proof of equivalent definitions of analytic sets for Polish spaces

Let $A$ be a nonempty subset of a Polish space $X$. Then, letting $𝒩$ denote Baire space and $Y$ be any uncountable Polish space, we show that the following are equivalent.

1. $A$ is $ℱ$-analytic (http://planetmath.org/AnalyticSet2).

2. $A$ is the projection (http://planetmath.org/GeneralizedCartesianProduct) of a closed subset of $X\times 𝒩$ 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 $f:𝒩\to X$.

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 $ℱ$ be the paving consisting of closed subsets of $X$. The collection of $𝒦$-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 $ℱ$-analytic.Then, there is a closed $S\subseteq 𝒩$ and a function $\theta :\mathbb{N}\to ℱ$ 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 $s\in 𝒩$ with $s_n=m$. Then, we can rearrange the above expression to get $A={\pi }_X(B)$ where ${\pi }_X:X\times 𝒩\to X$ 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 $X\times 𝒩$ 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 $X\times 𝒩$, where ${\pi }_X:X\times 𝒩\to X$ 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 $h:𝒩\to Z$. The result follows by taking $f=g\circ h$.

(4) implies (5):Suppose that $A$ is the image of a continuous function $g:𝒩\to X$. Since uncountable Polish spaces are all Borel isomorphic (see Polish spaces up to Borel isomorphism), there is a Borel isomorphism $h:Y\to 𝒩$. 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.