# 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. 1.

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

2. 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. 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. 4.

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

5. 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. 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. 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 ProofOfEquivalentDefinitionsOfAnalyticSetsForPavedSpaces 2013-03-22 18:48:36 2013-03-22 18:48:36 gel (22282) gel (22282) 4 gel (22282) Proof msc 28A05