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:

$d(x,z)<r,d(y,z)<s$

Define:

$t:=\min\{r-d(x,z),s-d(y,z)\}$

Then, $t>0$. We want to claim that $B(z,t)$ lies completely inside $B(x,r)\cap B(y,s)$.

Let's prove this. Suppose $p\in B(z.t)$. Then:

$d(z,p)<t\leq r-d(x,z)$

By the triangle inequality and an application of the above we have:

$d(x,p)\leq d(x,z)+d(z,p)<d(x,z)+r-d(x,z)=r$

Thus, $p\in B(x.r)$. Analogously, $p\in B(y,s)$. This completes the proof.