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 topology the on $X$ induced by $d$. Also, a topological space $X$ is a pseudometrizable topological space if there exists a pseudometric $d$ on $X$ whose pseudometric topology coincides with the given topology for $X$ [1, 2].

Proposition 1.

$ℬ$ is a base for a topology.

Proof.

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

First, as $d(x,x)=0$ for all $x\in X$, it follows that $ℬ$ 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 $d$ is symmetric. Therefore, we have, in fact, also shown that any quasimetric induces a topology.

Title	pseudometric topology
Canonical name	PseudometricTopology
Date of creation	2013-03-22 14:40:47
Last modified on	2013-03-22 14:40:47
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	7
Author	matte (1858)
Entry type	Definition
Classification	msc 54E35
Defines	pseudometrizable
Defines	pseudometric topology
Defines	pseudo-metric
Defines	pseudometrizable topological space
Defines	pseudo-metrizable topological space