I've been working on problems dealing with the Souslin property. A topological space $X$ has the Souslin property if every pairwise disjoint family of non-empty open subsets of $X$ is countable.
I came across this problem, and I am having trouble with formulating a proof.
A metrizable space $X$ has the Souslin property if and only if it has a countable base.
Can anyone help? Thanks in advance!
Let $\langle X,d\rangle$ be a non-separable metric space. If $A\subseteq X$ is countable, then $\operatorname{cl}A\ne X$, so we can choose a point $x\in X\setminus A$. This means that we can recursively construct a set $\{x_\xi:\xi<\omega_1\}\subseteq X$ such that if $F_\eta=\operatorname{cl}\{x_\xi:\xi<\eta\}$ for each $\eta<\omega_1$, then $x_\xi\notin F_\xi$ for $\xi<\omega_1$. For each $\xi<\omega_1$ there is an $n(\xi)\in\omega$ such that $B\left(x_\xi,2^{-n(\xi)}\right)\cap F_\xi=\varnothing$. For $k\in\omega$ let $A_k=\{\xi<\omega_1:n(\xi)=k\}$; there is some $m\in\omega$ such that $A_m$ is uncountable. Show that $$\left\{B\left(x_\xi,2^{-(m+1)}\right):\xi\in A_m\right\}$$ is an uncountable family of pairwise disjoint open sets. Conclude that if a metrizable space has the Suslin property, it must be separable. (The other direction is trivial.)
I've been trying to come up with a simpler proof to answer your question( Brain's answer is of course completed and great but I still fancied a simpler proof). So here it is:
If $X$ has countable base, then it is easy to see that $X$ has Souslin property. Now let $X$ be a metric space with Souslin property. By the Bing metrization theorem $X$ has a $\sigma$-discrete base $\mathscr{B}=\bigcup_n\mathscr{B}_n$, where each $\mathscr{B}_n$ is discrete. Since $X$ has Souslin property , any discrete family of open sets in $X$ is countable, so each of the families $\mathscr{B}_n$ is countable, and therefore the base $\mathscr{B}$ is countable as well.
