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:

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

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.

  • 1
    $\begingroup$ In at.yorku.ca/p/a/c/a/02.pdf I show that all metric spaces are fully normal, and in an earlier lemma, that fully normal is equivalent to paracompactness. This is the historical way it was proved. It might help, as it is more written out. It also uses a well order on the index set. $\endgroup$ Sep 3, 2014 at 19:44
  • $\begingroup$ Imagine that you define the $D_{\alpha n}$ inductively. For $n = 0$, the condition 2 is void. Then, when you define the $D_{\alpha 1}$, you know which points belong to some $D_{\beta 0}$, and you don't consider these points anymore as centres of the balls. Continue. $\endgroup$ Sep 3, 2014 at 20:30

2 Answers 2


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

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

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.

  • $\begingroup$ Thank you for the breakdown; that helps a lot! Sometimes Rudin's proofs just go right over my head. $\endgroup$
    – graydad
    Sep 4, 2014 at 4:12
  • 3
    $\begingroup$ @graydad: Steven Watson (himself a pretty good general topologist) said the following about Mary Ellen Rudin: Reading the articles of Mary Ellen Rudin, studying them until there is no mystery takes hours and hours; but those hours are rewarded, the student obtains power to which few have access. They are not hard to read, they are just hard mathematics, that's all. $\endgroup$
    – user642796
    Sep 4, 2014 at 18:39
  • $\begingroup$ good to know I'm not the only one who feels like that :) $\endgroup$
    – graydad
    Sep 4, 2014 at 19:12
  • $\begingroup$ @ArthurFischer Sorry to resurrect this, it seems to me that $D_{\alpha 1}$ could be empty for all $\alpha$s, for example if all the $C_{\alpha}$ are small enough not to contain $S(x,3/2)$. Of course for $n$ high enough $S(x,2^{-n})$ is contained in some $C_{\alpha}$ and I can start the inductive process at some higher value, but there is also (1) to satisfy, is it obvious that induction can be started at all? $\endgroup$
    – GFR
    Aug 3, 2015 at 19:38
  • $\begingroup$ I can now see that there is some $n$ for which (1) and (3) are satisfied: since the indexing set of the cover is well-ordered, it has a first element $\bar\alpha$. Let $x\in C_{\bar\alpha}$. Since $ C_{\bar\alpha}$ is open, it contains an open ball $S(x,3\cdot 2^{-n})$ for some $n$. However it needs not to be the least one for which (1) and (3) hold. (continued) $\endgroup$
    – GFR
    Aug 3, 2015 at 20:15

To satisfy condition 2., we assume (using recursion) that all sets $D_{\beta j}$ are already defined for all $j < n$ (which is an empty condition for the first $n$...). So condition 2 is then checkable when defining what $x$ to consider as centres for the balls that constitute $D_{\alpha n}$ (the condition should really say that for all $\beta$ and all $ j < n$, $ x \notin D_{\beta j}$). The conditions just define at stage $n$ of the recursion what $x$ are considered. Then we union up all balls of radius $2^{-n}$ around those $x$ to get an open set, one for each index $\alpha$, at stage $n$. It is clear that this is then well defined. Of course why one would pick these conditions in the first place is the clever bit of the proof...

The proof of local finiteness seems straightforward. For a fixed $x$ she defines a $n,j$ and radius $2^{-j-n}$ so that for the neighbourhood $O = S(x,2^{-j-n})$ of $x$:

  • for all $i \ge j +n$, $O$ intersects no $D_{\beta i}$
  • for all $i < j +n$, $O$ intersects at most 1 $D_{\beta i}$

This means that $O$ only intersects at most $j + n$, so finitely many, $D_{\beta i}$.

Maybe you could expand on what exactly you do not understand in the proof of these claims.

  • $\begingroup$ I think the confusing part in the linked article is the direction of the recursion: I think the right way to see it is as a recursion with respect to lexicographic order on $(n,\alpha)$, whilst at a glance (in absence of a more explicit explanation) it seems like the opposite lexicographic order $(\alpha,n)$ is used. It confused me as much as it did OP. $\endgroup$
    – tomasz
    Sep 3, 2014 at 20:43

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