A Souslin scheme on a set $X$ is family $(A_{s})_{s \in{\omega^{<\omega} }}$ of subsets of $X$. If $(X,d)$ is a metric space, we say again that $(A_s)$ has vanishing diameter if diam$(A_{x| n })$ as $n\rightarrow{\infty}$, for all $x \in {\mathcal{N}}$. Again, in this case, let $D=\{ x: \bigcap_{n} A_{x|n} \neq 0 \}$ and for $x \in D$, $\{f(x)\}=\bigcap_{n} A_{x|n}$. We call $f:D \to X$ the associated map.
Questions: If each $A_s$ is open and $A_s \subseteq \bigcup_{i}A_{s \hat{}i}$ for all $ s \in\omega^{<\omega}$, then $f$ is open.?
thanks.
For $s\in{^{<\omega}\omega}$ let $B_s=\{y\in D:s\subseteq y\}$; it suffices to show that $f[B_s]$ is open in $X$. Clearly $f[B_s]\subseteq A_s$, which is open, so in fact it suffices to show that $f[B_s]=A_s$.
Suppose that $x\in A_s$, where $s\in{^n\omega}$. Let $s_n=s$. Suppose that $x\in A_{s_m}$ for some $s_m\in{^m\omega}$ such that $s_m\upharpoonright k=s_k$ for $n\le k\le m$. By hypothesis there is an $i_m\in\omega$ such that $x\in A_{{s_m}^\frown i_m}$, and we set $s_{m+1}={s_m}^\frown i_m$. Let $y=\bigcup_{m\ge n}s_m\in\mathcal{N}$; clearly $y\upharpoonright m=s_m$ for $m\ge n$, so $x\in\bigcap_{m\in\omega}A_{y\upharpoonright m}$, $y\in B_s$, and $f(y)=x$. Thus, $A_s\subseteq f[B_s]$, and the result follows: $f$ is an open map.