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}


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

This entry was posted in math.GT and tagged , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s