proof of equivalent definitions of analytic sets for paved spaces

Let $(X,ℱ)$ be a paved space with $\mathrm{\varnothing }\in ℱ$, let $𝒩$ be Baire space, and let $Y$ be any uncountable Polish space. For a subset $A$ of $X$, we show that the following statements are equivalent.

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

2. There is a closed subset $S$ of $𝒩$ and $\theta :{\mathbb{N}}^2\to ℱ$ such that

   $$A=\bigcup _{s\in S}\bigcap _{n=1}^{\mathrm{\infty }}\theta (n,s_n).$$

3. There is a closed subset $S$ of $𝒩$ and $\theta :\mathbb{N}\to ℱ$ such that

   $$A=\bigcup _{s\in S}\bigcap _{n=1}^{\mathrm{\infty }}\theta \left(s_n\right).$$

4. $A$ is the result of a Souslin scheme on $ℱ$.

5. $A$ is the projection of a set in ${(ℱ\times 𝒦)}_{\sigma \delta }$ onto $X$, where $𝒦$ is the collection of compact subsets of $Y$.

6. $A$ is the projection of a set in ${(ℱ\times 𝒢)}_{\sigma \delta }$ onto $X$, where $𝒢$ is the collection of closed subsets of $Y$.

(1) implies (2):As $A$ is analytic, there exists a compact paved space $(K,𝒦)$ and a set $B\in {(ℱ\times 𝒦)}_{\sigma \delta }$ such that $A={\pi }_X(B)$, where ${\pi }_X:X\times K\to X$ is the projection map.Write

$$B=\bigcap _{n=1}^{\mathrm{\infty }}\bigcup _{m=1}^{\mathrm{\infty }}{A}_{n,m}\times K_{n,m}$$

for $A_{n,m}\in ℱ$ and $K_{n,m}\in 𝒦$.Rearranging this expression,

$$B=\bigcup _{s\in 𝒩}\bigcap _{n=1}^{\mathrm{\infty }}{A}_{n,s_n}\times K_{n,s_n}.$$

So, defining $S\subseteq 𝒩$ by

$$S=\{s\in 𝒩:\bigcap _{n=1}^{\mathrm{\infty }}{K}_{n,s_n}\ne \mathrm{\varnothing }\}.$$

gives

$$A={\pi }_X(B)=\bigcup _{s\in S}\bigcap _{n=1}^{\mathrm{\infty }}{A}_{n,s_n}.$$

Setting $\theta (n,m)=A_{n,m}$ gives the required expression, and it only remains to show that $S$ is closed.So, let $s^1,s^2,\mathrm{\dots }$ be a sequence in $S$ converging to a limit $s\in 𝒩$. For any $k\ge 0$ then $s_n^r=s_n$ for all $n\le k$ and large enough $r$. Hence,

$$\bigcap _{n\le k}{K}_{n,s_n}=\bigcap _{n\le k}{K}_{n,s_n^r}\supseteq \bigcap _{n=1}^{\mathrm{\infty }}{K}_{n,s_n}\ne \mathrm{\varnothing }.$$

So, the collection of sets $K_{n,s_n}$ for $n=1,2,\mathrm{\dots }$ satisfies the finite intersection property, and compactness (http://planetmath.org/PavedSpace) of the paving $𝒦$ gives

$$\bigcap _{n=1}^{\mathrm{\infty }}{K}_{n,s_n}\ne \mathrm{\varnothing },$$

showing that $s\in S$ and that $S$ is indeed closed.

(2) implies (3):Supposing that $A$ satisfies the required expression, choose any bijection $\varphi :\mathbb{N}\to {\mathbb{N}}^2$. Then define $\tilde{\theta }\equiv \theta \circ \varphi$ and $f:𝒩\to 𝒩$ by $f(s)=t$ where $t_n={\varphi }^{-1}(n,s_n)$. As $S$ is closed, it follows that $\tilde{S}=f(S)$ will also be closed and,

$$A=\bigcup _{s\in S}\bigcap _n\theta (n,s_n)=\bigcup _{s\in S}\bigcap _n\tilde{\theta }({\varphi }^{-1}(n,s_n))=\bigcup _{s\in \tilde{S}}\bigcap _n\tilde{\theta }(s_n)$$

as required.

(3) implies (4):Suppose that $A$ satisfies the required expression and define a Souslin scheme $(A_s)$ as follows. For any $n\ge 1$ and $s\in {\mathbb{N}}^n$ let us set

Then, for $s\in 𝒩$,

Here, if $s\notin S$, we have used the fact that $S$ is closed to deduce that for large $n$, there is no $t\in S$ with ${t|}_n={s|}_n$ and, therefore, $A_{{s|}_n}=\mathrm{\varnothing }$.The result of the Souslin scheme $(A_s)$ is then

$$\bigcup _{s\in 𝒩}\bigcap _{n=1}^{\mathrm{\infty }}{A}_{{s|}_n}=\bigcup _{s\in S}\bigcap _{n=1}^{\mathrm{\infty }}\theta (s_n)=A$$

as required.

(4) implies (5):Suppose that $A$ is the result of a Souslin scheme $(A_s)$. Let us first consider the case where $Y$ is Cantor space, $𝒞={\{0,1\}}^{\mathbb{N}}$, which is a compact Polish space.Then, for any $s\in {\mathbb{N}}^n$, let $K_s$ be the set of $t\in 𝒞$ such that $t_k=1$ if $k=s_1+\mathrm{\cdots }+s_m$ for some $m\le n$ and $t_k=0$ for all other .These are closed and, therefore, compact sets.

Given any sequence $s^1\in {\mathbb{N}}^1,s^2\in {\mathbb{N}}^2,\mathrm{\dots }$ it is easily seen that ${\bigcap }_n{K}_{s^n}$ is nonempty if and only if there is an $s\in 𝒩$ such that ${s|}_n=s^n$ for all $n$.Define the set $B$ in ${(ℱ\times 𝒦)}_{\sigma \delta }$ by

The projection of $B$ onto $X$ is then

$${\pi }_X(S)=\bigcup _{s\in 𝒩}\bigcap _{n=1}^{\mathrm{\infty }}{A}_{{s|}_n},$$

which is the result $A$ of the scheme $(A_s)$ as required.The result then generalizes to any uncountable Polish space $Y$, as all such spaces contain Cantor space (http://planetmath.org/UncountablePolishSpacesContainCantorSpace).

(5) implies (6):This is trivial, since all compact sets are closed.

(6) implies (1):This is a consequence of the result that projections of analytic sets are analytic.