Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why are nonnormal spaces not metrizable?

It's a statement I found on wikipedia that I need to understand. When looking at the metrization theorems I don't see any obvious connections. Is this result trivial or not?

share|cite|improve this question
In other words: Why are metric spaces normal? The question seems simpler that way. – Michael Hardy Nov 9 '12 at 0:32
Also see this thread: Is a metric space perfectly normal? – user48839 Nov 9 '12 at 23:16
up vote 6 down vote accepted

It’s an immediate consequence of the fact that every metric space is normal, which follows readily from the non-trivial fact that every metric space is paracompact.

Added: I just realized that one can avoid paracompactness quite easily. Let $\langle X,d\rangle$ be a metric space, and let $H$ and $K$ be disjoint, non-empty closed subsets of $X$. Now let

$$f:X\to\Bbb R:x\mapsto\frac{d(x,H)}{d(x,H)+d(x,K)}\;;$$

this is a continuous real-valued function separating $H$ and $K$, showing that $X$ is not just normal, but perfectly normal.

share|cite|improve this answer
Is that the most common route (through paracompact) to showing that a metric space is normal? – user41728 Nov 9 '12 at 0:20
@user41728: Yes, at least in my experience. – Brian M. Scott Nov 9 '12 at 0:22
@user41728: But there are other, more elementary ways, and I just thought of one. – Brian M. Scott Nov 9 '12 at 0:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.