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 $X$.

Proof

To prove that the subsets form a basis for a topology, we need to prove the following fact: the intersection of two open balls is a union of open balls. Equivalently, given two open balls $B(x,r)$ and $B(y,s)$, and $z\in B(x,r)\cap B(y,s)$, then there exists some radius $t$ such that $B(z,t)\subset B(x,r)\cap B(y,s)$.

It turns out that the following works for $t$:

$t:=\min \left(r-d(x,z),r-d(y,z)\right)$

This essentially follows from the triangle inequality.