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Let $X$ be a separable compact Hausdorff topological space. Then every closed subset of $X$ is a $G_{\delta}$ set (i.e. $X$ is perfectly normal).

I've been skimming through some topology textbooks recently. As we know, every closed subset of a metric space is a $G_{\delta}$ set. But I am not sure whether the above proposition is correct because I did not find any counterexample. Can anyone give a proof or counterexample? Thanks.

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Let $X = \{0,1\}^{\mathbb{R}}$. Since $\{0,1\}$ is separable and $\operatorname{card} (\mathbb{R}) \leqslant 2^{\aleph_0}$, $X$ is separable. Let

$$F = \{0\} = \{ x \in X : t \in \mathbb{R} \implies x(t) = 0\}.$$

$F$ is a closed subset of $X$ that is not a $G_{\delta}$-set. For if $U$ is an open subset of $X$ containing $F$, then there is a finite subset $I = I(U)$ of $\mathbb{R}$ such that

$$V_I = \{ x \in X : t \in I \implies x(t) = 0\} \subset U.$$

Then if we have a countable family $\{ U_n : n \in \mathbb{N}\}$ of open subsets containing $F$, the union

$$K := \bigcup_{n\in \mathbb{N}} I(U_n)$$

is a countable, thus proper, subset of $\mathbb{R}$, and hence

$$\bigcap_{n\in \mathbb{N}} U_n \supset \bigcap_{n \in \mathbb{N}} V_{I(U_n)} = V_K \supsetneqq F.$$

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  • $\begingroup$ Why if $U$ is an open subset of $X$ containing $F$, then there is a finite subset $I = I(U)$ of $\mathbb{R}$ such that $$V_I = \{ x \in X : t \in I \implies x(t) = 0\} \subset U?$$ $\endgroup$ – Jay Apr 2 '16 at 13:47
  • $\begingroup$ I've simplified the example, it was needlessly complicated, and now the existence of such a $V_I \subset U$ follows easily from the definition of the product topology. I'm not sure it actually did in the original formulation. (Though the original $F$ was not a $G_\delta$, the proof may have needed to be different.) $\endgroup$ – Daniel Fischer Apr 2 '16 at 13:49
  • $\begingroup$ @JacobZhang Not sure whether you have seen the updated answer. In the updated version, note that a base of the product topology is formed by "open rectangles" $R = \prod_{t\in \mathbb{R}} W_t$, where $\bigl\{ t\in \mathbb{R} : W_t \neq \{0,1\}\bigr\}$ is finite, and $W_t \neq \varnothing$ for all $t$, by definition of the product topology. For such a rectangle, we have $0 \in R$ if and only if there is a finite $I$ with $R = V_I$. $\endgroup$ – Daniel Fischer Apr 2 '16 at 13:53
  • $\begingroup$ I have seen the undated answer. Your construction is skillful. Thanks. $\endgroup$ – Jay Apr 2 '16 at 13:57
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Although Henno’s general result renders specific examples a bit superfluous, I’ll note that $\beta\Bbb\omega$ is another relatively easy example. It is certainly compact, Hausdorff, and separable. The ultrafilter construction of $\beta\Bbb\omega$ makes it obvious that it has a base of cardinality $2^\omega=\mathfrak{c}$: if for each $A\subseteq\omega$ we set

$$\widehat A=\{p\in\beta\omega:A\in p\}\;,$$

then $\mathscr{B}=\left\{\widehat A:A\subseteq\omega\right\}$ is a base for $\beta\omega$.

If $p\in\beta\omega$, and $\{p\}$ is a $G_\delta$, there is a countable $\mathscr{B}_p\subseteq\mathscr{B}$ such that $\bigcap\mathscr{B}_p=\{p\}$. $\mathscr{B}$ has only $\left(2^\omega\right)^\omega=2^\omega$ countable subsets, so there can be at most $2^\omega$ points $p\in\beta\omega$ such that $\{p\}$ is a $G_\delta$. But $|\beta\omega|=2^{\mathfrak{c}}$, so most singletons in $\beta\omega$ are closed sets that are not $G_\delta$ sets.

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