I have difficulty understanding the proof of Theorem 12.6 in Kechris's Classical Descriptive Set Theory that if $X$ is Polish then the Effros Borel space of $F(X)$ is standard. $F(X)$ consists of all closed sets in $X$, and I am not giving the definition of Effros Borel space since it is probably not so related to my confusion. The proof proceeds as follows:

  1. Let $\overline{X}$ be a compactification of $X$.
  2. Identify $F(X)$ with a subset $G$ of $K(\overline{X})$, the collection of compact sets in $\overline{X}$.
  3. Prove that $G$ is $G_\delta$ in $K(\overline{X})$, hence Polish.
  4. Carry the topology on $G$ back to $F(X)$. Prove that Effros Borel space coincides with the topology.

I guess in 3 the author implicitly assumes that $K(\overline{X})$ is Polish, which seems to depend on the metrizability of $\overline{X}$. But why is it metrizable? Certainly not any compactification works, for example the Stone–Čech compactification of $\mathbb{N}$ is not even first countable; I think the one-point compactification of $\mathbb{R}^{\mathbb{N}}$ is not first countable either. Can every Polish space be embedded into some compact metric space?


1 Answer 1


The citations below refer, as in the OP, to Kechris' book.

This is what Theorem $4.14$ does:

Every separable metrizable space is homeomorphic to a subset of the Hilbert cube.

This gives the desired embedding result. Note that Kechris uses a nonstandard (in my experience) notion of "compactification," in Definition $4.15$: a compactification of a separable metrizable space $X$ is a compact metrizable space $Y$ such that $X$ is homeomorphic to a dense subset of $Y$. So the "pick a compactification" language in the proof in question is unproblematic ... if we grant Kechris' use of the term.


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