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:

${\begin{aligned}\quad \|\mathbf {x} -\mathbf {y} \|={\sqrt {(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+...+(x_{n}-y_{n})^{2}}}\end{aligned}}$

Then it can be shown similarly that the following set is dense in $\mathbb {R} ^{n}$:

${\begin{aligned}\quad \mathbb {Q} ^{n}=\{(x_{1},x_{2},...,x_{n})\in \mathbb {R} ^{n}:x_{1},x_{2},...,x_{n}\in \mathbb {Q} \}\end{aligned}}$

Licensing

Content obtained and/or adapted from:

• Dense Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license
• Separable Metric Spaces, mathonline.wikidot.com under a CC BY-SA license