pseudometric topology

Let $(X,d)$ be a pseudometric space. As in a metric space, we define

for $x\in X$, $\epsilon >0$.

In the below, we show that the collection of sets

$$ℬ=\{B_{\epsilon }(x)\mid \epsilon >0,x\in X\}$$

form a base for a topology for $X$. We call this topologythe on $X$induced by $d$. Also,a topological space $X$ is a pseudometrizable topological spaceif there exists a pseudometric $d$ on $X$ whosepseudometric topology coincides with the given topologyfor $X$ [1, 2].

Proposition 1.

$ℬ$ is a base for a topology.

Proof.

We shall use the http://planetmath.org/node/5845this resultto prove that $ℬ$ is a base.

First, as $d(x,x)=0$ for all $x\in X$, it followsthat $ℬ$ is a cover.Second, suppose $B_1,B_2\in ℬ$ and $z\in B_1\cap B_2$.We claim that there exists a $B_3\in ℬ$ such that

$z$$\in$$B_3\subseteq B_1\cap B_2.$(1)

By definition, $B_1=B_{{\epsilon }_1}(x_1)$and $B_2=B_{{\epsilon }_2}(x_2)$ for some $x_1,x_2\in X$and ${\epsilon }_1,{\epsilon }_2>0$. Then

Now we can define $\delta =\min \{{\epsilon }_1-d(x_1,z),{\epsilon }_2-d(x_2,z)\}>0$, and put

$$B_3=B_{\delta }(z).$$

If $y\in B_3$, then for $k=1,2$, we have by the triangle inequality

$d(x_k,y)$$\le$$d(x_k,z)+d(z,y)$

$d(x_k,z)+\delta$

$\le$${\epsilon }_k,$

so $B_3\subseteq B_k$ and condition 1 holds.∎

Remark

In the proof, we have not used the fact that $d$ issymmetric. Therefore, we have, in fact, also shown that anyquasimetric induces a topology.