Michael's theorem on paracompact spaces

Template:Short description In mathematics, Michael's theorem gives sufficient conditions for a regular topological space (in fact, for a T₁-space) to be paracompact.

Statement

A family $E_i$ of subsets of a topological space is said to be closure-preserving if for every subfamily $E_{i_j}$,

$\overline{\bigcup E_{i_j}}=\bigcup \overline{E_{i_j}}$.

For example, a locally finite family of subsets has this property. With this terminology, the theorem states:

Frequently, the theorem is stated in the following form:

In particular, a regular-Hausdorff Lindelöf space is paracompact. The proof of the theorem uses the following result which does not need regularity:

Proof of Theorem from Proposition

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

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

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

Examples

Here are some examples of refinements satisfying the condition of the corollary.

• A countable open refinement trivially satisfies the condition of the corollary. Thus, a regular (Hausdorff) Lindelöf space is paracompact.
• Let $X$ be a metric space and $U_a,a\in A$ an open cover indexed by a well-ordered set $A$. Then let

  $U_{a,n}=\{x\mid d(x,X-U_a)>2^{-n}\},$

  $T_{a,n}=U_{a,n}\cap ({\cap }_{b<a}(X-U_b)).$

Then $d(T_{a,n},T_{b,n})\ge 2^{-n}$ for $b\ne a$ and, from that, for each $n$, one can choose neighborhoods $V_{a,n}\supset T_{a,n}$ inside $U_{a,n}$ such that $\{V_{a,n}\mid a\}$ is locally finite. Then $\{V_{a,n}\mid a,n\}$ is a refinement satisfying the condition of the corollary, and we conclude $X$ is paracompact.

Proof

Lemmas

The proof of the proposition uses the following general lemmas:

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

$W(A):=\cap \{X-F\mid F\in ℱ,F\cap A=\mathrm{\varnothing }\}$

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

$A^{\prime }=U(A)\cap W(A),$

which is open. Then $\{A^{\prime }\mid A\in 𝒜\}$ is a locally finite open refinement of $𝒰$ (use $ℱ$ is a locally finite cover). $◻$

Proof: Let $𝒰={\cup }_1^{\mathrm{\infty }}{𝒰}_n$ be an open cover that is a countable union of locally finite collections ${𝒰}_n$. Then let $U_n=\cup {𝒰}_n$, and then, for each $n$ and $U\in {𝒰}_n$, let

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

Then $𝒞:=\{C(n,U)\mid n,U\}$ is a locally-closed cover; indeed, if $n$ is the least integer such $x\in U_n$, then $x$ is in some $U\in {𝒰}_n$ and $x\in ̸U_k,k<n$. For "locally finite", let $x$ be in $X$; say, $x\in C(n,U)$ for some $n,U$. Since each ${𝒰}_k$ is locally finite, there is a neighborhood $V$ of $x$ such that $V$ intersects only finitely many sets in ${\cup }_{k\le n}{𝒰}_k$. Then $U\cap V$ intersects only finitely many sets in $𝒞$, since $U$ is disjoint from the sets $C(k,W),k>n$. The "also" part is proved in essentially the same way; namely, given $𝒱={\cup }_0^{\mathrm{\infty }}{𝒱}_n$ and $V_n=\cup {𝒱}_n$, let

$D(n,V)=\bar{V}-{\cup }_{k<n}{V}_k.$

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

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

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

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

$G_i=\bigcup \{F\in ℱ\mid F\subset U_i\}.$

Then this cover $G_i$ has required properties.

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

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

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

It is an open cover: if $a$ is the least element with the property $x\in F_{a,n-1}\subset U_a$, then $x\in ̸F_{b,n-1},b<a$ and so $x\in U_{a,n}$. By (*), let $F_{a,n}\subset U_{a,n}$ be a closure-preserving closed cover.

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

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

which is open since $\{F_{a,n}\mid a,n\}$ is closure-preserving. We claim $\{V_{a,n}\mid a,n\}$ is an open cover. Let $x\in X$ and for each $n$, let $a_n$ be the least element such that $x\in F_{a_n,n}$. We have $a_1\ge a_2\ge \cdots$ since $F_{a,n}\cap F_{b,n-1}=\mathrm{\varnothing },b<a$. Since $A$ is well-ordered, the sequence $a_n$ cannot be strictly decreasing all the way; i.e., $a_{n-1}=a_n$ for some $n$ and then similarly we see $x\in ̸F_{b,n},b\ne a_n.$ That is, $x$ is in $V_{a_n,n}$. Note also $V_{a,n}\cap V_{b,n}=\mathrm{\varnothing },b\ne a.$

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

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

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

$W_{a^{\prime },n}=W_{a,n}\cap {U_n}^{\prime }.$

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

Similar result

There is a similar result due to A. H. Stone (which was actually used to prove a metric space is paracompact).

Given a set $X$ and a cover $𝒞$ of it, for each element $x\in X$, we write

$\mathrm{St}(x,𝒞)=\bigcup \{C\in 𝒞\mid x\in C\},$

and call it the star over $x$ with respect to $𝒞$. A collection $𝒟$ of subsets of $X$ is then called a barycentric refinement of $𝒞$ if $\{\mathrm{St}(x,𝒟)\mid x\in X\}$ is a refinement of $𝒞$. A barycentric refinement is in particular a refinement.

Lemma 1 is then strengthened to

• If each open cover has a locally finite closed refinement, then each open cover has a locally finite barycentric open refinement.

The proof of the other direction of the theorem uses the following idea: construct a sequence of successive barycentric refinements

${𝒰}_0\succ {𝒰}_1\succ {𝒰}_2\succ \cdots$

where ${𝒰}_0:=𝒰$ is the given open cover $𝒰$ and we wrote $\prec$ for a barycentric refinement. Very roughly, this puts us in a setup similar to the metric-space one discussed at Template:Section link and we finish the proof similarly (the actual details are more involved).

References

• E. Michael, A note on paracompact spaces, Proc. Amer. Math. Soc. vol. 4 (1953) pp. 831-838.
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Further reading

• Michael's Theorem in Ncatlab