# Dowker’s Theorem

Let $\mathbb{I}$ be the closed unit interval $[0,1]$. In 1951, Dowker proved that for normal spaces $X$, $X \times \mathbb{I}$ is normal if and only if $X$ is countably paracompact. A space $X$ is countably paracompact if every countable open cover of $X$ has a locally finite open refinement. Dowker’s theorem is a fundamental result on products of normal spaces. With this theorem, a question was raised about the existence of a normal but not countably paracompact space (such a space became known as a Dowker space). For a detailed discussion on Dowker spaces, see the survey paper [Rudin]. The focus here is on presenting a proof for the Dowker’s theorem, laying the groundwork for future discussion.

An open cover $\mathcal{V}=\lbrace{V_\alpha:\alpha \in S}\rbrace$ of a space $X$ is shrinkable if there exists an open cover$\mathcal{W}=\lbrace{W_\alpha:\alpha \in S}\rbrace$ such that $\overline{W_\alpha} \subset V_\alpha$ for each $\alpha \in S$. The open cover $\mathcal{W}$ is called a shrinking of $\mathcal{V}$. We use the following lemma in proving Dowker’s theorem. Go here to see a proof of this lemma.

Lemma
A space $X$ is normal if and only if every point-finite open cover of $X$ is shrinkable.

Dowker’s Theorem
For a normal space $X$, the following conditions are equivalent:

1. $X$ is a countably paracompact space.
2. If $\lbrace{U_n:n<\omega}\rbrace$ is an open cover of $X$, then there is a locally finite open refinement $\lbrace{V_n:n<\omega}\rbrace$ such that $\overline{V_n} \subset U_n$ for each $n$.
3. $X \times Y$ is normal for any compact metrizable space $Y$.
4. $X \times \mathbb{I}$ is normal.
5. For each sequence of closed sets such that $A_0 \supset A_1 \supset ...$ and $\cap_n A_n=\phi$, there exists open sets $B_n \supset A_n$ such that $\cap_n B_n=\phi$.

Proof
$1 \Longrightarrow 2$
Suppose $X$ is countably paracompact. Let $\mathcal{U}=\lbrace{U_n:n<\omega}\rbrace$ be an open cover of $X$. Then $\mathcal{U}$ has a locally open refinement $\mathcal{V}$. For each $n<\omega$, let $W_n=\bigcup \lbrace{V \in \mathcal{V}:V \subset U_n}\rbrace$. Note that $\lbrace{W_n:n<\omega}\rbrace$ is still locally finite (thus point-finite). By the lemma, it has a shrinking $\lbrace{V_n:n<\omega}\rbrace$.

$2 \Longrightarrow 3$
Let $Y$ be a compact metrizable space. Let $\lbrace{E_0,E_1,E_2,...}\rbrace$ be a base of $Y$ such that it is closed under finite unions. Let $H,K$ be two disjoint closed subsets of $X \times Y$. The goal here is to find an open set $\mathcal{V}$ such that $H \subset \mathcal{V}$ and $\overline{\mathcal{V}}$ misses $K$.

For each $x \in X$, consider the following:
$S_x=\lbrace{y \in Y: (x,y) \in H}\rbrace$
$T_x=\lbrace{y \in Y: (x,y) \in K}\rbrace$.

For each $n \in \omega$, let $O_n$ be defined by:
$O_n=\lbrace{x \in X:S_x \subset E_n}$ and $T_x \subset Y-\overline{E_n}\rbrace$.

Claim 1.
Each $x \in X$ is an element of some $O_n$.

Both $S_x$ and $T_x$ are compact. Thus we can find some $E_n$ such that $S_x \subset E_n$ and $T_x \subset Y-\overline{E_n}$.

Claim 2.
Each $O_n$ is open in $X$.

Fix $z \in O_n$. For each $y \in Y-E_n$, $(z,y) \notin H$. Choose open set $A_y \times B_y$ such that $(z,y) \in A_y \times B_y$ and $A_y \times B_y$ misses $H$. The set of all $B_y$ is a cover of $Y-E_n$. We can find a finite subcover covering $Y-E_n$, say $B_{y(0)},B_{y(1)},...,B_{y(k)}$. Let $A=\bigcap_i A_{y(i)}$.

For each $a \in \overline{E_n}$, $(z,a) \notin K$. Choose open set $C_a \times D_a$ such that $(z,a) \in C_a \times D_a$ and $C_a \times D_a$ misses $K$. The set of all $D_a$ is a cover of $\overline{E_n}$. We can find a finite subcover covering $\overline{E_n}$, say $D_{a(0)},D_{a(1)},...,D_{a(j)}$. Let $C=\bigcap_i C_{a(i)}$.

Now, let $O=A \cap C$. Clearly $z \in O$. To show that $O \subset O_n$, pick $x \in O$. We want to show $S_x \subset E_n$ and $T_x \subset Y-\overline{E_n}$. Suppose we have $y \in S_x$ and $y \in Y-E_n$. Then $y \in B_{y(i)}$ for some $i$. As a result, $(x,y) \in A_{y(i)} \times B_{y(i)}$. This would mean that $(x,y) \notin H$. But this contradicts with $y \in S_x$. So we have $S_x \subset E_n$. On the other hand, suppose we have $y \in T_x$ and $y \in \overline{E_n}$. Then $y \in D_{a(i)}$ for some $i$. As a result, $(x,y) \in C_{a(i)} \times D_{a(i)}$. This means that $(x,y) \notin K$. This contradicts with $y \in T_x$. So we have $T_x \subset Y-\overline{E_n}$. It follows that $O \subset O_n$ and $O_n$ is open in $X$.

Claim 3.
We can find an open set $\mathcal{V}$ such that $H \subset \mathcal{V}$ and $\overline{\mathcal{V}}$ misses $K$.

Let $\mathcal{O}=\bigcup_n O_n \times \overline{E_n}$. Note that $H \subset \mathcal{O}$ and $\mathcal{O}$ misses $K$. The open set $\mathcal{V}$ being constructed will satisfiy $H \subset \mathcal{V} \subset \overline{\mathcal{V}} \subset \mathcal{O}$. The open cover $\lbrace{O_n}\rbrace$ has a locally finite open refinement $\lbrace{V_n}\rbrace$ such that $\overline{V_n} \subset O_n$ for each $n$. Now define $\mathcal{V}=\bigcup_n V_n \times E_n$. Note that $H \subset \mathcal{V}$.

We also have $\overline{\mathcal{V}}=\overline{\bigcup_n V_n \times E_n}=\bigcup_n \overline{V_n \times E_n}$. We have a closure preserving situation because the open sets $V_n$ are from a locally finite collection. Continue the derivation and we have:

$=\bigcup_n \overline{V_n} \times \overline{E_n} \subset \bigcup_n O_n \times \overline{E_n}=\mathcal{O}$.

$3 \Longrightarrow 4$ is obvious.

$4 \Longrightarrow 5$
Let $A_0 \supset A_1 \supset ...$ be closed sets with empty intersection. We show that $A_n$ can be expanded by open sets $B_n$ such that $A_n \subset B_n$ and $\bigcap_n B_n=\phi$.

Choose $p \in \mathbb{I}$ and a sequence of distinct points $p_n \in \mathbb{I}$ converging to $p$. Let $H=\cup \lbrace{A_n \times \lbrace{p_n}\rbrace:n \in \omega}\rbrace$ and $K=X \times \lbrace{p}\rbrace$. These are disjoint closed sets. Since $X \times \mathbb{I}$ is normal, we can find open set $V \subset X \times \mathbb{I}$ such that $H \subset V$ and $\overline{V}$ misses $K$.

Let $B_n=\lbrace{x \in X:(x,p_n) \in V}\rbrace$. Note that each $B_n$ is open in $X$. Also, $A_n \subset B_n$. We want to show that $\bigcap_n B_n=\phi$. Let $x \in X$. The point $(x,p) \in K$ and $(x,p) \notin \overline{V}$. There exist open $E \subset X$ and open $F \subset Y$ such that $(x,p) \in E \times F$ and $(E \times F) \cap \overline{V}=\phi$. Then $p_n \in F$ for some $n$. Note that $(x,p_n) \notin V$. Thus $x \notin B_n$. It follows that $\bigcap_n B_n=\phi$.

$5 \Longrightarrow 1$
Let $\lbrace{T_n:n \in \omega}\rbrace$ be an open cover of $X$. Let $E_n=\bigcup_{i \leq n} T_i$. Let $A_n=X-E_n$. Each $A_n$ is closed and $\bigcap_n A_n=\phi$. So there exist open sets $B_n \supset A_n$ such that $\bigcap_n B_n=\phi$.

Since $X$ is normal, there exists open $W_n$ such that $A_n \subset W_n \subset \overline{W_n} \subset B_n$. It follows that $\bigcap_n \overline{W_n}=\phi$. Let $O_n=X-\bigcap_{i \leq n} \overline{W_n}$.

Claim. $\lbrace{O_n:n \in \omega}\rbrace$ is an open cover of $X$. Furthermore, $\overline{O_n} \subset \bigcup_{i \leq n}T_i$.
Let $x \in X$. Since $\bigcap_n B_n=\phi$, $x \notin B_n$ for some $n$. Then $x \notin \overline{W_n}$ and $x \in O_n$. To show the second half of the claim, let $y \notin \bigcup_{i \leq n}T_i$. Then $y \notin E_i$ for each $i \leq n$. This means $y \in A_i$ and $y \in W_i$ for each $i \leq n$. Then $\bigcap_{i \leq n}W_i$ is an open set containing $y$ that contains no point of $O_n$. Thus we have $\overline{O_n} \subset \bigcup_{i \leq n}T_i$.

Let $S_n=T_n-\bigcup_{i < n} \overline{O_i}$. We show that $\lbrace{S_n:n \in \omega}\rbrace$ is a locally finite open refinement of $\lbrace{T_n:n \in \omega}\rbrace$. Clearly $S_n \subset T_n$. To show that the open sets $S_n$ form a cover, let $x \in X$. Choose least $n$ such that $x \in T_n$. Then for each $i, $x \notin E_i$. This means $x \in A_i$ and $x \in \overline{W_i}$ for each $i. It follows that $x \notin O_i$ for each $i. This implies $x \in S_n$. To see that the open sets $S_n$ form a locally finite collection, note that each $x \in X$ belong to some $O_n$. The open set $O_n$ misses $S_m$ for all $m>n$. Thus $O_n$ can only meet $S_i$ for $i \leq n$. We just prove that $X$ is countably paracompact.

Note. In $4 \Longrightarrow 5$, all we need is that the factor $Y$ has a non-trivial convergent sequence. That is, if $X \times Y$ is normal and $Y$ has a non-trivial convergent sequence, then $X$ satisfies condition $5$.

Reference
[Dowker] Dowker, C. H. [1951], On Countably Paracompact Spaces, Canad. J. Math. 3, 219-224.

[Rudin] Rudin, M. E., [1984], Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 761-780.

____________________________________________________________________
$\copyright \ 2009-2016 \text{ by Dan Ma}$
Revised November 27, 2016