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A space is normal iff every pair of disjoint closed subsets have disjoint closed neighbourhoods.

Given space $X$ and two disjoint closed subsets $A$ and $B$.

I have shown necessity: If X is normal then by Urysohn's lemma there exists continuous map $f:X \to [0,1]$ such that $f(A)=0$ and $f(B)=1$, then $f^{-1}(0)$ and $f^{-1}(1)$ are two disjoint closed neighbourhoods of A and B.

But how to show the sufficiency?

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If $A$ and $B$ have disjoint closed neighborhoods $U$ and $V$, then by definition of neighborhood we know that $A\subseteq \mathrm{int}(U)$ and $B\subseteq\mathrm{int}(V)$. Now, the interior of $U$ is an open neighborhood of $A$, the interior of $V$ is an open neighborhood of $B$, and so $A$ and $B$ have disjoint open neighborhoods.

Thus, if a space satisfies your requirement, then it is normal.

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There’s a flaw in your argument for necessity: $f^{-1}[\{0\}]$ and $f^{-1}[\{1\}]$ may not be nbhds of $A$ and $B$, respectively. Consider, for example, $X=[0,1]$, $A=\{0\}$, $B=\{1\}$, and $f:X\to[0,1]:x\mapsto x$: $f^{-1}[\{0\}]=\{0\}$, which is not a nbhd of $\{0\}$, and $f^{-1}[\{1\}]=\{1\}$, which is not a nbhd of $\{1\}$.

You can fix it by using $f^{-1}\left[\left[0,\frac13\right]\right]$ and $f^{-1}\left[\left[\frac23,1\right]\right]$ instead: these really are closed nbhds of $A$ and $B$, respectively, since they contain the open nbhds $f^{-1}\left[\left[0,\frac13\right)\right]$ and $f^{-1}\left[\left(\frac23,1\right]\right]$.

Arturo has already taken care of the other half of the argument.

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  • $\begingroup$ Thanks! I was not sure about the definition of nbd of subset, now it's clear. $\endgroup$ – user31899 Jun 13 '12 at 4:05
  • $\begingroup$ @user31899: A set $U$ is a neighborhood of a point $x$ if and only if $x$ is in the interior of $U$. A set $U$ is a neighborhood of a set $X$ if and only if it is a neighborhood of every point $x\in X$. $\endgroup$ – Arturo Magidin Jun 13 '12 at 4:10

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