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If $X$ is a normal space having disjoint closed subspaces $A$ and $B$, is there a continuous function $f:X \to [0,1]$ such that ${f^{ - 1}}(0) = A$ and ${f^{ - 1}}(1) = B$? (We have already known there is a $f$ such that $f(A) = 0$ and $f(B) = 1$ )

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This isn't true in general. A space in which your condition holds for all closed subspaces $A$ and $B$ is called perfectly normal, a strictly stronger condition than being normal. As wiki states, a normal space is perfectly normal iff every closed set is $G_\delta$.

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    For completeness you could add a specific counterexample. A simple one is to take $\mathbb{R}$ as the underlying set of $X$, make every non-zero point isolated, and say that $V$ is a nbhd of $0$ iff $0\in V$ and $|\mathbb{R}\setminus V|$ is countable. It’s easy to check that $X$ is normal and $\{0\}$ is closed, but if $f:X\to\mathbb{R}$ is continuous and $f(0)=0$, then each $f^{-1}[(-2^{-n},2^{-n})]=\mathbb{R}\setminus C_n$ for some countable $C_n$, and $f^{-1}[\{0\}]=\bigcap_nf^{-1}[(-2^{-n},2^{-n})]=\mathbb{R}\setminus\bigcup_nC_n\ne\{0\}$, since $\bigcup_nC_n$ is countable.2011-12-09