## General Shrinking Lemma for Normal Spaces

I’ve spent the last week or so thinking a lot about paracompact and fully normal topological spaces. The impetus for this effort is the Bing Metrization Theorem, proved by R.H. Bing in the 1951 article “Metrization of Topological Spaces”. In the proof of one the theorems–I forget which–Bing references the 1948 article “Paracompactness and Product Spaces” by A.H. Stone for the proof of a result that Bing uses. Stone’s article is of some importance to General Topology, for Stone proves his theorem that a Hausdorff space is paracompact if and only if it is fully normal. I am not going to prove Stone’s theorem today–this post will be long enough without it. Instead, I want to prove a technical “shrinking lemma” for normal spaces.

Recall that a topological space ${X}$ is normal if for disjoint closed sets ${A}$ and ${B}$, there exist disjoint open sets ${U,V}$ such that ${A\subset U}$ and ${B\subset V}$. Some authors say that the set ${U}$ and ${V}$ separate ${A}$ and ${B}$, or that ${A}$ and ${B}$ are separated. An equivalent definition of normality is stated in the following lemma:

Lemma 1 A topological space ${X}$ is normal if and only if given a closed set ${A}$ and an open set ${U\supset A}$, there exists an open set ${V}$ such that ${A\subset V\subset\overline{V}\subset U}$.

Quick note on notation: I use to ${\overline{A}}$ to denote the topological closure in ${X}$ of a subset ${A}$. Proof: Suppose ${X}$ is normal. ${A}$ and ${U^{c}}$ are disjoint closed sets, hence separated by disjoint open sets ${V}$ and ${W}$, respectively. Observe that ${W^{c}}$ is a closed set containing ${V}$, hence ${\overline{V}\subset W^{c}}$

For the converse, let ${A}$ and ${B}$ be disjoint closed sets. ${B^{c}}$ is an open set containing ${A}$, hence there is an open ${V}$ such that ${A\subset V\subset\overline{V}\subset B^{c}}$. Thus, ${A}$ and ${B}$ are separated by open sets ${V}$ and ${\overline{V}^{c}}$, respectively. $\Box$

Being able to separate disjoint, closed sets is a nice property of a space, especially when dealing with open covers. We can use normality to take an existing cover ${\mathcal{U}}$ and find a refinement ${\mathcal{V}}$ of the same cardinality, such the closure of each element ${V_{\alpha}}$ is contained in the corresponding ${U_{\alpha}}$. For finite collections, we can perform this refinement on any open cover. For countably infinite collections, we’ll need an technical hypothesis. And for uncountably infinite collections, we’ll need the Axiom of Choice.

Lemma 2 Suppose ${U}$ and ${V}$ form an open cover of a normal space ${X}$. Then there exists an open set ${W}$ such that ${\overline{W}\subset U}$ and ${\left\{W,V\right\}}$ is an open cover of ${X}$.

Proof: ${U^{c}}$ and ${V^{c}}$ are disjoint, closed sets. Since ${X}$ is normal, there exist disjoint open sets ${W'}$ and ${W}$ containing ${U^{c}}$ and ${V^{c}}$, respectively. Observe that ${x\notin V\Rightarrow x\in W}$, so ${\left\{W,V\right\}}$ is an open cover. Furthermore, ${W\subset X\setminus W'\subset U}$, which implies ${\overline{W}\subset U}$ since the set in the middle inclusion is closed. $\Box$

We use the preceding lemma together with induction to extend the result to an arbitrary finite open cover of ${X}$.

Lemma 3 Suppose ${\left\{U_{i}\right\}_{i=1}^{n}}$ is an open cover of a normal space ${X}$. Then there exists an open cover ${\left\{W_{i}\right\}_{i=1}^{n}}$ such that ${\overline{W_{i}}\subset U_{i}}$ for each ${i}$.

Proof: Set ${V^{1}=\bigcup_{i=1}^{n-1}U_{i}}$. Then ${\left\{U,V^{1}\right\}}$ forms an open cover of ${X}$. By Lemma 2, there exists an open set ${W_{1}}$ such that ${\left\{W_{1},V^{1}\right\}}$ and ${\overline{W_{1}}\subset U_{1}}$. It follows trivially that ${\left\{W_{1},U_{2},\ldots,U_{n}\right\}}$ is an open cover of ${X}$.

Suppose we have an open cover ${\left\{W_{1},\ldots,W_{k-1},U_{k},\cdots,U_{n}\right\}}$ such that ${\overline{W_{i}}\subset U_{i}}$, for ${i=1,\ldots,k-1}$. Set ${V^{k}=\bigcup_{i=1}^{k-1}W_{i}\cup\bigcup_{i=k+1}^{n}U_{i}}$. Then ${\left\{U_{k},V^{k}\right\}}$ forms an open cover of ${X}$. By Lemma 2, there exists an open set ${W_{k}}$ such that ${\left\{W_{k},V^{k}\right\}}$ is an open cover of ${X}$ and ${\overline{W_{k}}\subset U_{k}}$. It follows that ${\left\{W_{1},\ldots,W_{k},U_{k+1},\ldots,U_{n}\right\}}$ is an open cover of ${X}$. $\Box$

You may think that the argument used in the preceding proof holds for any countable open cover of a normal space, since the arbitrary union of open sets is open; I naively did after finishing the proof. But you would be wrong. Suppose that you have a countable open cover ${\left\{U_{i}\right\}_{i=1}^{\infty}}$ of a normal space ${X}$. The preceding lemma tells us that for any ${n\in\mathbb{N}}$, there exist open sets ${W_{1},\ldots,W_{n}}$ such that ${\overline{W_{i}}\subset U_{i}}$ and ${\left\{W_{i}\right\}_{i=1}^{n}\cup\left\{U_{i}\right\}_{i=n+1}^{\infty}}$ is an open cover of ${X}$. The problem is that it’s not clear that ${\bigcup_{i=1}^{\infty}W_{i}=X}$, and therefore ${\left\{W_{i}\right\}_{i=1}^{\infty}}$ is an open cover of ${X}$. All we know is that at some ${n}$, if ${x\notin\bigcup_{i=1}^{n}W_{i}}$, then ${x\in U_{m}}$, for some ${m\geq n+1}$.

It turns out that, assuming the Axiom of Choice, there does exist a normal space and a countable open cover which has no refinement as described in the shrinking lemma. Amer Beslagic made this observation in the 1985 article “A Dowker Product”. The example is a Dowker space after the mathematician C.H. Dowker, and it is the only known example of a Dowker space. Mary Ellen Rudin, the wife of the late textbook author Walter Rudin, constructed itin the 1971 article “A normal space ${X}$ for which ${X\times I}$ is not normal”. Many thanks to Henno Brandsma for pointing this out in his thoughtful response to a Math.StackExchange question.

If we knew that ${x}$ only belonged to finitely many of the ${U_{i}}$, then there would be index ${m}$ such that ${x\in W_{m}}$. So, if we require that ${\left\{U_{i}\right\}_{i=1}^{\infty}}$ be point-finite, meaning any ${x\in X}$ belongs to ${U_{i}}$ for only finitely many ${i}$, then we obtain the following result. The proof of this lemma is actually Exercise 4 in Chapter 4, Section 36 of James Munkres’s Topology.

Lemma 4 Suppose ${\left\{U_{i}\right\}_{i=1}^{\infty}}$ is a point-finite open cover of a normal space ${X}$. Then there exists an open cover ${\left\{W_{i}\right\}_{i=1}^{\infty}}$ such that ${\overline{W_{i}}\subset U_{i}}$.

To extend the lemma to possibly uncountable open covers, we’ll need the Axiom of Choice in the form of Zorn’s Lemma. An alternative proof also uses the Well-Ordering Theorem, which is equivalent to the Axiom of Choice, together with transfinite induction. It can be found here.

Lemma 5 Suppose ${\left\{U_{\alpha}\right\}_{\alpha\in I}}$ is a point-finite ${I}$-indexed open cover of a normal space ${X}$. Then there exists an open cover ${\left\{W_{\alpha}\right\}_{\alpha\in I}}$ such that ${\overline{W_{\alpha}}\subset U_{\alpha}}$.

Proof: We will use Zorn’s lemma to prove the existence of the open cover ${\left\{W_{\alpha}\right\}}$. Let ${{S}}$ denote the set of collections ${\mathcal{V}_{J}=\left\{V_{\alpha,J}\right\}}$ of open sets in ${X}$ of the following form: for a subset of indices ${J\subset I}$, ${\overline{V_{\alpha,J}}\subset U_{\alpha}}$ for all ${\alpha \in J}$, ${V_{\alpha,J}=U_{\alpha}}$ for all ${\alpha\in I\setminus J}$, and ${\mathcal{V}_{J}}$ is an open cover of ${X}$. Note that ${{S}}$ is nonempty by hypothesis that ${X}$ is normal and the argument used for finitely many indices. We define a partial order ${\leq}$ on ${{S}}$ by ${\mathcal{V}_{J}\leq\mathcal{V}_{K}}$ if and only if ${J\subset K}$ and ${V_{\alpha,J}=V_{\alpha,K}}$ for each ${\alpha\in J}$.

I claim that every totally ordered subset ${T}$ has an upper bound. Indeed, define an indexing set ${K}$ by

$\displaystyle K=\bigcup\left\{J:\mathcal{V}_{J}\in T\right\}$

We define a collection of open sets ${\mathcal{V}_{K}}$ as follows:

$\displaystyle \forall \alpha \in K,\indent V_{\alpha,K}=V_{\alpha,J}\text{ for some set }J\ni \alpha$

Since ${S}$ is totally ordered, ${V_{\alpha,K}}$ is well-defined and satisfies ${\overline{V_{\alpha,K}}\subset U_{\alpha}}$. We need to show that ${\mathcal{V}_{K}}$ is an open cover of ${X}$ in order to prove that ${\mathcal{V}_{K}\in S}$. We have already considered the case where ${K}$ is finite, so assume ${K}$ is infinite. Since ${\left\{U_{\alpha}\right\}}$ is point-finite, a fortiori ${\mathcal{V}_{K}}$ is point-finite. Hence, given ${x\in X}$, there exists a finite set of indices ${\left\{\alpha_{1},\ldots,\alpha_{k}\right\}}$ such that for any ${\mathcal{V}_{J}\in T}$ with ${J\supset\left\{\alpha_{1},\ldots,\alpha_{k}\right\}}$, ${x\in\bigcup_{\alpha \in J}V_{\alpha,J}}$.

By Zorn’s lemma, ${S}$ has a maximal element ${\mathcal{W}=\left\{W_{\alpha}\right\}}$. The indexing set of ${\mathcal{W}}$ must be ${I}$, otherwise we can use the normality of ${X}$ to enlarge the indexing set by one additional element, contradicting the maximality of ${\mathcal{W}}$. Tautologically, ${\mathcal{W}}$ is the desired open cover. $\Box$

1. A. Bešlagić, A Dowker Product, Trans. Amer. Math. Soc. 292 (1985), 519-530.
2. J.R. Munkres, Topology: A First Course, Prentice-Hall, 1975.
3. M. Rudin, A normal space X for which X× I is not normal, Fundamenta Mathematicae 73 (1971), 179-186.