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.

Let $(X,d)$ be a metric space. Suppose that $(X,d)$ has the property that every family of non-empty, pairwise disjoint open sets in $X$ is countable. Is $X$ second countable then?

share|improve this question
    
Look at math.stackexchange.com/questions/90427/… –  William Jun 8 '12 at 5:38
    
In the metrizable space $X$, the following conditions are equivalent, cf. General Topology by Engelking Page 255: > $X$ is seperable > $X$ is second countable > $X$ is lindelof > $X$ has countable extent > $X$ is star countable > $X$ is CCC –  Paul May 20 '13 at 2:57

2 Answers 2

You’re asking whether every ccc metric space is separable; the answer is yes.

Let $X$ be a ccc metric space. 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 $X$ is ccc, 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
    
The proof of the question is short and nice. –  Paul May 20 '13 at 2:54

I'll give an answer without using a metrization theorem.

First show that $X$, a ccc metric space, is separable. To see this, for every $x \in X, r > 0$, let $B(x, r)$ be the open ball of radius $r$ around $x$. For every $n$, Zorn's lemma implies that the family of all open balls of radius $\frac{1}{n}$ has a maximal (in the sense of inclusion) pairwise disjoint subfamily $\mathcal{B}_n = \{ B(x_i, \frac{1}{n}): i \in I_n, x_i \in X \}$. This family is countable, so all $I_n$ are countable index sets.

I claim that $D = \cup_{n \in N} \{ x_i: i \in I_n \}$ is dense in $(X,d)$. To see this, we only need to see that every $B(x,r)$ intersects $D$; let $n$ be such that $\frac{2}{n} < r$, if $B(x,r)$ would miss $I_n$, all points $x_i \in I_n$ would have distance $r > \frac{2}{n}$ from $x$, and so we could have added $B(x, \frac{1}{n})$ to $\mathcal{B}_n$ and contradicted maximality. So $B(x,r)$ intersects $I_n$ and so $D$ as well.

Now, it is well-known that a separable metric space has a countable base, and one can show this quite easily: if $D$ is a countable dense subset, then all balls with radius in $\mathbb{Q}$ and centre in $D$ form a countable open base.

share|improve this answer

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.