Michael's theorem on paracompact spaces

Proof of Theorem from Proposition

Here is a proof sketch of the theorem from the proposition.

(1) ${\displaystyle \Rightarrow }$ (2) trivial. (2) ${\displaystyle \Rightarrow }$ (3) Since ${\displaystyle X}$ is regular, we can find an open cover ${\displaystyle {\mathcal {V}}=\{V_{x}\mid x\in X\}}$ such that ${\displaystyle \{{\overline {V_{x}}}\mid x\in X\}}$ is a refinement of a given open cover ${\displaystyle {\mathcal {U}}}$. By (2), ${\displaystyle {\mathcal {V}}}$ has a closure-preserving refinement ${\displaystyle {\mathcal {C}}}$. Then ${\displaystyle \{{\overline {C}}\mid C\in {\mathcal {C}}\}}$ is a refinement of ${\displaystyle {\mathcal {U}}}$ and is clearly closure-preserving.

(3) ${\displaystyle \Rightarrow }$ (4) by Proposition, and (4) ${\displaystyle \Rightarrow }$ (1) again by Proposition with the aid of Lemma 2 below. ${\displaystyle \square }$

Lemmas

The proof of the proposition uses the following general lemmas:

Proof: Given an open cover ${\displaystyle {\mathcal {U}}}$, let ${\displaystyle {\mathcal {A}}}$ be a locally finite closed refinement of it. By "locally finite", we can find an open cover ${\displaystyle \{V_{x}\mid x\in X\}}$ of ${\displaystyle X}$ such that each ${\displaystyle V_{x}}$ intersects only finitely many sets in ${\displaystyle {\mathcal {A}}}$. Then let ${\displaystyle {\mathcal {F}}}$ be a locally finite closed refinement of ${\displaystyle \{V_{x}\mid x\}}$. Then, for each ${\displaystyle A\in {\mathcal {A}}}$,

${\displaystyle W(A):=\cap \{X-F\mid F\in {\mathcal {F}},\,F\cap A=\emptyset \}}$

is an open set since ${\displaystyle {\mathcal {F}}}$ is locally finite; thus, closure-preserving. Let ${\displaystyle U:{\mathcal {A}}\to {\mathcal {U}}}$ be a function such that ${\displaystyle A\subset U(A)}$ and then let

${\displaystyle A'=U(A)\cap W(A),}$

which is open. Then ${\displaystyle \{A'\mid A\in {\mathcal {A}}\}}$ is a locally finite open refinement of ${\displaystyle {\mathcal {U}}}$ (use ${\displaystyle {\mathcal {F}}}$ is a locally finite cover). ${\displaystyle \square }$

Proof: Let ${\displaystyle {\mathcal {U}}=\cup _{1}^{\infty }{\mathcal {U}}_{n}}$ be an open cover that is a countable union of locally finite collections ${\displaystyle {\mathcal {U}}_{n}}$. Then let ${\displaystyle U_{n}=\cup {\mathcal {U}}_{n}}$, and then, for each ${\displaystyle n}$ and ${\displaystyle U\in {\mathcal {U}}_{n}}$, let

${\displaystyle C(n,U)=U-\cup _{k<n}U_{k}.}$

Then ${\displaystyle {\mathcal {C}}:=\{C(n,U)\mid n,U\}}$ is a locally-closed cover; indeed, if ${\displaystyle n}$ is the least integer such ${\displaystyle x\in U_{n}}$, then ${\displaystyle x}$ is in some ${\displaystyle U\in {\mathcal {U}}_{n}}$ and ${\displaystyle x\not \in U_{k},k<n}$. For "locally finite", let ${\displaystyle x}$ be in ${\displaystyle X}$; say, ${\displaystyle x\in C(n,U)}$ for some ${\displaystyle n,U}$. Since each ${\displaystyle {\mathcal {U}}_{k}}$ is locally finite, there is a neighborhood ${\displaystyle V}$ of ${\displaystyle x}$ such that ${\displaystyle V}$ intersects only finitely many sets in ${\displaystyle \cup _{k\leq n}{\mathcal {U}}_{k}}$. Then ${\displaystyle U\cap V}$ intersects only finitely many sets in ${\displaystyle {\mathcal {C}}}$, since ${\displaystyle U}$ is disjoint from the sets ${\displaystyle C(k,W),k>n}$. The "also" part is proved in essentially the same way; namely, given ${\displaystyle {\mathcal {V}}=\cup _{0}^{\infty }{\mathcal {V}}_{n}}$ and ${\displaystyle V_{n}=\cup {\mathcal {V}}_{n}}$, let

${\displaystyle D(n,V)={\overline {V}}-\cup _{k<n}V_{k}.}$

Then ${\displaystyle \{D(n,V)\mid n,V\}}$ is a required refinement. ${\displaystyle \square }$

Proof of Proposition

Let X be a T₁-space in which each open cover admits a closure-preserving closed refinement.

We first note that the assumption on open covers can be strengthen to the following^[9]

(*) For each open cover ${\displaystyle U_{i},i\in I}$, there exists a closure-preserving closed cover ${\displaystyle F_{i},i\in I}$ indexed by the same ${\displaystyle I}$ such that ${\displaystyle F_{i}\subset U_{i}.}$

Indeed, given such an open cover ${\displaystyle U_{i}}$, by assumption, we have a closure-preserving closed refinement ${\displaystyle {\mathcal {F}}}$ of it. For each ${\displaystyle i\in I}$, let

${\displaystyle G_{i}=\bigcup \{F\in {\mathcal {F}}\mid F\subset U_{i}\}.}$

Then this cover ${\displaystyle G_{i}}$ has required properties.

Second, we shall note that ${\displaystyle X}$ is normal.^[10] Indeed, given disjoint closed subsets ${\displaystyle F,G}$, the open cover ${\displaystyle \{X-F,X-G\}}$ has a closed refinement ${\displaystyle \{X-U,X-V\}}$; say, ${\displaystyle X-U\subset X-F}$ and then ${\displaystyle U,V}$ are disjoint neighborhoods of ${\displaystyle F,G}$. In particular, ${\displaystyle X}$ is Hausdorff since ${\displaystyle X}$ is T₁.

We shall now verify the hypothesis of Lemma 1. Let ${\displaystyle U_{a},a\in A}$ be an open cover indexed by a well-ordered set ${\displaystyle A}$. First, we shall recurviely construct a closure-preserving closed cover ${\displaystyle F_{a,n}}$ as follows. For ${\displaystyle n=1}$, let ${\displaystyle F_{a,1}\subset U_{a}}$ be a closed cover given by (*). Assuming we have constructed ${\displaystyle F_{a,n-1}}$, let

${\displaystyle U_{a,n}=U_{a}-\cup _{b<a}F_{b,n-1}.}$

It is an open cover: if ${\displaystyle a}$ is the least element with the property ${\displaystyle x\in F_{a,{n-1}}\subset U_{a}}$, then ${\displaystyle x\not \in F_{b,n-1},b<a}$ and so ${\displaystyle x\in U_{a,n}}$. By (*), let ${\displaystyle F_{a,n}\subset U_{a,n}}$ be a closure-preserving closed cover.

Now, since ${\displaystyle X=F_{a,n}\cup (\cup _{b\neq a}F_{b.n})}$, we have:

${\displaystyle V_{a,n}:=X-(\cup _{b\neq a}F_{b,n})\subset F_{a,n},}$

which is open since ${\displaystyle \{F_{a,n}\mid a,n\}}$ is closure-preserving. We claim ${\displaystyle \{V_{a,n}\mid a,n\}}$ is an open cover. Let ${\displaystyle x\in X}$ and for each ${\displaystyle n}$, let ${\displaystyle a_{n}}$ be the least element such that ${\displaystyle x\in F_{a_{n},n}}$. We have ${\displaystyle a_{1}\geq a_{2}\geq \cdots }$ since ${\displaystyle F_{a,n}\cap F_{b,n-1}=\emptyset ,b<a}$. Since ${\displaystyle A}$ is well-ordered, the sequence ${\displaystyle a_{n}}$ cannot be strictly decreasing all the way; i.e., ${\displaystyle a_{n-1}=a_{n}}$ for some ${\displaystyle n}$ and then similarly we see ${\displaystyle x\not \in F_{b,n},b\neq a_{n}.}$ That is, ${\displaystyle x}$ is in ${\displaystyle V_{a_{n},n}}$. Note also ${\displaystyle V_{a,n}\cap V_{b,n}=\emptyset ,b\neq a.}$

Finally, by (*), we can find a closure-preserving closed refinement ${\displaystyle D_{a,n}\subset V_{a,n}}$. By normality, we then find open sets ${\displaystyle W_{a,n}}$ such that

${\displaystyle D_{a,n}\subset W_{a,n}\subset {\overline {W_{a,n}}}\subset V_{a,n}.}$

For each ${\displaystyle n}$, let ${\displaystyle U_{n}}$ be the union of all open sets, each of which intersects only finitely many sets in ${\displaystyle \{W_{a,n}\mid a\}}$. It is a neighborhood of ${\displaystyle D_{n}:=\cup _{a}D_{a,n}}$ since ${\displaystyle \{W_{a,n}\mid a\}}$ is a disjoint collection. By normality, we have neighborhoods ${\displaystyle U'_{n}}$ of ${\displaystyle D_{n}}$ with closures lying in ${\displaystyle U_{n}}$. Then let

${\displaystyle W_{a,n}'=W_{a,n}\cap U_{n}'.}$

Then ${\displaystyle \{W_{a,n}'\mid a\}}$ is locally finite. Thus, by Lemma 2, we find a locally finite locally-closed refinement ${\displaystyle {\mathcal {C}}}$ of ${\displaystyle \{W_{a,n}'\mid a,n\}.}$ Then, since passing to closure doesn’t spoil “locally finite", ${\displaystyle \{{\overline {C}}\mid C\in {\mathcal {C}}\}}$ is a refinement satisfying the hypothesis of Lemma 1. (The argument after "finally" except Lemma 2 is essentially due to Dowkey.^[11]) ${\displaystyle \square }$