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Let $X$ be a nonempty perfect Polish space and let $Q$ be a countable dense subset of $X$. Then $Q$ is $F_{\sigma}$ but not $G_{\delta}$.

My question is about the $G_{\delta}$ part; specifically, I am unsure why the hypothesis that $X$ is perfect is included. Below is my (attempted) proof of why $Q$ is not $G_{\delta}$.

To show $Q$ is not $G_{\delta}$, suppose that it is. So then $Q$ is Polish, because it is a $G_{\delta}$ subset of a Polish space. Then $Q$ is also a Baire space, because it is completely metrizable. Baire spaces are not meager, because they are themselves open. However, $Q$ is meager, because it is $\bigcup_{q \in Q} \{ q \}$ (countable union because $Q$ is countable), where each singleton is nowhere dense. But $Q$ cannot be both meager and nonmeager, so the assumption that $Q$ is $G_{\delta}$ causes a contradiction and is therefore false.

Edit: It seems I need the perfectness to show that each singleton $\{q\}$ is actually nowhere dense (if $q$ were an isolated point, it wouldn't be nowhere dense) so that $Q$ is meager.

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Bear in mind that the discrete topology on $\mathbb{N}$ is Polish (use a metric where each pair of distinct points is at a distance $1$ from each other). In this space, each singleton is an open set.

The mistaken step in your proof is where you say that each singleton must be nowhere dense (and this is why the statement included the assumption that $X$ is perfect).

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  • $\begingroup$ To be clear, the statement that the singletons are nowhere dense isn't wrong, it's just that I need the perfect-ness to prove it? $\endgroup$ – Reed Oei Sep 25 '19 at 20:09
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    $\begingroup$ All of the singletons in $\mathbb{N}$ are both closed and open, so none of them are nowhere dense. For a Hausdorff space $X$ "every singleton in $X$ is nowhere dense" is equivalent to "$X$ is perfect". You can also have Polish spaces like $\{0\} \cup \{\frac{1}{n} \mid n \in \mathbb{N} \} \subseteq \mathbb{R}$ where the singleton $\{0\}$ is nowhere dense, but the others are open (so neither perfect nor discrete). $\endgroup$ – Robert Furber Sep 25 '19 at 20:12
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Missing argument:

$\{q\}$ is not isolated in $Q$, for if it were we could write it as $\{q\}= Q\cap O$ for some open $O$ of $X$, but then $O\setminus \{q\}$ would be an open set of $X$ that would miss $Q$, so $O \setminus \{q\} =\emptyset$ (as $Q$ is dense) and $q$ would be an isolated point of $X$, contradiction (to perfectness of $X$).

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