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.

First, suppose that we have a countable cover ${\left\{U_i\right\}}$. Then the idea is to consider the differences ${U_i - \bigcup_{j, which form a point-finite cover of ${X}$ (i.e. each point is contained in finitely many of the differences). Then, expand these slightly using the metric to make them open.

However, this naive approach as to be modified. First, we will have to generalize to arbitrary index sets, not just the natural numbers. Second, we need local finiteness, not just point-finiteness.

So, for starters, well-order the index set ${A}$ in which ${\alpha}$ takes values. This is where we use the axiom of choice; for separable metric spaces, this would not be necessary, since they have a countable basis, and every open cover can be replaced by a countable subcover.

Now we shrink the ${U_{\alpha}}$ slightly. Namely, we write

$\displaystyle V^n_{\alpha} = \left\{x: d(x, X - U_\alpha) \geq 2^{-n} \right\}.$

In other words, to say that a point belongs to ${V^n_\alpha}$ is to say that it belongs to ${U_\alpha}$ and is not too far from the boundary. Note that ${\bigcup_n V^n_\alpha = U_\alpha}$.

We define the sets

$\displaystyle W^n_\alpha = V^n_\alpha - \bigcup_{\beta > \alpha }V^{n+1}_{\beta}$

The point of this is to excise out redundancies when possible. Note that the ${W^n_\alpha}$ form a cover of ${X}$. Indeed, if ${x \in X}$, choose the smallest ${\alpha}$ with ${x \in U_\alpha}$. Then ${x \in W^n_\alpha}$ for ${n}$ sufficiently large. The good news is that we have excised out redundancies, but the bad news is that the ${W^n_\alpha}$ are not open. So set

$\displaystyle Z^n_\alpha = \left\{ x: d(x, W^n_\alpha)< 2^{-n-3} \right\}.$

These are small neighborhoods of the ${W^n_\alpha}$ and are consequently open. Moreover, the ${Z^n_\alpha}$ are subsets of ${U_\alpha}$ and consequently form a refinement of ${\left\{U_\alpha\right\}}$. Thus, if we can show that each ${\left\{Z^n_\alpha, \alpha \in A\right\}}$ is locally finite, then we will have a refinement

$\displaystyle \bigcup_{n, \alpha} Z^n_\alpha$

of ${\mathfrak{A}}$ which can be decomposed into a countable collection of locally finite families.

So, that’s the plan. We will actually show that for each ${n}$, there is a ${\delta = \delta_n}$ such that ${d(Z^{n}_\alpha, Z^n_\beta) \geq \delta}$ if ${\alpha \neq \beta}$. Indeed, suppose ${x \in Z^n_\alpha, y \in Z^n_\beta}$, and ${\alpha \neq \beta}$. Without loss of generality, ${\beta < \alpha}$. Then ${x}$ is within ${2^{-n-3}}$ of a point ${z \in W^n_\alpha}$ and ${y}$ is within ${2^{-n-3}}$ of a point ${z'}$ in ${W^n_\beta}$. Now the distance of ${z'}$ to ${X-U_\beta}$ is, by definition, at least ${2^{-n}}$, because ${z' \in V^n_\beta}$. However, ${z}$ is not in ${V^{n+1}_\beta}$ because of the way the ${W^n_\alpha}$ were defined, so its distance to ${X - U_\beta}$ is at most ${2^{-n-1}}$. In particular,

$\displaystyle d(z, z') \geq 2^{-n-1}$

so it follows that ${d(x,y) \geq 2^{-n-1} - 2(2^{-n-3}) = 2^{-n-2}}$.

In particular, the ${Z^n_\alpha}$ are mildly separated from each other, which means that any ${2^{-n-3}}$ neighborhood of any point can intersect at most one of the ${Z^{n}_\alpha}$ (where ${n}$ of course is fixed). In particular, the ${Z^n_\alpha}$ for each ${n}$ are locally finite. This proves the claim we wanted. This also establishes the full result.