1\in [\kappa ]^{\kappa }$ and $(\zeta ,n)\in \mu \times \omega$, the following set is stationary:

  $$\begin{aligned} G(S^\zeta _n,B_0,B_1):=\{\alpha \in S^\zeta _n\mid \sup (A_\alpha \cap B_0)=\sup (A_\alpha \cap B_1)=\alpha \}. \end{aligned}$$

For every nonempty $Z\subseteq \mu$, we shall want to define a topological space ${\mathbb {X}}^Z$. To this end, fix a nonempty $Z\subseteq \mu$. For every $n<\omega$, let $S^Z_{n+1}:=\biguplus _{\zeta \in Z}S^\zeta _{n+1}$, and then let $S^Z_0:=\kappa \setminus \biguplus _{n<\omega } S^Z_{n+1}$. For every $\alpha <\kappa$, let $n^Z(\alpha )$ denote the unique $n<\omega$ such that $\alpha \in S^Z_n$. For each $n<\omega$, let $W^Z_n:=\bigcup _{i\le n}S^Z_i$. Then, define a sequence $\vec {L^{Z}}=\langle L^Z_\alpha \mid \alpha <\kappa \rangle$ via

$$ \begin{aligned} L^Z_\alpha :={\left\{ \begin{array}{ll} W^Z_{n^Z(\alpha )-1}\cap A_\alpha ,&{}\text {if }n^Z(\alpha )>0\ \& \ \sup (W^Z_{n^Z(\alpha )-1}\cap A_\alpha )=\alpha ;\\ \emptyset ,&{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

Denote $S^Z:=\{ \alpha \in {{\,\textrm{acc}\,}}(\kappa )\mid \sup (L^Z_\alpha )=\alpha \}$. Finally, let ${\mathbb {X}}^Z=(\kappa ,\tau ^Z)$ be the ladder-system space determined by $\vec {L^{Z}}$, that is, a subset $U\subseteq \kappa$ is $\tau ^Z$-open iff, for every $\alpha \in U\cap S^Z$, $\sup (L^Z_\alpha \setminus U)<\alpha$.

Claim A.1.1

Let Z and $Z'$ be nonempty subsets of $\mu$. Then

• (1) for all $n<\omega$ and $\alpha \in S^Z_{n+1}$, $L^Z_\alpha \subseteq W^Z_n$;

• (2) if $Z\setminus Z'$ is nonempty, then $S^Z\setminus S^{Z'}$ is stationary;

• (3) for all $\alpha \ne \alpha '$ from $S^Z$, $\sup (L^Z_\alpha \cap L^Z_{\alpha '})<\alpha$;

• (4) for all $B_0,B_1\in [\kappa ]^\kappa$, there exists $m<\omega$ such that, for every $n\in \omega \setminus m$, the following set is stationary:

  $$\begin{aligned} \{ \alpha \in S^Z_n \mid \sup (L^Z_\alpha \cap B_0)=\sup (L^Z_\alpha \cap B_1)=\alpha \};\end{aligned}$$

• (5) $S^Z$ is a nonreflecting stationary set.

Proof

(1) Clear.

(2) Suppose that $Z\setminus Z'\ne \emptyset$, and pick $\zeta \in Z{\setminus } Z'$. As $W^Z_0=S^Z_0\supseteq S^\zeta _0$, the former set is cofinal. So, $S^Z\setminus S^{Z'}$ covers the stationary set $G(S^\zeta _1,W^Z_0,\kappa )$.

(3) For all $\alpha \ne \alpha '$ from $S^Z$, $\sup (L^Z_\alpha \cap L^Z_{\alpha '})\le \sup (A_\alpha \cap A_{\alpha '})<\alpha$.

(4) Pick $\zeta \in Z$. Given two cofinal subsets $B_0,B_1$ of $\kappa$, find $m_0,m_1<\omega$ such that $|B_0\cap S^Z_{m_0}|=|B_1\cap S^Z_{m_1}|=\kappa$. Set $m:=\max \{m_0,m_1\}+1$. Then, for every $n\in \omega \setminus m$,

$$\begin{aligned} G(S_n^\zeta ,B_0\cap S^Z_{m_0},B_1\cap S^Z_{m_1})\subseteq \{ \alpha \in S^Z_n \mid \sup (L^Z_\alpha \cap B_0)=\sup (L^Z_\alpha \cap B_1)=\alpha \} \end{aligned}$$

and hence the latter set is stationary.

(5) By Clause (4), $S^Z$ is stationary. As $S^Z\subseteq \bigcup _{n<\omega }S^Z_{n+1}\subseteq \bigcup {\mathcal {S}}$, and since $\bigcup {\mathcal {S}}$ is a nonreflecting stationary set, so is $S^Z$. $\square$

By the preceding claim, and the results of [12, §3], for every nonempty $Z\subseteq \mu$, ${\mathbb {X}}^Z$ is a Dowker space. Thus, what is left is to prove the following:

Claim A.1.2

Suppose that Z and $Z'$ are two distinct nonempty subsets of $\mu$. Then ${\mathbb {X}}^Z$ and ${\mathbb {X}}^{Z'}$ are not homeomorphic.

Proof

Without loss of generality, we may pick $\zeta \in Z\setminus Z'$. Toward a contradiction, suppose that $f:\kappa \leftrightarrow \kappa$ forms a homeomorphism from ${\mathbb {X}}^Z$ to ${\mathbb {X}}^{Z'}$. As f is a bijection, there are club many $\alpha <\kappa$ such that $f^{-1}[\alpha ]=\alpha$. By Claim A.1.1(2), then, we may pick some $\alpha \in S^Z\setminus S^{Z'}$ such that $f^{-1}[\alpha ]=\alpha$. Set $\beta :=f(\alpha )$.

$\blacktriangleright$ If $\beta \notin S^{Z'}$, then $U:=\{\beta \}$ is a $\tau ^{Z'}$-open neighborhood of $\beta$.

$\blacktriangleright$ If $\beta \in S^{Z'}$, then $\beta >\alpha +1$ and the ordinal interval $U:=[\alpha +1,\beta +1]$ is a $\tau ^{Z'}$-open neighborhood of $\beta$.

In both cases, $U\subseteq \kappa {\setminus }\alpha$, so that $f^{-1}[U]\subseteq f^{-1}[\kappa {\setminus }\alpha ]=\kappa {\setminus }\alpha$. As f is continuous and U is a $\tau ^{Z'}$-open neighborhood of $f(\alpha )$, $f^{-1}[U]$ must be a $\tau ^Z$-open neighborhood of $\alpha$, contradicting the fact that $f^{-1}[U]$ is disjoint from $L^Z_\alpha$. $\square$

This completes the proof. $\square$