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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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    Also see this thread: [Is a metric space perfectly normal?](http://math.stackexchange.com/q/73609)2012-11-09

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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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    @user41728: But there are other, more elementary ways, and I just thought of one.2012-11-09