# Existence of schemes Souslin in separable spaces.

If $X$ is nonempty separable we can build a Souslin scheme $U_{s}$ such that $U_{\emptyset}=X$, $U_{s}$ open nonempty, $\overline{U}_{s\hat {}i}\subseteq U_{s}$, $U=\bigcup_{i}U_{s\hat {}i}$ and diam$(U_{s}) \leq 2 ^{-length(s)}$ if $s\neq \emptyset$.?

Where 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.

So suppose $(X,d)$ is a separable metric space.
Suppose we have defined $U_s$ for all $s$ with length$(s) = |s| = n$ for some $n \ge 0$. Fix $s$ of length $n$ in $\omega^{<\omega}$, we then have $U_s$ defined already, and this is in itself a separable metric space. For each $x \in U_s$ we take a small ball $B(x,r(x))$ such that $\overline{B(x,r(x))} \subseteq U_s$ and $\operatorname{diam}(B(x,r(x))) \le 2^{-(n+1)}$. This can be done by regularity of $U_s$ and using that the diameter of a ball of radius $r$ is at most $2r$ so we can always choose the radius $r$ small enough.
The cover $\{B(x,r): x \in U_s \}$ of $U_s$ has a countable subcover (as a separable metric space is Lindelöf) say $\{B(x_n, r(x_n)): n \in \mathbb{N}\}$. Then define $U_{s||i} = B(x_i, r(x_i))$. This obeys all the requirements: they together cover $U_s$ and we define all $U_t$ with $|t|=n+1$ this way (as these are unique extensions of a sequence of length one less). This defines a Souslin scheme by recursion.