# Analytic sets have perfect set property (Kechris)

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

• What is a "Lusin scheme?" Commented Oct 21, 2018 at 17:50
• @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.
– Ryan
Commented Oct 21, 2018 at 18:13
• Ah, apologies. I’m on a mobile device. I hope the idea is clear (can you edit comments? Not on mobile it seems)
– Ryan
Commented Oct 21, 2018 at 19:03
• Let me fix that for you. Commented Oct 21, 2018 at 19:04

## 1 Answer

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.