# Nonnormal spaces cannot be metrizable.

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?

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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

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.

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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