## A.H. Stone’s Theorem I

A.H. Stone’s theorem says that a topological space is paracompact (and Hausdorff) if and only if it is fully normal. Today, I am going to give a brief introduction to the notions of paracompactness and full normality, prove some basic results, and then prove one direction of Stone’s theorem–that paracompact and Hausdorff ${\Rightarrow}$ fully normal. This direction is easier of the two. I don’t know if both directions are attributed to Stone, but in any case, he proves both directions in his 1948 article “Paracompactness and Product Spaces”. By the way, the Stone here is not the well-known Stone of the Stone-Weierstrass theorem or the Stone-Von Neumann theorem. According to Wikipedia, Arthur Harold“A.H.” Stone was a British mathematician and of no relation to the American mathematician Marshall Harvey “M.H.” Stone.

A topological space ${X}$ is paracompact if its Hausdorff (most authors don’t assume Hausdorff) and every open covering of ${X}$ has an open locally-finite refinement. A covering ${\mathcal{U}=\left\{U_{\alpha}\right\}}$ is locally finite if, for every point ${x\in X}$, there is a neighborhood ${U\ni x}$ which intersects finitely many sets in ${\mathcal{U}}$.

For any open covering ${\mathcal{W}=\left\{W_{\alpha}\right\}}$, the star refinement ${(S,\mathcal{W})}$ of a set ${S}$ is defined to be the subcovering ${\bigcup\left\{W_{\alpha}: W_{\alpha}\cap S\neq\emptyset\right\}}$. If ${S}$ is a singleton ${\left\{x\right\}}$, we denote the star refinement of the singleton by ${(x,\mathcal{W})}$. An open cover ${\mathcal{W}}$ is an open star refinement for an open cover ${\mathcal{U}}$ if, the collection of stars ${\left\{(x,\mathcal{W})\right\}}$ forms an open refinement of ${\mathcal{U}}$. We say that a topological space ${X}$ is fully normal if every open cover has an open star refinement.

Full normality is a stronger property than normality, as the following lemma shows.

Lemma 1 Every fully normal space ${X}$ is normal.

Proof: Let ${C_{1}, C_{2}}$ be disjoint closed sets in ${X}$. Define open sets ${U_{1}=X\setminus C_{2}, U_{2}=X\setminus C_{1}}$. Then ${\left\{U_{i}\right\}}$ form an open cover of ${X}$ By full normality, there exists an open star refinement ${\mathcal{W}=\left\{W_{\alpha}\right\}}$. Define open sets ${V_{1}}$ and ${V_{2}}$ by

$\displaystyle V_{1}=\bigcup_{x\in C_{1}}(x,\mathcal{W})\subset U_{1},\indent V_{2}=\bigcup_{x\in C_{2}}(x,\mathcal{W})\subset U_{2}$

I claim that ${V_{1}}$ and ${V_{2}}$ are disjoint. Indeed,
$\displaystyle x\in V_{1}\cap V_{2}\Rightarrow\exists W_{\alpha},W_{\beta}\in\mathcal{W}\text{ s.t. }W_{\alpha}\cap C_{1}\neq\emptyset,W_{\beta}\cap C_{2}\neq\emptyset\text{ and }x\in W_{\alpha}\cap W_{\beta}$

Since ${\mathcal{W}}$ is a star refinement, we have that ${(x,\mathcal{W})\subset U_{1}\Leftrightarrow (x,\mathcal{W})\cap C_{2}=\emptyset}$ or ${(x,\mathcal{W})\subset U_{2}\Leftrightarrow(x,\mathcal{W})\cap C_{1}=\emptyset}$. But ${W_{\alpha}\cup W_{\beta}\subset(x,\mathcal{W})}$, which implies that ${(x,\mathcal{W})\cap C_{1}\neq\emptyset}$ and ${(x,\mathcal{W})\cap C_{2}\neq\emptyset}$–this is a contradiction. $\Box$

Every normal space is Hausdorff (take the disjoint closed sets to be singletons), so we obtain the following corollary. Every fully normal space is Hausdorff.

There are plenty of fully normal spaces because every metric space or metrizable space is fully normal. There are examples of fully normal spaces which are not metrizable, such as the Sorgenfrey Line.

Proposition 2 Every metric space ${(X,d)}$ is fully normal.

Proof: Let ${\mathcal{U}=\left\{U_{\alpha}\right\}}$ be an open cover of ${X}$. Given any ${x\in X}$, there is some index ${\alpha=\alpha(x)}$ such that ${x\in U_{\alpha}}$. Since ${U_{\alpha}}$ is open, there exists some ${\varepsilon=\varepsilon(x,\alpha)>0}$ such that ${B_{\varepsilon}(x)\subset U_{\alpha}}$. Define an open cover ${\mathcal{V}=\left\{B_{\varepsilon/2}(x):x\in X\right\}}$. I claim that ${\mathcal{V}}$ is a star refinement of ${\mathcal{U}}$. Indeed, fix ${x\in X}$ and suppose that ${x\in B_{\varepsilon/2}(y)}$, where $y\in X$. Then by the triangle inequality,

$\displaystyle d(x,z)\leq d(x,y)+d(y,z)<\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon\Rightarrow z\in B_{\varepsilon}(x),\forall z\in B_{\varepsilon/2}(y)$

Hence, ${B_{\varepsilon/2}(y)\subset B_{\varepsilon}(x)\subset U_{\alpha}}$. Noting that

$\displaystyle(x,\mathcal{W})=\bigcup\left\{B_{\varepsilon/2}(y):x\in B_{\varepsilon/2}(y)\right\}$

completes the proof of the claim. $\Box$

So far, we’e proved some basic results about full normality. We’ll now prove a few basic results about paracompactness, which we will use to prove the main theorem of this post, paracompactness $\Rightarrow$ full normality.

Given an infinite collection of sets $\left\{E_{\alpha}\right\}$, it is generally not true that $\overline{\bigcup_{\alpha}E_{\alpha}}=\bigcup_{\alpha}\overline{E_{\alpha}}$; the right-hand side of the equality need not be closed. For example, consider the open intervals $E_{n}=(1/n,(n-1)/n)$, for ${n>1}$. Then ${\bigcup_{n}\overline{E_{n}}=[1/n,(n-1)/n]=(0,1)}$, whereas ${\overline{\bigcup_{n}E_{n}}=\overline{(0,1)}=[0,1]}$.

Lemma 3 If ${\left\{E_{\alpha}\right\}}$ is a locally finite collection, then ${\bigcup_{\alpha}\overline{E_{\alpha}}=\overline{\bigcup_{\alpha}E_{\alpha}}}$.

Proof: If ${x\in\bigcup_{\alpha}\overline{E_{\alpha}}}$. then ${x\in\overline{E_{\alpha}}}$ for some ${\alpha}$. For any open neighborhood ${U\ni x}$, ${\emptyset\neq U\cap E_{\alpha}\subset U\cap\bigcup_{\alpha}E_{\alpha}}$, which implies that ${x\in\overline{\bigcup_{\alpha}E_{\alpha}}}$. Note that this direction always holds, regardless of the assumptions we impose on ${\left\{E_{\alpha}\right\}}$.

For the reverse inclusion, let ${x\in\overline{\bigcup_{\alpha}E_{\alpha}}}$. We will show that ${x}$ belongs to the closure of finitely many ${E_{\alpha}}$. By local finiteness, there exists a neighborhood ${U_{x}\ni x}$ so that the set ${\left\{E_{\alpha}: U_{x}\cap E_{\alpha}\neq\emptyset\right\}}$ is finite. So for any open neighborhood ${V\ni x}$, ${V_{x}:=U_{x}\cap V\subset U_{x},V}$ is an open neighborhood containing ${x}$. Hence, the set ${\left\{E_{\alpha}: V_{x}\cap E_{\alpha}\neq\emptyset\right\}\subset\left\{E_{\alpha}:U\cap E_{\alpha}\neq\emptyset\right\}}$ is nonempty and finite. We conclude that

$\displaystyle x\in\bigcup\left\{\overline{E_{\alpha}}:U\cap E_{\alpha}\neq\emptyset\right\}\subset\bigcup_{\alpha}\overline{E_{\alpha}}$

$\Box$

As is the case with compact spaces, closed subspaces of paracompact spaces are also paracompact.

Proposition 4 Every closed subset of a paracompact space is paracompact.

Proof: Let ${S}$ be a paracompact space, and let ${C\subset S}$ be a closed subset. Let ${\mathcal{U}=\left\{U_{\alpha}:\alpha\in I\right\}}$ be an open cover of ${C}$. Then ${\mathcal{U}\cup\left\{S\setminus C\right\}}$ is an open cover of ${S}$. By paracompactness, there is a locally finite open refinement ${\mathcal{V}=\left\{V_{\alpha}:\alpha\in J\right\}}$. Any set in ${\mathcal{V}}$ satisfies ${V_{\alpha}\subset U_{\beta}}$, for some ${\beta\in I}$, or ${V_{\alpha}\subset S\setminus C}$. We don’t care about the sets that satisfy the latter condition, so consider the collection ${\mathcal{V}'=\left\{V_{\alpha}:\alpha\in J, V_{\alpha}\not\subset S\setminus C\right\}}$. ${\mathcal{V}'}$ is an open refinement of ${\mathcal{U}}$ which covers ${C}$; local finiteness follows from that of ${\mathcal{V}}$. $\Box$

If we can show that paracompact spaces are normal, then they a fortiori regular. However, we’ll need use regularity as a stepping stone to proving normality.

Lemma 5 Any paracompact space is regular.

Proof: Let ${X}$ be a paracompact space, let ${A\subset X}$ be closed, and let ${x_{0}\in X\setminus A}$. For each ${x\in A}$, there is an open neighborhood ${U_{x}\ni x}$ such that ${x_{0}\notin \overline{U_{x}}}$. This construction is possible since ${X}$ is Hausdorff. The sets ${\left\{U_{x}:x\in A\right\}\cup\left\{A^{c}\right\}}$ form an open cover of ${X}$ and therefore have a locally finite open refinement ${\left\{V_{\alpha}:\alpha\in I\right\}}$. Define an indexing set ${J=\left\{\alpha\in I: V_{\alpha}\not\subset A^{c}\right\}}$, and define an open set ${V=\bigcup_{\alpha\in J}V_{\alpha}\supset A}$. I claim that ${x_{0}\notin\overline{V}}$. Indeed, by Lemma 3, ${\overline{V}=\bigcup_{\alpha\in J}\overline{V_{\alpha}}}$, which contains ${x_{0}}$ in no set in the union. Hence, ${\overline{V}^{c}}$ is an open neighborhood of ${x_{0}}$ tautologically disjoint from ${V}$. Hence, ${U}$ and ${V=\overline{U}^{c}}$ are disjoint open sets containing ${A}$ and ${B}$, respectively. $\Box$

We use the preceding lemma to show that a paracompact space ${X}$ is normal.

Lemma 6 Every paracompact space is normal.

Proof: Let ${A}$ and ${B}$ be disjoint, closed subsets of ${X}$. Lemma 3 tells us that every ${a\in X}$ can be separated from by ${B}$ by disjoint open sets, which implies that there is an open neighborhood ${U_{a}\ni a}$ such that ${B\cap\overline{U_{a}}=\emptyset}$. Observe that ${\left\{U_{a}:a\in A\right\}\cup\left\{A^{c}\right\}}$ forms an open cover of ${X}$ and therefore has a locally finite open refinement ${\left\{V_{\alpha}:\alpha \in I\right\}}$. If ${J=\left\{\alpha\in I:V_{\alpha}\not\subset A^{c}\right\}}$, then ${\left\{V_{\alpha}:\alpha\in J\right\}}$ forms a locally finite open cover of ${A}$. Since, for every ${\alpha\in J}$, ${V_{\alpha}\subset U_{a}}$, for some ${a\in A}$, we have that ${B\cap\overline{V_{\alpha}}=\emptyset}$. Define an open set ${V:=\bigcup_{\alpha\in J}V_{\alpha}\supset A}$. By Lemma 2, ${\overline{V}=\bigcup_{\alpha\in J}\overline{V_{\alpha}}}$, which implies that ${\overline{V}\cap B=\emptyset}$, $\Box$

We now have all the tools ready to prove today’s main theorem. As mentioned in the introduction, I don’t know if Theorem 7 is attributed to A.H. Stone. A proof of the result is contained in his 1948 article, but given the difficulty of the proof relative to that of the converse, I would guess that it was proved earlier.

Theorem 7 Every paracompact space is fully normal.

Proof: Let ${S}$ be a paracompact space, and let ${\mathcal{U}=\left\{U_{\alpha}\right\}}$ be an open cover of ${S}$. Replacing ${\mathcal{U}}$ with a locally finite refinement, if necessary, we may assume that ${\mathcal{U}}$ is locally finite. We need to show that ${\mathcal{U}}$ has an open star refinement.

By the shrinking lemma, there exists, for each ${\alpha}$, an open set ${X_{\alpha}}$ such that ${\overline{X_{\alpha}}\subset U_{\alpha}}$ and ${\bigcup_{\alpha}X_{\alpha}=S}$. Since ${S}$ is locally finite, each ${x\in S}$ has an open neighborhood ${V_{x}}$ such that the set of indices ${A_{x}=\left\{\alpha:V_{x}\cap U_{\alpha}\neq\emptyset\right\}}$ is finite. Define subsets ${B_{x}=\left\{\alpha\in A_{x}:x\in U_{\alpha}\right\}}$ and ${C_{x}=\left\{\alpha\in A_{x}:x\notin\overline{X_{\alpha}}\right\}}$. Note that ${B_{x}}$ is nonempty, since ${\mathcal{U}}$ is an open cover. Also note that ${B_{x}\cup C_{x}=A_{x}}$, but ${B_{x}}$ and ${C_{x}}$ are not necessarily disjoint, for an element ${x}$ could belong to ${U_{\alpha}\setminus\overline{X_{\alpha}}}$.

The intuition for these definitions is that we seek an open cover ${\mathcal{W}}$ such that the stars ${(x,\mathcal{W})}$ are contained in some ${U_{\alpha}}$. Observe that ${\bigcap_{B_{x}}U_{\alpha}}$ and ${\bigcap_{C_{x}}S\setminus\overline{X_{\alpha}}}$ are both finite intersections of open sets containing ${x}$, hence are both open sets containing ${x}$. We want to use local finiteness, so if we interesect the intersections with ${V_{x}}$, we obtain an open neighborhood of ${x}$.

$\displaystyle W_{x}=V_{x}\cap\bigcap_{\alpha\in B_{x}}U_{\alpha}\cap\bigcap_{\alpha\in C_{x}}\left(S\setminus\overline{X_{\alpha}}\right)$

The collection ${\mathcal{W}=\left\{W_{x}:x\in S\right\}}$ forms an open cover ${S}$. Fix ${y\in S}$. Since ${\left\{X_{\alpha}\right\}}$ is an open cover, ${y\in X_{\beta}}$, for some ${\beta}$. If ${y\in W_{x}}$, for some ${x}$, then the intersection ${U_{\beta}\cap W_{x}\subset U_{\beta}\cap V_{x}}$ is nonempty and contains $y$, which implies that ${\beta\in B_{y}}$. Hence, ${W_{x}\subset U_{\beta}}$. $\Box$