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 be a metric space, and let be an open cover of indexed by ordinals (possible by Well-Ordering Theorem). Let be a metric on , and let be the open ball of radius centered at . By induction on , define a collection of an open sets by setting to be the union of all open balls satisfying

- is the smallest ordinal with
- , for

For any given , a priori we don’t know that is nonempty; but since as , there exists a positive integer such that there is a point with .

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 . Try to visualize what properties (1)-(3) look like in this space.

We will show that is an locally finite refinement of which covers . To see that is a refinement covering , observe that for any , by definition there is an such that . For any , there is a smallest ordinal such that . For sufficiently large , . If , for some and some , then , for all , a contradiction.

The most involved step of the proof is showing that is locally finite. Let , and let be the minimal ordinal such that , for some . Let be sufficiently large so that .

- if , then for all ;
- if , then for at most one .

Let , and let , for some ordinal . Clearly, . We can write as the union , for a set of centers. By property (2) in the definition of , each does not belong to . Hence, , which implies that

We conclude that , for every ; equivalently, .

Let , and suppose and for ordinals . Without loss of generality, assume that . Let and be the centers of the open balls of radius which contain and , respectively. By property (1), . Since by property (3) of the definition, we have that ; hence,

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.”

- 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. - A.H. Stone, Paracompactness and product spaces,
*Bull. Amer. Math. Soc.***54**(1948), 977-982. - M.E. Rudin, A new proof that metric spaces are paracompact,
*Proc. Amer. Math. Soc.***20**(1969), 603.