Metric induces topology

Statement

Suppose $(X,d)$ is a metric space. Then, the collection of subsets:

$B(x,r):=\{y\in X\mid d(x,y)<r\}$

form a basis for a topology on $X$ . These are often called the open balls of $X$ .

Definitions used

Metric space

Further information: metric space

A metric space $(X,d)$ is a set $X$ with a function $d:X\times X\to \mathbb {R}$ satisfying the following:

$d(x,y)\geq 0\ \forall \ x,y\in X$ (non-negativity)
$d(x,y)=0\iff x=y$ (identity of indiscernibles)
$d(x,y)=d(y,x)$ (symmetry)
$d(x,y)+d(y,z)\geq d(x,z)\ \forall \ x,y,z\in X$ (triangle inequality)

Basis for a topological space

Further information: Basis for a topological space

A collection of subsets $\{U_{i}\}_{i\in I}$ of a set $X$ is said to form a basis for a topological space if the following two conditions are satisfied:

$\bigcap _{i\in I}U_{i}=X$
For any $i,j\in I$ , and any $p\in U_{i}\cap U_{j}$ , there exists $U_{k}\subset U_{i}\cap U_{j}$ such that $p\in U_{k}$ .

Note that this is the definition for a collection of subsets that can form the basis for some topology.

Proof

It suffices to show the following two things:

The space $X$ is the union of subsets of the form $B(x,r)$
Given two sets $B(x,r)$ and $B(y,s)$ , and any $z\in B(x,r)\cap B(y,s)$ , there exists $t>0$ such that $B(z,t)\subset B(x,r)\cap B(y,s)$ (this suffices because $z\in B(z,t)$ for any $t$ , since $d(z,z)=0$ ).

Proof that the union is the whole space

For any $x\in X$ , and any $r>0$ , we have, because $d(x,x)=0$ :

$x\in B(x,r)$

Thus, in particular, we have:

$x\in \bigcup _{r>0}B(x,r)$

Taking the union over all $x$ , we get:

$X=\bigcup _{x\in X,r>0}B(x,r)$

Proof for intersection of two

Consider two balls $B(x,r)$ and $B(y,s)$ , where $r,s>0$ and $x,y\in X$ . (Note that $x,y$ may be equal). Suppose $z\in B(x,r)\cap B(y,s)$ . Then, by definition of the balls, we have: