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## Characterization of separability

Prove that a metric space ${X}$ is separable if and only if there does not exist an uncountable set ${S \subset X}$ such that ${d(x,y)\geq c>0}$ for every ${x,y \in S,\ x \neq y}$.

This is a useful property which makes trivial the proof of the fact that any subset of a separable space is separable.

Proof: One implication is trivial. If ${X}$ is separable, then there cannot exist an uncountable set with pairwise “big” distances. The argument would be that if ${X}$ is separable, then in each ball ${B(x,c/2)}$ there should be an element of the countable dense set, and this is impossible, since this is an uncountable family of disjoint balls.

Suppose now that there does not exist an uncountable set with big enough pairwise distances. That means that whenever we find a set ${S}$ such that ${d(x,y) \geq c>0}$ for every ${x \neq y \in S}$ we have that ${S}$ is at most countable.

Pick now ${c>0}$ and denote

$\displaystyle \mathcal{F}_c =\{ S \subset X : d(x,y)\geq c,\ \forall x\neq y \in S\}.$

For every ${c}$ we know that ${\mathcal{F}_c}$ os not empty (pick for example a two point set, and we will prove using Zorn’s lemma that each ${\mathcal{F}_c}$ has maximal elements. Indeed, order each ${\mathcal{F}_c}$ by inclusion and consider a totally ordered subfamily ${\mathcal{C}=(S_i)_{i \in I} \subset \mathcal{F}_c}$. Pick now ${S = \cup_{i \in I}S_i}$. Obviously ${S}$ is an upper bound for every ${S_i}$, and moreover, ${S \in \mathcal{F}_c}$, since if ${x,y \in S}$ then there exist ${i,j \in I}$ such that ${x \in S_i,y \in S_j}$ and by total order assumption we have, for example ${S_i \subset S_j}$, which means that ${d(x,y) \geq c}$. Since each chain has an upper bound, it follows by Zorn’s lemma that the set ${\mathcal{F}_c}$ admits maximal elements. (a maximal element ${M}$ has the property that whenever ${M\subset N}$ we necessarily have ${M=N}$; note that maximal elements are not necessarily unique.)

For each ${c>0}$ rational, consider a maximal element ${M_c}$ of ${\mathcal{F}_c}$. Then ${M=\cup_{c \in \Bbb{Q}^+} M_c}$ is a countable subset of ${X}$ which I claim is dense. To prove this pick ${x_0 \in X}$. Since ${M_c}$ is maximal, there exists at least one ${x_c \in M_c}$ such that ${d(x_c,x_0) < c}$. We do this for every positive rational ${c}$ and we find a sequence ${(x_c) \in M}$ with the property that ${d(x_c,x_0). Taking ${c \rightarrow 0}$ we see that ${x_c \rightarrow x_0}$. By the arbitrairiness of ${x_0}$ we can say that ${M}$ is dense in ${X}$ and therefore ${X}$ is separable.