Let $X$ be some compact Hausdorff space (or even normal space). Is it true that each closed subset $X'$ is $f^{-1}(0)$ for some $f\in C(X,\mathbb{R})$? I know that there is Urysohn's lemma which gives us an opportunity to continue each function $X'\longrightarrow \mathbb{R}$ to a function $X\longrightarrow \mathbb{R}$, but it doesn't seem to be useful in my case.
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Yes, that's a little bit too crude of a tool to use. How about the metric case; can we prove it then? – hardmath Nov 12 '13 at 17:28
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1Related would be my answer here. – user642796 Nov 12 '13 at 17:30
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Incidentally, every compact Hausdorff space is normal Hausdorff. – Cameron Buie Nov 12 '13 at 17:31
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No. The compact ordinal space $\omega_1+1$ with the order topology is an easy counterexmple, as is $\beta\Bbb N$: each has at least one point $x$ such that $\{x\}$ is not a $G_\delta$-set and therefore cannot be $f^{-1}[\{0\}]$ for any continuous real-valued $f$.
You’re asking for spaces in which every closed set is a zero-set (or as Engelking somewhat idiosyncratically calls it, a functionally closed set). Among $T_1$ spaces these are precisely the perfectly normal spaces, i.e., the normal spaces in which every closed set is a $G_\delta$.
Brian M. Scott
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