Today, I shall use the theorem of Michael discussed earlier to prove that a metric space is paracompact.

Theorem 1 (Stone) A metric space is paracompact.

This theorem seems to use the axiom of choice, or some version thereof, in all proofs.

1. Proof of Stone’s theorem

Suppose given a cover ${\mathfrak{A}=\left\{U_\alpha\right\}}$ of the metric space ${X}$ (with metric ${d}$, say). We will show that there is a refinement of ${\mathfrak{A}}$ that can be decomposed into a countable collection of locally finite families. Thanks to Michael’s theorem, this will prove the result.
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Today I want to discuss an equivalent and seemingly weaker condition of paracompactness due to Ernest Michael (1953, Proc. of the AMS).

Theorem 1 (Michael) Suppose ${X}$ is regular and every open cover ${\mathfrak{A}}$ of ${X}$ has a refinement that can be decomposed into a countable collection of locally finite ${\mathfrak{A}_i}$ collections of open sets. Then ${X}$ is paracompact.

Note that the ${\mathfrak{A}_i}$‘s are not required to be covers, only locally finite! This is a significant strengthening of the usual definition of paracompactness.

Following Michael’s original paper, I shall discuss the proof of this result. First, I shall give an auxiliary result, of independent result, that yields yet another variation on the theme of paracompactness: we don’t have to require the locally finite refinements to be open.

I’ve been reviewing some basic general topology as of late. I will post some of this material here. Apologies to readers who prefer more advanced topics; my current focus is on foundational material.

Often, we’d like to prove that a given subset ${A \subset X}$ of a topological space ${X}$ has some given property, e.g. that it is open or closed. In many cases, however, the big space ${X}$ may not be easily understandable, but local pieces of it may be. For instance, ${X}$ might be a manifold, and we might not know what the global structure of ${X}$ is, but we do know that ${X}$ is locally homeomorphic (or diffeomorphic) to a ball in ${\mathbb{R}^n}$. So we need a way to go from local results to global results.

Proposition 1 Let ${\left\{U_\alpha\right\}}$ be an open cover of the topological space ${X}$. Suppose ${W \subset X}$ and ${W \cap U_{\alpha}}$ is open in ${U_{\alpha}}$ for each ${\alpha}$. Then ${W}$ is open in ${X}$.

So openness is a local property.

This is the easy result. Indeed, since ${U_{\alpha}}$ is an open set, each ${W \cap U_{\alpha} }$ is open in ${X}$. But $\displaystyle W = \bigcup W \cap U_{\alpha}$

so that ${W}$ is a union of open sets, hence open.

Similarly, we can deduce the corresponding result for closed sets:

Corollary 2 Suppose ${U_{\alpha}}$ is an open cover of ${X}$. Let ${W \subset X}$. Suppose ${W \cap U_{\alpha}}$ is closed in ${ U_{\alpha}}$ for each ${U_{\alpha}}$. Then ${W}$ is closed in ${X}$.

Indeed, this follows from the previous result, with ${W}$ replaced with ${X - W}$.

However, the analogous result is no longer true if we look at closed covers. Consider for instance the closed cover of ${\mathbb{R}^2}$ by vertical lines. The set ${W}$, defined as the intersection of the graph ${y=x}$ with the upper right quadrant, is not closed, though its intersection with each vertical line is closed (in fact, is a point). So we need something more. The problem, as we will see, is that there are too many lines. (more…)