Separable Metric Spaces

Dense Sets in a Metric Space

We will now look at a new concept regarding metric spaces known as dense sets which we define below.

Definition: Let $(M,d)$ be a metric space and let $S\subseteq M$ . Then $S$ is said to be Dense in $M$ if for every $x\in M$ and for every $r>0$ we have that $B(x,r)\cap S\neq \emptyset$ , i.e., every open ball in $(M,d)$ contains a point of $S$ .

In any metric space $(M,d)$ the whole set $M$ is always dense in $M$ . Furthermore, the empty set $\emptyset$ is not dense in $M$ .

For a less trivial example, consider the metric space $(\mathbb {R} ,d)$ where $d$ is the usual Euclidean metric defined for all $x,y\in \mathbb {R}$ by $d(x,y)=\mid x-y\mid$ , and consider the subset $\mathbb {Q} \subset \mathbb {R}$ of rational numbers.

The set $\mathbb {Q}$ is dense in $\mathbb {R}$ because for any open ball, i.e., for any $x\in \mathbb {R}$ and for any $r>0$ we have that the open interval $(x-r,x+r)$ contains a rational number.

For a counterexample, consider the set $\mathbb {Z} \subset \mathbb {R}$ of integers. We claim that $\mathbb {Z}$ is not dense in $\mathbb {R}$ . To show this, consider the following ball:

{\begin{aligned}\quad B\left({\frac {1}{2}},{\frac {1}{2}}\right)=(0,1)\end{aligned}}

Clearly $\mathbb {Z} \cap (0,1)=\emptyset$ and so $\mathbb {Z}$ is not dense in $\mathbb {R}$ .

We will now look at a nice theorem which tells us that for a metric space $(M,d)$ a set $S\subseteq M$ is dense in $M$ if and only if its closure equals $M$ .

Theorem 1: Let $(M,d)$ be a metric space and let $S\subseteq M$ . Then, $S$ is dense in $M$ if and only if ${\bar {S}}=M$ .

Recall that ${\bar {S}}$ denotes the closure of $S$ , and we defined the closure of $S$ to be the set of adherent points of $S$ .

Proof: $\Rightarrow$ Suppose that $S$ is dense in $M$ . Then for all $x\in M$ and all $r>0$ we have that:

{\begin{aligned}\quad B(x,r)\cap S\neq \emptyset \end{aligned}}

So every $x\in M$ is an adherent point of $S$ . The set of all adherent points of $S$ is the closure of $S$ , so ${\bar {S}}=M$ .

$\Leftarrow$ Suppose that ${\bar {S}}=M$ . Then every point of $M$ is an adherent point of $S$ , i.e., for all $x\in M$ and for all $r>0$ we have that:

{\begin{aligned}\quad B(x,r)\cap S\neq \emptyset \end{aligned}}

Therefore $S$ is dense in $M$ . $\blacksquare$

Separable Metric Spaces

Recall that if $(M,d)$ is a metric space then a subset $S\subseteq M$ is said to be dense in $M$ if for every $x\in M$ and for all $r>0$ we have that:

{\begin{aligned}\quad B(x,r)\cap S\neq \emptyset \end{aligned}}

In other words, $S$ is dense in $M$ if every open ball contains a point of $S$ .

We will now look at a special type of metric space known as a separable metric space which we define below.

Definition: A metric space $(M,d)$ is said to be Separable if there exists a countable dense subset $S$ of $M$ .

For example, consider the metric space $(\mathbb {R} ,d)$ where $d$ is the usual Euclidean metric defined for all $x,y\in \mathbb {R}$ by $d(x,y)=\mid x-y\mid$ . Then the subset $\mathbb {Q}$ is dense in $\mathbb {R}$ since every open interval contains rational numbers.

In fact, in general, the metric space $(\mathbb {R} ^{n},d)$ where $d$ is the usual Euclidean metric defined for all $\mathbf {x} =(x_{1},x_{2},...,x_{n}),\mathbf {y} =(y_{1},y_{2},...,y_{n})\in \mathbb {R} ^{n}$ by: