TITLE: Mary Ellen Rudin's proof that all metric space are paracompact
QUESTION [7 upvotes]: Given a metric space $(X,d)$, show that the space is paracompact. I have no idea where to begin on this, and the proofs of this I have seen have been difficult for me to understand. Can anyone offer a proof or pointers on this? The proof I have been looking at is from

Mary Ellen Rudin, A new proof that metric spaces are paracompact, Proc. Amer. Math. Soc. vol.20 (1969), p.603, link


Here is the start of the proof.

Assume that $X$ is a metric space and that $\{ C_\alpha \}$ is an open cover of 
  $X$ indexed by ordinals. Let $\rho$ be a metric on $X$ and let $S(x,r)$ be the
  open sphere with center $x$ and radius $r$. For each positive integer $n$
  define $D_{\alpha n}$ (by induction on $n$) to be the union of all spheres $S(x,2^{-n})$
  such that:

$\alpha$ is the smallest ordinal with $x \in C_\alpha$,
$x \notin D_{\beta j}$ if $j < n$,
$S(x , 3 \cdot 2^{-n} ) \subset C_\alpha$.


I am confused on condition 2 of the definition of $D_{\alpha n}$. Certainly it is necessary for paracompactness, but I don't understand how we know such a property exists in an arbitrary metric space. 
I am also confused about the demonstration that $\{ D_{\alpha n} \}$ is locally finite.

REPLY [4 votes]: Condition (2) basically says the following: Suppose $x \in X$, and $\alpha$ is least such that $x \in C_\alpha$.

If $x$ happens to belong to $D_{\beta j}$ (for some $\beta$ and $j$), then $x$ will never be the center of a sphere used in the construction of $D_{\alpha n}$ where $n > j$.
In conjunction with condition (3) we get that if $m$ is least such that $S(x,3 \cdot 2^{-m} ) \subseteq C_\alpha$ and $x \in D_{\beta j}$ (for some $j < m$ and some $\beta$), then $x$ is never the center of a sphere used in the construction of $D_{\alpha n}$ for any $n$.

We can take this a little more slowly, and instead define sets $A_{\alpha n}$ (the centers of the spheres used to construct $D_{\alpha n}$) by induction on $n$ (but for all $\alpha$ "simultaneously") as follows: $A_{\alpha n}$ is the set of all $x \in X$ such that

$\alpha$ is least such that $x \in C_\alpha$;
for all $j < n$, $\rho (x,y) \geq 2^{-j}$ for all $y \in \bigcup_\beta A_{\beta j}$; and
$S(x,3 \cdot 2^{-n} ) \subseteq C_\alpha$.

If $A_{\alpha n}$ is defined, then $D_{\alpha n} = \bigcup \{ S(x,2^{-n}) : x \in A_{\alpha n} \}$.
An important point about condition (2) is that the ordinal $\alpha$ plays absolutely no part. This means that as long as every $A_{\beta j}$ is available for each $j < n$, then we can check whether any $x \in X$ satisfies condition (2) for that $n$. As conditions (1) and (3) present no problems, let's just see how we can inductively make these checks.

Note that for each $x \in X$ condition (2) is vacuous for $n=1$. It follows that there is no problem in defining the $A_{\alpha 1}$ sets "simultaneously".

To check whether any $x \in X$ satisfies condition (2) for $n=2$, we need only ensure that $\rho (x,y) \geq 2^{-1}$ for all $y \in \bigcup_\beta A_{\beta 1}$. As all the $A_{\beta 1}$ are available, there is no problem here to define the $A_{\alpha 2}$ sets "simultaneously".

To check whether any $x \in X$ satisfies condition (2) for $n=3$, we need only ensure that $\rho (x,y) \geq 2^{-1}$ for all $y \in \bigcup_\beta A_{\beta 1}$ and $\rho (x,z) \geq 2^{-2}$ for all $z \in \bigcup_\beta A_{\beta 2}$. Again, all the required sets are already available to us, so there is no problem to define the $A_{\alpha 3}$ sets "simultaneously".


And we continue in this manner.