proof of equivalent definitions of analytic sets for Polish spaces
Let $A$ be a nonempty subset of a Polish space^{} $X$. Then, letting $\mathcal{N}$ denote Baire space^{} and $Y$ be any uncountable Polish space, we show that the following are equivalent^{}.

1.
$A$ is $\mathcal{F}$analytic^{} (http://planetmath.org/AnalyticSet2).

2.
$A$ is the projection (http://planetmath.org/GeneralizedCartesianProduct) of a closed subset of $X\times \mathcal{N}$ onto $X$.

3.
$A$ is the image (http://planetmath.org/DirectImage) of a continuous function^{} $f:Z\to X$ for some Polish space $Z$.

4.
$A$ is the image of a continuous function $f:\mathcal{N}\to X$.

5.
$A$ is the image of a Borel measurable function $f:Y\to X$.

6.
$A$ is the projection of a Borel subset of $X\times Y$ onto $X$.
(1) implies (2): Let $\mathcal{F}$ be the paving consisting of closed subsets of $X$. The collection^{} of $\mathcal{K}$analytic sets^{} contains the Borel $\sigma $algebra of $X$ (see countable unions and intersections of analytic sets are analytic) and, as the analytic sets are given by a closure operator^{} (http://planetmath.org/AnalyticSetsDefineAClosureOperator) it follows that it contains all analytic subsets of $X$. So, any analytic subset $A$ of $X$ is $\mathcal{F}$analytic. Then, there is a closed $S\subseteq \mathcal{N}$ and a function $\theta :\mathbb{N}\to \mathcal{F}$ such that
$$A=\bigcup _{s\in S}\bigcap _{n=1}^{\mathrm{\infty}}\theta ({s}_{n})$$ 
(see proof of equivalent definitions of analytic sets for paved spaces). For each $m,n\in \mathbb{N}$ let ${K}_{m,n}$ denote the closed subset of $s\in \mathcal{N}$ with ${s}_{n}=m$. Then, we can rearrange the above expression to get $A={\pi}_{X}(B)$ where ${\pi}_{X}:X\times \mathcal{N}\to X$ is the projection map and
$$B=(X\times S)\cap \bigcap _{n=1}^{\mathrm{\infty}}\bigcup _{m=1}^{\mathrm{\infty}}\theta (m)\times {K}_{m,n}.$$ 
It is easily seen that ${\bigcup}_{m}\theta (m)\times {K}_{m,n}$ is a closed subset of $X\times \mathcal{N}$ for each $n$, and therefore $B$ is closed, as required.
(2) implies (3): Suppose that $A={\pi}_{X}(S)$ for a closed subset $S$ of $X\times \mathcal{N}$, where ${\pi}_{X}:X\times \mathcal{N}\to X$ is the projection map. As the product^{} of Polish spaces is Polish, and every closed subset of a Polish space is Polish, then $S$ will be a Polish space under the subspace topology. So, we can take $Z=S$ and let $f:Z\to X$ be the restriction^{} of ${\pi}_{X}$ to $Z$.
(3) implies (4): Suppose that $A$ is the image of a continuous function $g:Z\to X$, for a Polish space $Z$. As Baire space is universal for Polish spaces, there exists a continuous and onto (http://planetmath.org/Surjective^{}) function $h:\mathcal{N}\to Z$. The result follows by taking $f=g\circ h$.
(4) implies (5): Suppose that $A$ is the image of a continuous function $g:\mathcal{N}\to X$. Since uncountable Polish spaces are all Borel isomorphic (see Polish spaces up to Borel isomorphism), there is a Borel isomorphism $h:Y\to \mathcal{N}$. The result follows by taking $f=g\circ h$.
(5) implies (6): Suppose that $A$ is the image of a Borel measurable function $f:Y\to X$, and let $\mathrm{\Gamma}$ be its graph (http://planetmath.org/Graph2)
$$\mathrm{\Gamma}\equiv \{(f(y),y):y\in Y\}\subseteq X\times Y.$$ 
The projection of $\mathrm{\Gamma}$ onto $X$ is equal to $f(Y)=A$, so the result will follow once it is shown that $\mathrm{\Gamma}$ is a Borel set.
Choose a countable^{} and dense subset $\{{x}_{1},{x}_{2},\mathrm{\dots}\}$ of $X$, and let $d$ be a metric generating the topology^{} on $X$. Then, for integers $m,n\ge 1$, denote the open ball^{} about ${x}_{m}$ of radius $1/n$ by ${B}_{m,n}$. Since the ${x}_{m}$ form a dense set, ${\bigcup}_{m}{B}_{m,n}=X$ for each $n$. Let us define
$${\mathrm{\Gamma}}_{n}\equiv \bigcup _{m=1}^{\mathrm{\infty}}{B}_{m,n}\times {f}^{1}({B}_{m,n})\subseteq X\times Y,$$ 
which contains $\mathrm{\Gamma}$. Furthermore, since ${f}^{1}({B}_{m,n})$ are Borel, ${\mathrm{\Gamma}}_{n}$ are Borel sets. Suppose that $(x,y)\in {\bigcap}_{n}{\mathrm{\Gamma}}_{n}$. Then, for each $n$, there is an $m$ such that $x\in {B}_{m,n}$ and $y\in {f}^{1}({B}_{m,n})$. So,
$$d(x,f(y))\le d(x,{x}_{m})+d({x}_{m},f(y))\le 2/n.$$ 
This holds for all $n$, showing that $y=f(x)$ and so $(x,y)\in \mathrm{\Gamma}$. We have shown that $\mathrm{\Gamma}={\bigcap}_{n}{\mathrm{\Gamma}}_{n}$ is Borel, as required.
(6) implies (1): This is an immediate consequence of the result that projections of analytic sets are analytic.
Title  proof of equivalent definitions of analytic sets for Polish spaces 

Canonical name  ProofOfEquivalentDefinitionsOfAnalyticSetsForPolishSpaces 
Date of creation  20130322 18:48:48 
Last modified on  20130322 18:48:48 
Owner  gel (22282) 
Last modified by  gel (22282) 
Numerical id  5 
Author  gel (22282) 
Entry type  Proof 
Classification  msc 28A05 