## A.H. Stone’s Theorem II

Stone’s Theorem. Every fully normal ${T_{1}}$ space is paracompact.

In the first installment, we proved that paracompact and Hausdorff implies fully normal, the “easy direction” of Stone’s theorem. We now prove the converse, which I consider to be the “hard direction” of Stone’s theorem. In the case reader has forgotten, I have provided the definitions of paracompact and fully normal below, along with the relevant auxiliary definitions.

A topological space ${S}$ is paracompact if its Hausdorff and every open covering of ${S}$ has an open locally-finite refinement. A covering ${\mathcal{U}=\left\{U_{\alpha}\right\}}$ is locally finite if, for every point ${x\in S}$, there is a neighborhood ${U\ni x}$ which intersects finitely many sets in ${\mathcal{U}}$.

For any open cover ${\mathcal{W}=\left\{W_{\alpha}\right\}}$, the star ${(S,\mathcal{W})}$ of a set ${S}$ is defined to be the set ${\bigcup\left\{W_{\alpha}: W_{\alpha}\cap S\neq\emptyset\right\}}$–in words, ${(S,\mathcal{W})}$ is the union of all sets which intersect ${S}$. If ${S}$ is a singleton ${\left\{x\right\}}$, we denote the star of ${S}$ by ${(x,\mathcal{W})}$. An open cover ${\mathcal{W}}$ is an 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 a star refinement.

Proof of Hard Direction: Let ${S}$ be a fully normal ${T_{1}}$ space, and let ${\mathcal{U}=\left\{U_{\alpha}\right\}}$ be an open covering of ${S}$. We will construct a locally finite refinement of ${\mathcal{U}}$.

By full normality, there exists an open cover ${\mathcal{U}^{1}=\left\{U^{1}\right\}}$ which star-refines ${\mathcal{U}}$. By induction, we may construct a sequence of open covers ${\left\{\mathcal{U}^{n}\right\}_{n=1}^{\infty}}$ such that

$\displaystyle\mathcal{U}^{n+1}\text{ star refines }\mathcal{U}^{n}$

For notational convenience, we denote the star of ${X\subset S}$ in ${\mathcal{U}^{n}}$ by ${(X,n)}$, and define

$\displaystyle (X,-n)=S\setminus\left(S\setminus X,n\right)\indent\forall n\in\mathbb{N}$

In the proof of Stone’s theorem, we will make frequent use of the following identities. Some of them are quite trivial; I encourage the reader to try to prove them on his or her own to become comfortable with the definitions of stars and star refinements.

Lemma 1

1. ${(X,-n)=\left\{x:(x,n)\subset X\right\}}$
2. ${((X,-n),n)\subset X}$
3. ${((X,n+1),n+1)\subset (X,n)}$
4. ${X\subset Y\Rightarrow (X,n)\subset (Y,n)}$
5. ${m\geq n\Rightarrow (X,m)\subset (X,n)}$
6. ${\overline{X}\subset (X,n)}$
7. ${y\in (x,n)\Leftrightarrow x\in (y,n)}$

Proof: 1.

$\displaystyle x\in (X,-n)\Leftrightarrow x\notin (S\setminus X,n)\Leftrightarrow x\notin\bigcup\left\{U_{\alpha}^{n}:(S\setminus X) \cap U_{\alpha}^{n}\neq\emptyset\right\}$

Since ${\mathcal{U}^{n}}$ is an open cover of ${S}$, there is some ${\alpha}$ such that ${x\in U_{\alpha}^{n}\subset X}$.

2. We use (1) to obtain ${((X,-n),n)=\left\{x:(x,n)\subset (X,-n)\right\}}$. Since ${(X,-n)\subset X}$, we conclude that ${(x,n)\subset X}$, where ${x}$ is an element in the aforementioned set.

3. If ${x\in ((X,n+1),n+1)}$, then ${x\in U_{\alpha}^{n+1}}$, for ${\alpha}$, where ${U_{\alpha}^{n+1}\cap (X,n+1)\neq\emptyset}$. Noting that

$\displaystyle U_{\alpha}^{n+1}\cap (X,n+1)\neq\emptyset\Leftrightarrow\exists \beta \text{ s.t. }U_{\beta}^{n+1}\cap X\neq\emptyset\text{ and }U_{\beta}^{n+1}\cap U_{\alpha}^{n+1}\neq\emptyset$,

let ${y\in U_{\beta}^{n+1}\cap U_{\alpha}^{n+1}}$ so that ${x\in (y,n+1)}$. By definition of star refinement, ${(y,n+1)\subset U_{\gamma}^{n}\in\mathcal{U}^{n}}$, for some ${\gamma}$. Since ${U_{\beta}^{n+1}}$ meets ${X}$ and is a subset of ${(y,n+1)}$, we conclude that ${U_{\gamma}^{n}}$ meets ${X}$; hence, ${x\in U_{\gamma}^{n}\subset (X,n)}$.

4. and 5. are obvious.

6. Observe that

$\displaystyle x\in\overline{X}\Rightarrow \forall U_{\alpha}^{n}\ni x, U_{\alpha}\cap X\neq\emptyset\Rightarrow x\in U_{\alpha}\subset(X,n)$

7. Observe that

$\begin{array}{lcl}\displaystyle y\in (x,n)=\bigcup\left\{U_{\alpha}^{n}: x\in U_{\alpha}^{n}\right\}&\Leftrightarrow&\exists\alpha\text{ s.t. }x,y\in U_{\alpha}^{n}\\[2 em]&\Leftrightarrow&x\in (y,n)=\bigcup\left\{U_{\alpha}^{n}: y\in U_{\alpha}^{n}\right\}\end{array}$

$\Box$

A property of star refinements is that if two sets $U_{\alpha}^{k+1},U_{\beta}^{k+1}\in\mathcal{U}^{k+1}$ intersect, then their the union is contained in some member of the collection $\mathcal{U}^{k}$. To see this, observe that if $x\in U_{\alpha}^{k+1}\cap U_{\beta}^{k+1}$, then both sets are contained in the star $(x,k+1)$. Since $\mathcal{U}^{k+1}$ star refines $\mathcal{U}^{k}$, we see that $(x,k+1)\subset U^{k}$ for some set $U^{k}\in\mathcal{U}^{k}$.

Invoking the Well-Ordering Theorem, we may assume that the elements of ${\mathcal{U}}$ are well-ordered.

For each ${\alpha}$, define a sequence of sets ${\left\{V_{\alpha}^{n}\right\}_{n=1}^{\infty}}$ by

$\displaystyle V_{\alpha}^{1}=(U_{\alpha},-1),\indent V_{\alpha}^{n}=(V_{\alpha}^{n-1},n) \ \forall n\geq 2$

It is evident that ${V_{\alpha}^{n}\subset V_{\alpha}^{n+1}}$, and ${V_{\alpha}^{n}}$ is open for ${n\geq 2}$. Hence, ${\bigcup_{n=1}^{\infty}V_{\alpha}^{n}=\bigcup_{n=2}^{\infty}V_{\alpha}^{n}}$ is open, being the union of open sets. I claim that ${(V_{\alpha}^{n},n)\subset U_{\alpha}}$. Indeed, ${V_{\alpha}^{1}\subset U_{\alpha}}$ by Lemma 1 identity (2). If ${(V_{\alpha}^{n},n)\subset U_{\alpha}}$, then

$\displaystyle (V_{\alpha}^{n+1},n+1)=\left((V_{\alpha}^{n},n+1),n+1\right)\subset (V_{\alpha}^{n},n)$

by Lemma 1 identity (3) and the definition of ${V_{\alpha}^{n+1}}$. Induction completes the proof of the claim. Since ${V_{\alpha}^{n}\subset (V_{\alpha}^{n},n)}$, we also obtain ${V_{\alpha}^{n}\subset U_{\alpha}}$ for all ${n}$; whence, ${V_{\alpha}\subset U_{\alpha}}$.

I claim that ${\bigcup V_{\alpha}=S}$. It suffices to prove that ${S\subset\bigcup V_{\alpha}}$. For ${x\in S}$, the star ${(x,1)\subset U_{\alpha}}$, for some ${\alpha}$, by the star-refinement construction. Identity (1) in Lemma 1 implies that ${x\in V_{\alpha}^{1}\subset V_{\alpha}}$. Furthermore, I claim that for ${x\in V_{\alpha}}$, there exists ${n>0}$ such that ${(x,n)\subset V_{\alpha}}$. Indeed, by definition of ${V_{\alpha}}$, there exists ${n\geq 2}$ such that ${x\in V_{\alpha}^{n-1}}$. It follows from the definition of ${V_{\alpha}^{n}}$ that ${(x,n)\subset V_{\alpha}^{n}\subset V_{\alpha}}$.

We define a transfinite sequence of closed sets ${H_{n\alpha}}$ by

$\displaystyle H_{n1}=(V_{1},-n),\indent H_{n\alpha}=\left(V_{\alpha}\setminus\bigcup_{\beta<\alpha}H_{n\beta},-n\right)$

where we make use of the well ordering. The motivation for this definition is that we don’t want a set ${U^{n}\in\mathcal{U}^{n}}$ to intersect both ${H_{n\alpha}}$ and ${H_{n\gamma}}$ when ${\alpha\neq\gamma}$. To see that we succeed, suppose that if ${\gamma<\alpha}$ and ${x\in U^{n}\cap H_{n\alpha}\neq\emptyset}$. Then ${U^{n}\subset (x,n)\subset V_{\alpha}\setminus\bigcup_{\beta<\alpha}H_{n\beta}}$, which is disjoint from ${H_{n\gamma}}$.

I claim that ${\bigcup_{n,\alpha}H_{n\alpha}=S}$. It suffices to prove the inclusion ${S\subset\bigcup_{n,\alpha}H_{n\alpha}}$. Since the set ${\left\{V_{\beta}: x\in V_{\beta}\right\}}$ is nonempty, our well ordering gives the existence of minimal element ${\alpha}$ such that ${x\in V_{\alpha}}$. An earlier result shows that there exists an ${n>0}$ such that ${(x,n)\subset V_{\alpha}}$. I claim that ${x\in H_{\alpha n}}$. If not, then there exists ${y\in (x,n)}$ such that ${y\notin V_{\alpha}\setminus\bigcup_{\beta<\alpha}H_{n\beta}}$, which implies that ${y\in H_{n\beta}}$, for some ${\beta<\alpha}$. But then by identity (3) of Lemma 1

$\displaystyle x\in(H_{n\beta},n)\subset\left((V_{\beta},-n),n\right)\subset V_{\beta}$

which contradicts the minimality of ${\alpha}$. This completes the proof of the claim.

Define sets ${E_{n\alpha}}$ and ${G_{n\alpha}}$ by

$\displaystyle E_{n\alpha}=(H_{n\alpha},n+3),\indent G_{n\alpha}=(H_{n\alpha},n+2)$

We have the trivial inclusions ${H_{n\alpha}\subset E_{n\alpha}\subset\overline{E_{n\alpha}}\subset G_{n\alpha}}$. After much head banging against the wall, I claim that no $U^{n+2}\in\mathcal{U}^{n+2}$ intersects both $G_{n\alpha}$ and $G_{n\gamma}$ when $\alpha\neq\gamma$. Indeed, suppose $U_{\alpha}^{n+2}\cap G_{n\alpha}\neq\emptyset$ and $U_{\alpha}^{n+2}\cap G_{n\gamma}\neq\emptyset$, where $\alpha\neq\gamma$, then there are sets $U_{\beta_{1}}^{n+2}, U_{\beta_{2}}^{n+2}$ nontrivially intersecting $U_{\alpha}^{n+2}, H_{n\alpha}$ and $U_{\gamma}^{n+2}, H_{n\gamma}$, respectively. Recall the observation that if $U_{\alpha}^{k+1}\cap U_{\beta}^{k+1}\neq\emptyset$, then the union $U_{\alpha}^{k+1}\cup U_{\beta}^{k+1}$ is contained in some set $U^{k}\in\mathcal{U}^{k}$. Therefore there are sets $U_{\alpha'}^{n+1}\supset U_{\alpha}^{n+2}\cup U_{\beta_{1}}^{n+2}$ and $U_{\gamma'}^{n+1}\supset U_{\alpha}^{n+2}\cup U_{\beta_{2}}^{n+2}$. Since $U_{\alpha}^{n+2}\subset U_{\alpha'}^{n+1}\cap U_{\gamma'}^{n+1}$, we obtain that there is a set $U^{n}\in\mathcal{U}^{n}$ containing their union. But by construction, $U^{n+2}\cap H_{n\alpha}$ and $U^{n+2}\cap H_{n\gamma}$ are both nonempty, which is a contradiction.

I claim that ${F_{n}=\bigcup_{\alpha}\overline{E_{n\alpha}}}$ is closed, despite being the possibly uncountable union of closed sets. If ${x\in\overline{F_{n}}}$, then any open neighborhood ${N\ni x}$ intersects ${F_{n}\Leftrightarrow}$ ${N}$ intersects ${E_{n\alpha}}$ for some ${\alpha}$. Be careful: if we change the neighborhood ${N}$, the index ${\alpha}$ may change, so we can’t yet conclude that ${x\in\overline{E_{n\alpha}}\subset F_{n}}$; we need to show that the ${\alpha}$ is unique. Replacing ${N}$ with a smaller neighborhood, if necessary, we may assume that ${N\subset (x,n+2)}$. But then since no set in ${\mathcal{U}^{n+2}}$ meets two distinct ${G_{(n+2)\beta}, G_{(n+2)\gamma}}$, the neighborhood ${N}$ can intersect at most one set ${E_{n\alpha}}$. Hence, ${x\in\overline{E_{n\alpha}}\subset F_{n}}$.

Define open sets ${W_{n\alpha}}$ by

$\displaystyle W_{1\alpha}=G_{1\alpha},\indent W_{n\alpha}=G_{n\alpha}\setminus\bigcup_{i=1}^{n-1}F_{i}\indent\forall n>1$

We complete the proof of Stone’s theorem by showing that ${\mathcal{W}=\left\{W_{n\alpha}:n,\alpha\right\}}$ is the desired locally finite refinement.

First, I claim that ${\bigcup_{n,\alpha}W_{n\alpha}=S}$. It suffices to prove the inclusion ${S\subset\bigcup_{n,\alpha}W_{n\alpha}}$. If ${x\in S}$, then there exists ${n>0}$ and ${\alpha}$ such that ${x\in H_{n\alpha}\subset\overline{E_{n\alpha}}}$. Taking ${n}$ smaller, if necessary, we may assume that ${n}$ is the minimal integer ${m}$ such that ${x\in\overline{E_{m\beta}}}$. Then ${x\in G_{n\alpha}}$ and ${x\notin\bigcup_{i=1}^{n-1}F_{i}}$, which implies that ${x\in W_{n\alpha}}$.

Second, I claim that ${\mathcal{W}}$ refines, though not necessarily star refines, ${\mathcal{U}}$. Indeed, by Lemma 1,

$\displaystyle W_{n\alpha}\subset G_{n\alpha}=\left(H_{n\alpha},n+2\right)\subset\left(H_{n\alpha},n\right)\subset\left((V_{\alpha},-n),n\right)\subset V_{\alpha}\subset U_{\alpha}$

Third, I claim that ${\mathcal{W}}$ is locally finite. For ${x\in S}$, we know that ${x\in H_{n\alpha}}$ for some ${n>0}$ and some ${\alpha}$. Hence, ${(x,n+3)\subset E_{n\alpha}\subset F_{n}}$, which satisfies ${F_{n}\cap W_{k\beta}=\emptyset}$ for any ${k>n}$. Observe that if ${k\leq n}$, then the star refinement assumption of the ${\left\{\mathcal{U}^{i}\right\}_{i=1}^{\infty}}$ implies that for any ${\alpha}$,

$\displaystyle (x,n+3)\subset U_{\alpha}^{n+2}\subset U_{\beta}^{k+2}$

for some ${\beta}$. If ${(x,n+3)}$ meets ${W_{k\beta}}$ and ${W_{k\gamma}}$, where ${\beta\neq\gamma}$, then ${U_{\beta}^{k+2}}$ meets both ${G_{k\beta}}$ and ${G_{k\gamma}}$, which is a contradiction. We conclude that ${(x,n+3)}$ is an open neighborhood of ${x}$ which at most intersects ${W_{1\beta},\ldots,W_{n\beta}}$. $\Box$