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I'm trying to solve the following problem:

Let $(X,T)$ be a second-countable, normal topological space, and let $U \subset X$ be an open set. Show that there exists a continuous function $f:X\rightarrow[0,1]$ such that $f(x) \neq 0$ iff $x \in U$.

Now, I know that in a metric space, every open set is $F_\sigma$. Does this holds also for a general second-countable space? I've got a hunch that the answer is positive, but I couldn't prove it.

If so, solving the problem would be easy, since $U=\bigcup_{i=1}^\infty F_i$, where each $F_i$ is closed, and so there exists a closed set $F$ such that $F \subset U \subset X$, and by applying Urysohn's lemma to $U^c$ and $F$ we are done.

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Yes, this is true. Suppose $B_n, n \in \mathbb{N}$ is a countable base for $X$, and $U$ is open. Then for each $x \in U$ we pick by regularity some open $V_x$ such that $x \in V_x \subseteq \overline{V_x} \subseteq U$. Then pick some $n(x)$ with $x \in B_{n(x)} \subseteq V_x$. Then all $\cup \{\overline{B_{n(x)}}: x \in U \}$ is equal to $U$ (all $x$ in $U$ are in their own $B_{n(x)}$; all of these have closures inside $U$ by construction, so their union is exaclty $U$) and shows $U$ is an $F_\sigma$ (as we have countably many $n(x)$ (many $x$ will give the same $n(x)$, but that does not matter, we have countably many (at most) different closed sets).

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By the Urysohn metrization theorem, a second-countable, normal space is metrizable, which solves your exercise.

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  • $\begingroup$ I'm not familiar yet with this theorem, is there any way to circumvent it? $\endgroup$ – Brassican Jan 5 '16 at 17:52
  • $\begingroup$ Let, each open subset of a topological space X is normal. Does this necessarily mean the space is hereditary normal? $\endgroup$ – math is love Jan 26 '17 at 20:54

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