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The Cantor-Bendixson theorem states that

If $X$ is Polish then any closed subset of $X$ can be written as the disjoint union of a perfect subset and an at most countable subset.

It seems that we can weaken the condition as follows:

If $X$ is second countable (even not $T_1$), then any closed subset $F$ of $X$ can be written as the disjoint union of a perfect subset and an at most countable subset

The canonical proof applies as follows:

Suppose $\mathcal B$ is a countable base. Let $P$ be the set of condensation points, since $F$ is closed, $P\subseteq F$. For each $x\in F\setminus P$, we assign $B_x\in\mathcal B$ such that $x\in B_x$, and $B_x\cap F$ is at most countable, then there's a map $\phi\colon F\setminus P\to\mathcal B,x\mapsto B_x$. Note that the preimage of each $B_x$ is at most countable, therefore $F\setminus P$ is countable.

It remains to show that $P$ is perfect. Every point $p\in P$ is a condensation point of $F$, hence $P$, since $F\setminus P$ is countable, hence a limit point of $P$. Conversely, if $x\not\in P$, there's an open set $\mathcal O$ contains $x$ such that $\mathcal O\cap F$ is at most countable, hence $\mathcal O\cap P=\emptyset$.

Is it right?

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  • $\begingroup$ Yes. There is no problem with this argument. $\endgroup$
    – PatrickR
    Oct 9, 2020 at 23:40

1 Answer 1

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It looks fine to me. If I had to hazard a guess about why it's not general stated in that generality it's that if your $X$ is Hausdorff then this is implied by the standard version by embedding $X$ in its completion, and that there aren't many non-Hausdorff spaces of interest in descriptive set theory.

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  • $\begingroup$ Even if it's Hausdorff, it's not necessarily metrizable, right? $\endgroup$
    – Yai0Phah
    Apr 6, 2014 at 12:30

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