# Topology induced by metric space

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.

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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 $X'$.