## Sunday, December 3, 2017

### Balogh's small Dowker space, part 1

A Dowker space is a normal space which is not countably paracompact. Here, a (Hausdorff) space $X$ is normal if for every pair of disjoint closed sets $H,K$, there exist disjoint open sets $U,V$ with $H\subseteq U$ and $K\subseteq V$. Normality is, according to M.E. Rudin, the boundary where point-set topology changes from analysis to set theory.

A space $X$ is countably paracompact if for every $\subseteq$-increasing sequence of open sets $\langle G_n:n<\omega\rangle$ with $\bigcup_n G_n=X$, there exist closed sets $H_n\subseteq G_n$ so that $\bigcup_n H_n=X$. An interesting history of the concept of paracompactness can be found in the Encyclopedia of General Topology.

It turns out that Dowker spaces are the spaces whose normality is not preserved by product with compact metric spaces. See Dan Ma's topology blog for more information about Dowker spaces and all of the topological concepts considered here.

Dowker spaces are considered to be quite rare, and the first ZFC example was due to M.E. Rudin in 1971. In 1996, Zoltan Balogh published an example of a Dowker space of size continuum in ZFC. The problem of whether there exists a Dowker space of size $\aleph_1$ is still open.

Balogh's example was based off an earlier example of Watson which used strongly compact cardinals, together with an earlier argument of Rudin using the submodel method. He was able to use the techniques applied in this Dowker space construction towards other Dowker spaces and the solutions of the second and third Morita conjectures. This post gives a carefully motivated presentation of Balogh's proof.

## 1 Basic construction

We start the construction of $X$ with an increasing sequence of open sets $\langle G_n:n<\omega\rangle$ witnessing the failure of paracompactness. So the underlying set of $X$ is $\omega\times\mathfrak{c}$, and we define $G_n=(n+1)\times\mathfrak{c}$. We will also consider the levels of the space, $L_n=\{n\}\times\mathfrak{c}$.

The levels $L_n$ will be relatively discrete, which helps in the normality proof later. The set of open neighborhoods of a fixed point $(n,\alpha)$ forms a filter. If $n=0$, a neighborhood base for $(n,\alpha)$ is just the singleton. If $n>0$, the filter concentrates on $\{(n,\alpha\}\cup G_{n-1}$. Following Watson's example, we will define a filter $\mathcal{F}_{(n,\alpha)}$ on $G_{n-1}$. Then, a set $U$ is open if and only if for every $(n,\alpha)\in U$, there is some $A\in \mathcal{F}_{(n,\alpha)}$ so that $A\subseteq U$.

Motivated by the normality argument in Watson's paper (which he credits to Rudin), we will end up working on consecutive pairs of levels. So the filter $\mathcal{F}_{(n,\alpha)}$ will be expressed as a filter $\mathcal{F}_\alpha$ on $\mathfrak{c}$ which does not depend on $n$. The filter will contain all cofinite subsets of $\mathfrak{c}$, which is necessary in a space where points are closed.

Let us find the properties of this filter system which we need to make $X$ Dowker.

## 2 Normality

Let us try to get normality. The first level $G_0$ is relatively normal, since it has the discrete topology. If we consider the first two levels, however, we need to separate arbitrary disjoint sets $A_0,A_1\subseteq L_1$ by disjoint open sets. We may as well assume that $A_1$ is the complement of $A_0$ in its level. In total, we are looking for:

$$(*) \textrm{For every }A\subseteq \mathfrak{c}, \textrm{ there exists } B \subseteq \mathfrak{c} \textrm{ such that }$$
$$B\in \mathcal{F}_\alpha \textrm{ if }\alpha\in A,$$
$$\mathfrak{c}\setminus B\in \mathcal{F}_\alpha \textrm{ if }\alpha\not\in A.$$

For the next few claims, assume $(*)$.  We will show that $X$ defined as above is normal.

First, we show this for subsets of single levels.

Claim 1. Suppose that $m\le n$ and $H\subseteq L_n$ and $K\subseteq L_m$ have disjoint closures. Then there are disjoint open sets $U,V$ separating them ($H\subseteq U$ and $K\subseteq V$). Moreover, we can arrange so that $L_n\setminus H\subseteq K$.

Proof of Claim 1: Using $(*)$ repeatedly, let $U',V'$ be disjoint open sets so that $L_m\setminus K\subseteq U'$ and $K\cap L_m\subseteq V'$. Then take $U=U'\cup(X\setminus (G_m\cup \bar{K}))$ and $V=V'$. $U$ is open since $U'$ and $X\setminus \bar{K}$ are open sets which are spliced together at a single level $m$, and $U'\cap L_m=X\setminus K=(X\setminus \bar{K})\cap L_m$. Clearly $K\subseteq V$, and $H\subseteq U$ since $H\cap \bar{K}=\emptyset$. $\square$

Now we can separate a subset of a single level from an arbitrary closed set. Suppose that $H,K$ are disjoint closed subsets of $X$. By taking only finite unions and intersections from the sets produced from Claim 1, we can show that there are disjoint open $U'_n,V'_n$ and disjoint open $U''_n,V''_n$, all subsets of $G_n$, so that
• $H\cap L_n\subseteq U'_n$, $K\cap G_n\subseteq V'_n$, and $L_n\setminus H\subseteq V'_n$.
• $H\cap G_n\subseteq U''_n$, $K\cap L_n\subseteq V''_n$, and $L_n\setminus K\subseteq U''_n$.
Finally, take $$U_n:=\bigcup_n (U'_n\setminus \bigcup_{k\le n}\mathrm{cl}(V''_k))$$ and $$V_n:=\bigcup_n (V''_n\setminus \bigcup_{k\le n}\mathrm{cl}(U'_k)).$$
These sets are clearly open, and disjoint by the subtraction done at each step.

For each $n$, $H\cap L_n\subseteq U'_n\setminus \bigcup_{k\le n}\mathrm{cl}(V''_k)$ and $K\cap L_n\subseteq V''_n\setminus \bigcup_{k\le n}\mathrm{cl}(U'_k)$. This is because $V^*_n:=V'_n\cup(X\setminus (G_n\cup H))$ and $U^*_n:=U''_n\cup(X\setminus (G_n\cup K))$ are open, by the splicing argument. Furthermore, we have $K\subseteq V^*_n$ and $H\subseteq U^*_n$ and $V^*_n\cap U'_n=U^*_n\cap V''_n =\emptyset$.

So we have reduced normality to the property $(*)$ of the filter system.

Remark: In fact, the proof shows that $X$ built from a system satisfying $(*)$ is hereditarily normal.

## 3 Non countable paracompactness

To get non-countable paracompactness, we need to show that every sequence of closed sets $\langle H_n:n<\omega\rangle$ with $H_n\subseteq G_n$ cannot have union $X$. The levels of each $F_n$ are small in the following sense. $H_n\cap L_n$ must be measure 0 according to $\mathcal{F}_\alpha$ (that is, the complement is in the filter) for each $\alpha\in\mathfrak{c}$. Otherwise, if $H_n\cap L_n$ is positive measure according to $\mathcal{F}_\alpha$, then $(n+1,\alpha)$ is in its closure, contradicting that $H_n\subseteq G_n$ closed. For each $n$, let us define the ideal $\mathcal{I}^1$ to be the ideal of sets which are measure 0 according to $\mathcal{F}_\alpha$ for every $\alpha\in\mathfrak{c}$.

Now if $n>0$, $H_n\cap L_{n-1}$ is measure 0 according to $\mathcal{F}_\alpha$ for all $\alpha$ except for a set in $\mathcal{I}^1$. Let $\mathcal{I}^2$ be the ideal of sets which have this property, and recursively define $\mathcal{I}^m$ for every $m<\omega$. Let $\mathcal{I}$ be the $\sigma$-complete ideal generated by $\bigcup_{m<\omega} \mathcal{I}^m$. For technical reasons, let $\mathcal{I}^0=\{\emptyset\}$.

We will be done if we can show that $\mathcal{I}$ is proper, since $\{\alpha:(0,\alpha)\in H_n\cap L_0\}\in\mathcal{I}$ is in $\mathcal{I}$.

Suppose otherwise. Then there is a partition of $\mathfrak{c}$ into countably many sets, each in $\mathcal{I}^m$ for some $m$. By partitioning further, we can assume for each class $C$ of the partition, $C\in \mathcal{I}_m$ is witnessed by a set $D\in \mathcal{I}_{m-1}$, so that $C\cap D=\emptyset$ and $C\in \mathcal{F}_\alpha$ for all $\alpha\not\in D$ (any set in $\mathcal{I}_m$ has such a $D$, and if $C\cap D$ is not empty, then it is in $\mathcal{I}_{m-1}$ and we can subtract off this part; repeat the procedure finitely many times to get the further partitioning). Now to each $\alpha\in C$, we can associate a set $A_\alpha \in \mathcal{F}_\alpha$ which is disjoint from $C$.

We want to ensure that there are no such partitions. So we want:

$(**)$ For any function $f:X\rightarrow \omega$ and any assignment $\alpha\mapsto A_\alpha\in \mathcal{F}_\alpha$, there are $\alpha\neq \beta$ so that $f(\alpha)=f(\beta)$ and $\beta\in A_\alpha$.

Then, $\alpha$ and $\beta$ will be in the same class $C$, but we will have $\beta\in A_\alpha$, contradicting that $A_\alpha$ was supposed to be disjoint from $C$.

We have reduced non countable paracompactness to $(**)$. Next time we will see how to get $(*)$ and $(**)$.

Note that in this part, $\mathfrak{c}$ here can be replaced by any infinite cardinal $\kappa$.

We will continue Balogh's proof in Part 2.

(These are notes from a seminar given at Bar-Ilan University on November 27, 2017.)