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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?

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  • $\begingroup$ Look at math.stackexchange.com/questions/90427/… $\endgroup$ – William Jun 8 '12 at 5:38
  • $\begingroup$ 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 $\endgroup$ – Paul May 20 '13 at 2:57
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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.

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  • $\begingroup$ The proof of the question is short and nice. $\endgroup$ – Paul May 20 '13 at 2:54
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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.

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