## Metric Spaces Are Paracompact: M.E. Rudin’s Elementary Proof

I promise to return to proving the “difficult” direction of Stone’s theorem, but please be patient with me while I go off on a tangent. Metric spaces are fully normal, a property which we proved in the previous post. Stone’s theorem therefore tells us that metric spaces are paracompact. However, using Stone’s theorem to prove paracompactness is overkill. An elementary proof by Mary Ellen Rudin published in 1969, 21 years after the publication of Stone’s theorem, dispenses with the language of fully normal in exchange for assuming the underlying topological space is metrizable–there’s no free lunch, afterall.

I have differed from Rudin in my notation, in particular my choice of letters for open sets. My apologies for any confusion; blame my idiosyncracies.

Let ${X}$ be a metric space, and let ${\left\{U_{\alpha}\right\}_{\alpha}}$ be an open cover of ${X}$ indexed by ordinals (possible by Well-Ordering Theorem). Let ${\rho}$ be a metric on ${X}$, and let ${B(x;r)}$ be the open ball of radius ${r}$ centered at ${x}$. By induction on ${n}$, define a collection of an open sets ${\left\{V_{\alpha,n}\right\}_{\alpha}}$ by setting ${V_{\alpha,n}}$ to be the union of all open balls ${B(x;2^{-n})}$ satisfying

1. ${\alpha}$ is the smallest ordinal with ${x\in U_{\alpha}}$
2. ${x\notin V_{\beta,j}}$, for ${j
3. ${B(x;3\cdot 2^{-n})\subset U_{\alpha}}$

For any given ${n}$, a priori we don’t know that ${V_{\alpha,n}}$ is nonempty; but since ${3\cdot 2^{-n}\rightarrow 0}$ as ${n\rightarrow\infty}$, there exists a positive integer ${n}$ such that there is a point ${x}$ with ${B(x;3\cdot 2^{-n})\subset U_{\alpha}}$.

Although I have yet to understand completely how one would arrive at this construction, but for some intution, consider the open unit interval endowed with the standard metric and the open cover ${\left\{(\frac{1}{m+1}, \frac{m}{m+1})\right\}_{m\geq 1}}$. Try to visualize what properties (1)-(3) look like in this space.

We will show that ${\left\{V_{\alpha,n}\right\}_{\alpha,n}}$ is an locally finite refinement of ${\left\{U_{\alpha}\right\}}$ which covers ${X}$. To see that ${\left\{V_{\alpha,n}\right\}}$ is a refinement covering ${X}$, observe that for any ${y\in V_{\alpha,n}}$, by definition there is an ${x=x(y)\in X}$ such that ${y\in B(x;2^{-n})\subset B(x;3\cdot 2^{-n})\subset U_{\alpha}}$. For any ${x\in X}$, there is a smallest ordinal ${\alpha}$ such that ${x\in U_{\alpha}}$. For sufficiently large ${n}$, ${B(x;3\cdot 2^{-n})\subset U_{\alpha}}$. If ${x\notin V_{\beta,j}}$, for some ${\beta}$ and some ${j\leq n}$, then ${B(x;3\cdot 2^{-j})\not\subset U_{\alpha}}$, for all ${j\leq n}$, a contradiction.

The most involved step of the proof is showing that ${\left\{V_{\alpha,n}\right\}}$ is locally finite. Let ${x\in X}$, and let ${\alpha}$ be the minimal ordinal such that ${x\in V_{\alpha,n}}$, for some ${n}$. Let ${j>1}$ be sufficiently large so that ${B(x;2^{-j})\subset V_{\alpha,n}}$.

1. if ${i\geq n+j}$, then ${B(x;2^{-n-j})\cap V_{\beta,i}=\emptyset}$ for all ${\beta}$;
2. if ${i, then ${B(x;2^{-n-j})\cap V_{\beta,i}\neq\emptyset}$ for at most one ${\beta}$.

Let ${i\geq n+j}$, and let ${y\in V_{\beta,i}}$, for some ordinal ${\beta}$. Clearly, ${i>n}$. We can write ${V_{\beta,i}}$ as the union ${\bigcup_{z\in E}B(z;2^{-i})}$, for a set ${E}$ of centers. By property (2) in the definition of ${V_{\beta,i}}$, each ${z\in E}$ does not belong to ${V_{\alpha,n}}$. Hence, ${\rho(z,x)\geq 2^{-j}}$, which implies that

$\displaystyle \rho(y,x)\geq\rho(z,x)-\rho(z,y)>2^{-j}-2^{-i}\geq 2^{-j}-2^{-n-j}\geq 2^{-n-j}$

We conclude that ${B(x;2^{-n-j})\cap B(z;2^{-i})}$, for every ${z\in E}$; equivalently, ${B(x;2^{-n-j})\cap V_{\beta,i}=\emptyset}$.

Let ${i, and suppose ${p\in V_{\beta,i}}$ and ${q\in V_{\gamma,i}}$ for ordinals ${\beta,\gamma}$. Without loss of generality, assume that ${\beta<\gamma}$. Let ${y}$ and ${z}$ be the centers of the open balls of radius ${2^{-i}}$ which contain ${p}$ and ${q}$, respectively. By property (1), ${z\notin U_{\beta}}$. Since ${B(y;3\cdot 2^{-i})\subset U_{\beta}}$ by property (3) of the definition, we have that ${\rho(y,z)>3\cdot 2^{-i}}$; hence,

$\displaystyle \rho(p,q)\geq\rho(y,z)-\rho(p,y)-\rho(q,z)>3\cdot 2^{-i}-2\cdot 2^{-i}>2^{-n-j}$

As an aside, we mention that the proof makes obvious use of the Axiom of the Choice through the Well-Ordering Theorem. I believe that the theorem metric spaces are paracompact is strictly weaker than the Axiom of Choice. For a proof of this fact, see the article “On Stone’s Theorem and the Axiom of Choice.”

1. C. Good, I.J. Tree, and W.S. Watson, On Stone’s theorem and the axiom of choice, Proc. Amer. Math. Soc. 126 (1998), 1211-1218.
2. A.H. Stone, Paracompactness and product spaces, Bull. Amer. Math. Soc. 54 (1948), 977-982.
3. M.E. Rudin, A new proof that metric spaces are paracompact, Proc. Amer. Math. Soc. 20 (1969), 603.
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