4
$\begingroup$

This a problem from Topology - James Munkres.

Let $X$ be a metric space with metric $d$. Let X' denote a space having the same underlying set as $X$. Show that if d: X'\times X' \rightarrow \mathbb{R} is continuous, then the topology of X' is finer than the topology of $X$.

Can anybody provide hints to solve this problem? Thanks in advance.

1 Answers 1

4

We say that $\tau$ is finer than \tau' if every open set of of \tau' is open in $\tau$.

The question, if so, simply asks to show that every open set of $X$ is open in X'. Of course it is enough to show this for basic open sets, in our case $B(x,\epsilon)=\{y\mid d(x,y)<\epsilon\}$ sort of sets.

Fix x\in X' and use the assumption that $f(y)=d(x,y)$ is a continuous function on X' to deduce that $B(x,\epsilon)$ is open in X'.