Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
see here and here (and remember that a metrizable space is separable if and only if it is second countable). –  t.b. Sep 3 '12 at 19:26
add comment

2 Answers

up vote 4 down vote accepted

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.)

share|improve this answer
    
Is it Suslin or Souslin? Google search inconclusive. –  Matt N. Sep 3 '12 at 19:55
3  
@Matt: It’s really Суслин; Suslin is the preferred transliteration nowadays, and Souslin is a traditional French-based transliteration. Both are acceptable. –  Brian M. Scott Sep 3 '12 at 19:59
1  
Thanks Brian, interesting. Perhaps knowing foreign languages other than English does come in handy from time to time. –  Matt N. Sep 4 '12 at 8:10
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.