As title says. I’m trying to learn some descriptive set theory but I don’t quite see this.

I want to use the following:

Given $X, Y$ Polish spaces, $f:X\to Y$ continuous, if $f(X)$ is uncountable there is a subset $K\subseteq X$ homeomorphic to Cantor space on which $f$ is injective.

I can reduce to the case where $f(U)$ uncountable for $U$ open in $X$, and Kechris says to show $\{K\in K(X) : f \text{ injective on K}\}$ is a dense $G_\delta$ set, in the Vietoris topology on $K(X)$ the compact subsets of X.

  1. How do I show this? I suspect “Lusin schemes” might be useful but I don’t really understand this technology. Other approaches are also welcome.

  2. Why does this give the result? Being $G_\delta$, this set is then Polish (right?), but why does this yield an uncountable K (which I understand would be sufficient)

Thank you

  • $\begingroup$ What is a "Lusin scheme?" $\endgroup$ – Noah Schweber Oct 21 '18 at 17:50
  • $\begingroup$ @NoahSchweber Kechris defines a Lusin scheme on p36, a collection of subsets indexed by $\mathbb N^{<\mathbb N}$ with $A_{s,i}$ disjoint from $A_{s,j}$, and $A_{s,i} \subseteq A_s$. I guess it’s not actually that complicated, maybe I got scared off by the name and other times he uses them. $\endgroup$ – Ryan Oct 21 '18 at 18:13
  • $\begingroup$ Ah, apologies. I’m on a mobile device. I hope the idea is clear (can you edit comments? Not on mobile it seems) $\endgroup$ – Ryan Oct 21 '18 at 19:03
  • $\begingroup$ Let me fix that for you. $\endgroup$ – Asaf Karagila Oct 21 '18 at 19:04

For 2. $K(X)$ is Polish, the intersection of countably many dense open sets is dense and hence nonempty, so there are plenty of $K$ on which $f$ is injective.

For 1. I would try something like this: take a countable base, $\mathcal{B}$, for the topology of $X$. For a pair $\langle B, C\rangle$ of elements of $\mathcal{B}$ with disjoint closures let $$U(B,C)=\{K \in K(X): f[K\cap B]\cap f[K\cap C]=\emptyset\}\text{.}$$

Then prove that $U(B,C)$ is open and dense.

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