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.
2. There is a closed subset $S$ of $\mathcal{N}$ and $\theta\colon \mathbb{N}^2\to\mathcal{F}$ such that \begin{equation*} A=\bigcup_{s\in S}\bigcap_{n=1}^\infty \theta\left(n,s_n\right). \end{equation*}
3. There is a closed subset $S$ of $\mathcal{N}$ and $\theta\colon \mathbb{N}\to\mathcal{F}$ such that \begin{equation*} A=\bigcup_{s\in S}\bigcap_{n=1}^\infty \theta\left(s_n\right). \end{equation*}
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 $Y$

() implies (): 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 \begin{equation*} B=\bigcap_{n=1}^\infty\bigcup_{m=1}^\infty A_{n,m}\times K_{n,m} \end{equation*}for $A_{n,m}\in\mathcal{F}$ and $K_{n,m}\in\mathcal{K}$ Rearranging this expression, \begin{equation*} B=\bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^\infty A_{n,s_n}\times K_{n,s_n}. \end{equation*}So, defining $S\subseteq\mathcal{N}$ by \begin{equation*} S=\left\{s\in\mathcal{N}\colon \bigcap_{n=1}^\infty K_{n,s_n}\not=\emptyset\right\}. \end{equation*}gives \begin{equation*} A=\pi_X(B)=\bigcup_{s\in S}\bigcap_{n=1}^\infty A_{n,s_n}. \end{equation*}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 $k\ge 0$ then $s^r_n=s_n$ for all $n\le k$ and large enough $r$ Hence, \begin{equation*} \bigcap_{n\le k}K_{n,s_n}=\bigcap_{n\le k}K_{n,s^r_n}\supseteq \bigcap_{n=1}^\infty K_{n,s_n}\not=\emptyset. \end{equation*}So, the collection of sets $K_{n,s_n}$ for $n=1,2,\ldots$ satisfies the finite intersection property, and compactness of the paving $\mathcal{K}$ gives \begin{equation*} \bigcap_{n=1}^\infty K_{n,s_n}\not=\emptyset, \end{equation*}showing that $s\in S$ and that $S$ is indeed closed.

() implies (): 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, \begin{equation*} 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) \end{equation*}as required.

() implies (): 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 \begin{equation*} A_s=\begin{cases} \theta(s_n),&\textrm{if $s=t|_n$ for some $t\in \mathcal{N}$},\\ \emptyset,&\textrm{otherwise}. \end{cases} \end{equation*}Then, for $s\in\mathcal{N}$ \begin{equation*} \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} \end{equation*}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 \begin{equation*} \bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^\infty A_{s|_n}=\bigcup_{s\in S}\bigcap_{n=1}^\infty\theta(s_n) = A \end{equation*}as required.

() implies (): 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 $m\le n$ 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 $\bigcap_nK_{s^n}$ 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{equation*}\begin{split} B&=\bigcap_{n=1}^\infty\bigcup_{s\in\mathbb{N}^n}A_s\times K_s\\ &=\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}\\ &=\bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^\infty A_{s|_n}\times K_{s|_n}. \end{split}\end{equation*}The projection of $B$ onto $X$ is then \begin{equation*} \pi_X(S)=\bigcup_{s\in\mathcal{N}}\bigcap_{n=1}^\infty A_{s|_n}, \end{equation*}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.

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

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