Balogh's small Dowker space, part 2

We are continuing the construction of Balogh's Dowker space of size continuum. From the paper, he says:

"The heart of the proof of Theorem 1.1 is the following combinatorial lemma."

 Lemma 1.2 [Balogh] Let $\lambda=2^{\mathfrak{c}}$, and let $\langle c_\xi:\xi<\lambda\rangle$ be a one-to-one enumeration of ${}^\mathfrak{c}2$. Then there is a sequence $\langle d_\xi:\xi<\lambda\rangle$ of functions $d_\xi:\mathfrak{c}\rightarrow 2$ in such a way that for every $g:\mathfrak{c}\rightarrow [\lambda]^{<\omega}$, $f:\mathfrak{c}\rightarrow \omega$ and $h:\mathfrak{c}\rightarrow [\mathfrak{c}]^{<\omega}$, there are $\alpha<\beta$ in $\mathfrak{c}$ such that $f(\alpha)=f(\beta)$, $\beta\not\in h(\alpha)$ and for every $\xi\in g(\alpha)$, $c_\xi(\alpha)=d_\xi(\beta)$.

1 Explanation

Quite a lot to take in! But let's break down how it works, given that we know we want to construct a system of filters satisfying $(*)$ and $(**)$ from last time.

First $(**)$: 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 $\beta\in A_\alpha$.

This is roughly what is expressed by Lemma 1.2. But we don't know how to build the filters yet.

We want to build them to satisfy $(*)$: For every $A\subseteq \mathfrak{c}$ there exists $B\subseteq \mathfrak{c}$ such that $B\in \mathcal{F}_\alpha \textrm{ if }\alpha\in A$ and $\mathfrak{c}\setminus B\in \mathcal{F}_\alpha \textrm{ if }\alpha\not\in A$.

Think of the sequence $\langle c_\xi:\xi<\lambda \rangle$ as enumerating the characteristic functions of all possible $A$. For each $A$, we will find a set $B$ as in $(*)$; its characteristic function is $d_\xi$.

We will use the $d_\xi$'s to generate our filters $\mathcal{F}_\alpha$. So for each $\xi$, $\{\beta:d_\xi(\beta)=1\}\in \mathcal{F}_\alpha$ if $c_\xi(\alpha)=1$ and $\{\beta:d_\xi(\beta)=0\}\in \mathcal{F}_\alpha$ if $c_\xi(\alpha)=0$.

What else has to be in the filter? We want to be able to take finite intersections of the sets given by $d_\xi$ or complement (depending on which side makes it into the filter). This is expressed by $g$. Also, the filter will contain all cofinite subsets of $\mathfrak{c}$, a necessary condition for neighborhood filters in a space where points are closed. This is expressed by $h$.

So Lemma 1.2 succintly expresses $(**)$ for filters satisfying $(*)$!

2 Proof of Lemma 1.2

Basic motivations

This section can be skipped if you are familiar with the use of elementary submodel arguments.

If we didn't care about the part involving the $d_\xi$'s the rest is very easy to arrange. By a standard argument, there is a closed unbounded set of $\beta<\mathfrak{c}$ closed under the function $h$; that is, for every $\alpha<\beta$, $h(\alpha)\subseteq \beta$. There is also a closed unbounded set of $\beta$ so that for every $n$ with $f^{-1}[n]$ unbounded, $f^{-1}[n]\cap \beta$ is unbounded in $\beta$. We will do similar, but more sophisticated arguments using the method of elementary submodels.

Let $\theta$ be a large enough regular cardinal ($(2^{2^{\mathfrak{c}}})^+$ suffices). One of the basic ideas of the elementary submodel method is that if $N\prec H(\theta)$, $N$ countable, and $\beta<\mathfrak{c}$ is greater than $\sup(N\cap \mathfrak{c})$ (an ordinal of countable cofinality), then formulas true of $\beta$ reflect to unboundedly many $\alpha$ in $N\cap\mathfrak{c}$. This is another way of stating what we did with clubs in the last paragraph.

But this idea is a bit more powerful. Suppose that we have $\beta$ and $N$ as above. Then there are unboundedly many $\alpha\in N\cap\mathfrak{c}$ which satisfy the desired properties for $f$ and $h$. Even more crucially, we can reflect some properties of $g(\beta)$, $d_\xi(\beta)$, and $c_\xi(\beta)$ down to these $\alpha$.

But we can't reflect all of those properties down, since there are $\lambda$ many $\xi$ to consider and not all of them will be in $N$, and also $g(\beta)$ is some subset of $\lambda$ that may not be in $N$. But we will reflect enough of this information down, given by what happens inside some smaller countable elementary submodel $M\in N$.

So we know what happens inside this $M$. But there is a part outside of $M$ as well. This, we will handle by knowing that there are many reflections---so many that we will be able to guess in advance what happens on at least one of them.

That's the basic idea, but we need to make sure the construction can be carried out in $\mathfrak{c}$ many steps, while we are constructing $\lambda$-many of these $d_\xi$. The construction outlined above depends in part on the choice of submodels, but not on the full information of $M$ and $N$---we shall see that there are only $\mathfrak{c}$ many choices for the crucial information here, and the construction is in some sense canonical. That is part of the magic of this technique.

Definition of the d's

We have the motivation in mind now. We will enumerate control triples $\langle (A_\beta,B_\beta,u_\beta):\beta<\mathfrak{c}\rangle$, which are ordered triples $(A,B,u)$ that satisfy the following properties:

1.  $A\in [\mathfrak{c}]^\omega$, $B\in [{}^A2]^{\le \omega}$,
2. $u$ is a function with $\mathrm{dom}(u)\in[A]^\omega$,
3. for every $\alpha\in\mathrm{dom}(u), u(\alpha)\in [{}^A2\setminus B]^{<\omega}$,
4. (disjoint images) if $\alpha\neq \beta$ in $\mathrm{dom}(u)$, then $u(\alpha)\cap u(\beta)=\emptyset$.

Furthermore, let us ensure in the enumeration that $\beta>\sup A_\beta$.

From the sketch above, $A$ corresponds to $N\cap \mathfrak{c}$, $B$ corresponds to the information we need from $M$, and $u$ captures the information that's not in $M$. But there are only $\mathfrak{c}$ such triples!

Suppose now that $\xi<\lambda$. We will define $d_\xi$. For each $\beta<\mathfrak{c}$, there are three cases.

1. If $c_\xi\upharpoonright A_\beta\in B_\beta$, then let $d_\xi(\beta)=c_\xi(\beta)$. 
2. If $c_\xi\upharpoonright A_\beta\in u_\beta(\alpha)$ for some $\alpha\in \mathrm{dom}(u_\beta)$, then let $d_\xi(\beta)=c_\xi(\alpha)$.
3. Otherwise, set $d_\xi(\beta)=0$.

Intuitively, Case 1 corresponds to the case when $\xi$ is in the part which is controlled by $M$, and Case 2 to when $u$ captures the information outside of $M$ of a reflected $\alpha$. Note that in Case 2, $c_\xi\upharpoonright A_\beta\not\in B_\beta$, and $\alpha$ is chosen uniquely by the restrictions on $u$ in the definition of a control triple.

Final proof

We will show that the sequence of $d_\xi$ we have defined above works. Let $f,g,h$ be functions as in the statement of Lemma 1.2. We will produce the required $\alpha,\beta$.

Let $M\in N$ be elementary submodels of $H(\theta)$ containing $\langle c_\xi:\xi<\lambda\rangle,  \langle d_\xi:\xi<\lambda\rangle, f,g,h$. Let $A=\mathfrak{c}\cap N$ and $B=\{c_\xi\upharpoonright A:\xi\in \lambda\cap M\}$.

We will construct $u:A\rightarrow[{}^A2\setminus B]^{<\omega}$ satisfying the requirements of the control triple such that whenever $v\in N$ is an infinite partial function $\mathfrak{c}\rightarrow [\lambda\setminus M]^{<\omega}$ and $\alpha\neq \alpha'$ in $\mathrm{dom}(v)$ implies that $v(\alpha)\cap v(\alpha')=\emptyset$, then there is $\alpha\in\mathrm{dom}(u)\cap\mathrm{dom}(v)$ such that 

$$u(\alpha)=\{c_\xi\upharpoonright A:\xi\in v(\alpha)\}.$$

Basically, $u$ agrees with $v$ on at least one point of its domain for any $v$ of the right shape. And here, agreement actually means $u(\alpha)=\{c_\xi\upharpoonright A:\xi\in v(\alpha)\}$, since $u$ cannot actually take the ordinals less than $\lambda$ in its range, as we were careful to only use $\mathfrak{c}$-many control triples.

This $u$ is easy to construct. Just enumerate the countably many such $v\in N$ and construct $u^*:\mathfrak{c}\rightarrow [\lambda\setminus M]^{<\omega}$ so that the disjoint images property holds. Then get $u$ by taking restrictions of the $c_\xi$ to $A$. The only thing that requires some argument is to make sure that by taking the restrictions, we do not accidentally break the disjoint images property for $u$. But we will not, since the relevant $\xi$ are in $N$ and if $\xi\neq\xi'$ then this is witnessed by something in $A=N\cap\mathfrak{c}$. 

Now pick $\beta$ so that $\langle A_\beta,B_\beta,u_\beta\rangle=\langle A,B,u\rangle$. Say that $\gamma$ reflects $\beta$ if

1. $f(\gamma)=f(\beta)$,
2. $g(\gamma)\cap M=g(\beta)\cap M$,
3. for every $\xi\in g(\gamma)\cap M$, $c_\xi(\gamma)=c_\xi(\beta)$.

Find a maximal $D$ which consists of $\gamma$ reflecting $\beta$ so that $\langle g(\gamma):\gamma\in D\rangle$ forms a $\Delta$-system with root $r:=g(\beta)\cap M$. Choose $D$ in $M$, possible since the definition of reflection only used parameters in $M$. This $D$ is uncountable since otherwise it would be a subset of $M$, but $\beta$ could be then be added to it, contradicting maximality. So there is also an infinite set $H$ of $\gamma\in D\cap N$ so that $g(\gamma)\setminus r$ is disjoint from the countable set $\lambda\cap M$.

Now define $v:H\rightarrow [\lambda\setminus M]^{<\omega}$ by $v(\gamma)=g(\gamma)\setminus r$. So $v\in N$ and there is $\alpha\in \mathrm{dom}(u)\cap\mathrm{dom}(v)$ with $u(\alpha)=\{c_\xi\upharpoonright A:\xi\in v(\alpha)\}$. This is the $\alpha$ we want.

Let's check: since $\alpha$ reflects $\beta$, $f(\alpha)=f(\beta)$. Since $\alpha\in N$, $h(\alpha)\subseteq N$. But $\beta>\sup(A_\beta)$, so $\beta\not\in h(\alpha)$. Finally, if $\xi\in g(\alpha)$, there are two cases depending on if $\xi$ is in the root of the $\Delta$-system or not. In the first case, $\xi\in r$, and then we defined $d_\xi(\beta)=c_\xi(\beta)=c_\xi(\alpha)$. In the second, $c_\xi\upharpoonright A\in u_\beta(\alpha)$, so we directly defined $d_\xi(\beta)=c_\xi(\alpha)$.


(These are notes from a seminar given at Bar-Ilan University on December 4, 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 countable 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 $\sigma$-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 $\{\alpha:(0,\alpha)\in H_n\cap L_0\}\in\mathcal{I}$.

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 $\mathfrak{c}\setminus C\in \mathcal{F}_\alpha$ 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 $A_\alpha=\mathfrak{c}\setminus C\in \mathcal{F}_\alpha$.

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.)