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I typically like to practice my knowledge on a specific concept by doing proofs using one definition of a term, and then doing the same proofs using an equivalent definition (without inducing the equivalence). However, I'm stuck on the following:

We say that a topological space $(S,T)$ is perfectly normal if for every closed set $F\subseteq S$ there is a continuous function $f:S\rightarrow\mathbb{R}$ with $F=\{x\in S:f(x)=0\}$.

Can we prove that a metric space is perfectly normal by this definition?



I've seen some posts (post 1, post 2) where this just follows as a corollary. More specifically, it follows by the equivalence of definitions and the results that metric spaces are normal and $G_\delta$. However, I'm wondering if there was a more direct approach, as opposed to such a roundabout method?

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Dustin Tran
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1 Answers1

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For all nonempty $A \subset X$ the function “least distance to $A$” given by $$x \mapsto \operatorname{dist}{(x,A)} = \inf\limits_{a \in A}\;d(x,a)$$ is continuous (even $1$-Lipschitz) and it is zero precisely on the closure of $A$, see this thread for some proofs of that fact.

If $A \neq \emptyset$ is closed then $A = \{x \in X\,:\,\operatorname{dist}(x,A) = 0\}$. If $A = \emptyset$ take a constant non-zero function.

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  • Great construction! I hadn't even considered doing a constructive proof since I thought it would be too hard. I was merely hoping that out of the definitions, I could somehow "pop out" the subtle existence of one. But this is much better! – Dustin Tran Oct 18 '11 at 10:05