# Why does metric space which has the countable chain condition implies separable?

I've looked around but all I could find is that if X is separable then X has ccc. Can anybody give me some help?

## 3 Answers

Let $X$ be a (non-empty) c.c.c. metric space. Let us temporarily fix a positive integer $n\geq 1$.

Let $\mathscr{F}_n$ be the family of classes of open balls of radius $1/n$ (in $X$) which are pairwise disjoint. In other words, an element of $\mathscr{F}_n$ is a class $\mathcal{C}$, whose elements are open balls, say $B^\mathcal{C}_i$, of radius $1/n$, and such that $B^\mathcal{C}_i\cap B^\mathcal{C}_j=\varnothing$ if $i\neq j$.

We can partially order $\mathscr{F}_n$ by inclusion. It's an easy exercise to verify that every chain in $\mathscr{F}_n$ has an upper bound (given, obviously, by the union of the elements of the chain) - and an even easier one to show that $\mathscr{F}_n\neq\varnothing$- . By Zorn's Lemma, $\mathscr{F}_n$ has a maximal element, say $\mathcal{C}_n$. Since open balls are open and $X$ is c.c.c., $\mathcal{C}_n$ is enumerable. Also, since it's elements are balls, we can find point $x^n_1,x^n_2,\ldots\in X$ such that $$\mathcal{C}_n=\left\{B_{1/n}(x^n_1),B_{1/n}(x^n_2),\ldots\right\}$$ (where $B_r(x)=\left\{y\in X:d(x,y)<r\right\}$, as usual)

Now, we can do the above procedure for each positive integer $n\geq 1$, thus finding maximal classes $\mathcal{C}_n$ of pairwise disjoint balls of radius $1/n$ and centers $x^n_1, x^n_2,\ldots$ Now, let $D$ be the set of all such $x^n_i$. Then $D$ is obviously enumerable, and we claim that it is dense in $X$.

Indeed, if it weren't, we could find a point $y\in X$ and a positive integer $N\geq 1$ such that $B_{2/N}(y)\cap D=\varnothing$. Hence, for all $x\in D$, we have $B_{1/N}(x)\cap B_{1/N}(y)=\varnothing$. Now, we can consider the set $C'_N=\left\{B_{1/N}(y)\right\}\cup C_N$. It is obvious then that $C'_N\in\mathscr{F}_N$ and $C'_N$ is strictly larger than $C_N$, contradicting it's maximality.

We can then conclude that $D$ is an enumerable dense set in $X$ which is therefore separable.

A proof not using $\sigma$-discrete bases: let $(X,d)$ be a metric space with ccc. Then for each $n \in \mathbb{N}$, let $D_n = \{ B(x, \frac{1}{n}) : x \in C_n \}$, be a maximal (by inclusion: we cannot enlarge this family) disjoint family of open balls in $X$, all of radius $\frac{1}{n}$. The set $C_n$ is here (by definition) the set of centres of these balls. A simple Zorn's lemma application shows that such a family exists, for each $n$. As $X$ has ccc, each $C_n$ is a countable set.

I claim that $D = \cup_n D_n$ is dense in $X$. To this end, we show that $D$ intersects each ball $B(x,\frac{1}{k})$. Suppose not. Then in particular, for each $y \in C_{2k} \subset D$, $d(x,y) \ge \frac{1}{k}$, which implies that the ball $B(x, \frac{1}{2k})$ is disjoint from all balls $B(y, \frac{1}{2k})$, where $y \in C_{2k}$, and this in turn means that we could have extended $D_{2k}$ by adding the ball $B(x, \frac{1}{2k})$ to it, contradicting the way we chose the $D_n$. This contradiction completes the proof, as $D$ is a countable dense set in $X$.

Let $X$ be a ccc metric space. Since $X$ is metric, it has a $\sigma$-discrete base $\mathscr{B}=\bigcup_{n\in\omega}\mathscr{B}_n$. In particular, each $\mathscr{B}_n$ is a pairwise disjoint family of open sets, hence countable, and $\mathscr{B}$ is therefore also countable. Thus, $X$ is second countable and hence separable.

More generally, if $c(X)=\kappa$, then each of the countably many discrete families of a $\sigma$-discrete base for $X$ has cardinality at most $\kappa$, so $w(X)\le\kappa$. On the other hand, it’s always true that $c(X)\le w(X)$, so if $X$ is metric we always have $c(X)=w(X)$. And since

$$c(X)\le d(X)\le hd(X)\le nw(X)\le w(X)$$

in any space, we get equality of cellularity, density, hereditary density, net weight, and weight in metric spaces.