Prove that countable intersection of analytic sets is an analytic set

Question

Analytic set is a continuous image of polish spaces. I don't know how to prove that the intersection of countable many analytic sets is analytic.

Answer 1

Let $A_1,A_2,...$ be analytic subsets of a Polish space $Y$. Let $f_n:X_n \to A_n$ be continuous and surjective, where $X_n$ is a Polish space for each $n$. Let $X =\prod_n X_n$. Let $Z$ consist of all sequences $\{(x_n)\}$ in $\prod_n X_n$ such that $f_1(x_1)=f_2(x_2)=...$. Then $Z$ is a closed subset of $\prod_n X_n$ and hence it is a Polish space. Define $f: Z \to Y$ by $f(z_n)=f_1(z_1)$. Then $f$ is continuous and the image under $f$ of $Z$ is precisely $\cap_n A_n$.