proof of equivalent definitions of analytic sets for paved spaces

Let $(X,\mathcal{F})$ be a paved space with $\emptyset\in\mathcal{F}$ , let $\mathcal{N}$ 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 $\mathcal{F}$ -analytic (http://planetmath.org/AnalyticSet2).
2.

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

$A=\bigcup_{s\in S}\bigcap_{n=1}^{\infty}\theta\left(n,s_{n}\right).$
3.

There is a closed subset $S$ of $\mathcal{N}$ and $\theta\colon\mathbb{N}\to\mathcal{F}$ such that

$A=\bigcup_{s\in S}\bigcap_{n=1}^{\infty}\theta\left(s_{n}\right).$
4.

$A$ is the result of a Souslin scheme on $\mathcal{F}$ .
5.

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

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

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

B=\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}A_{n,m}\times K_{n,m}

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

B=\bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^{\infty}A_{n,s_{n}}\times K_{n,s_{n}}.

So, defining $S\subseteq\mathcal{N}$ by

S=\left\{s\in\mathcal{N}\colon\bigcap_{n=1}^{\infty}K_{n,s_{n}}\not=\emptyset% \right\}.

gives

A=\pi_{X}(B)=\bigcup_{s\in S}\bigcap_{n=1}^{\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 $s^{1},s^{2},\ldots$ be a sequence in $S$ converging to a limit $s\in\mathcal{N}$ . For any $k\geq 0$ then $s^{r}_{n}=s_{n}$ for all $n\leq k$ and large enough $r$ . Hence,

\bigcap_{n\leq k}K_{n,s_{n}}=\bigcap_{n\leq k}K_{n,s^{r}_{n}}\supseteq\bigcap_% {n=1}^{\infty}K_{n,s_{n}}\not=\emptyset.

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

\bigcap_{n=1}^{\infty}K_{n,s_{n}}\not=\emptyset,

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

(2) implies (3): Supposing that $A$ satisfies the required expression, choose any bijection $\phi\colon\mathbb{N}\to\mathbb{N}^{2}$ . Then define $\tilde{\theta}\equiv\theta\circ\phi$ and $f\colon\mathcal{N}\to\mathcal{N}$ by $f(s)=t$ where $t_{n}=\phi^{-1}(n,s_{n})$ . 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}(\phi^{-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 $n\geq 1$ and $s\in\mathbb{N}^{n}$ let us set

A_{s}=\begin{cases}\theta(s_{n}),&\textrm{if $s=t|_{n}$ for some $t\in\mathcal% {N}$},\\ \emptyset,&\textrm{otherwise}.\end{cases}

Then, for $s\in\mathcal{N}$ ,

\bigcap_{n=1}^{\infty}A_{s|_{n}}=\begin{cases}\bigcap_{n=1}^{\infty}\theta(s_{% n}),&\textrm{if $s\in S$},\\ \emptyset,&\textrm{otherwise}.\end{cases}

Here, if $s\not\in S$ , 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, $A_{s|_{n}}=\emptyset$ . The result of the Souslin scheme $(A_{s})$ is then

\bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^{\infty}A_{s|_{n}}=\bigcup_{s\in S}% \bigcap_{n=1}^{\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, $\mathcal{C}=\{0,1\}^{\mathbb{N}}$ , which is a compact Polish space. Then, for any $s\in\mathbb{N}^{n}$ , let $K_{s}$ be the set of $t\in\mathcal{C}$ such that $t_{k}=1$ if $k=s_{1}+\cdots+s_{m}$ for some $m\leq n$ and $t_{k}=0$ for all other $k<s_{1}+\cdots+s_{n}$ . These are closed and, therefore, compact sets.

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

\begin{split}\displaystyle B&\displaystyle=\bigcap_{n=1}^{\infty}\bigcup_{s\in% \mathbb{N}^{n}}A_{s}\times K_{s}\\ &\displaystyle=\bigcup_{s^{1}\in\mathbb{N}^{1},s^{2}\in\mathbb{N}^{2},\ldots}% \bigcap_{n=1}^{\infty}A_{s^{n}}\times K_{s^{n}}\\ &\displaystyle=\bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^{\infty}A_{s|_{n}}\times K% _{s|_{n}}.\end{split}

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

\pi_{X}(S)=\bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^{\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.

Title	proof of equivalent definitions of analytic sets for paved spaces
Canonical name	ProofOfEquivalentDefinitionsOfAnalyticSetsForPavedSpaces
Date of creation	2013-03-22 18:48:36
Last modified on	2013-03-22 18:48:36
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	4
Author	gel (22282)
Entry type	Proof
Classification	msc 28A05