Let $X$ be a metric space with metric $d$.
(a) Show that $d:X\times X\to\Bbb R$ is continuous.
I've shown this already.
(b) Let $X'$ denote a space having the same underlying set as $X$. Show that if $d:X'\times X'\to\Bbb R:(x,y)\mapsto d(x,y)$ is continuous, then the topology of $X'$ is finer than the topology of $X$.
For $x_0\in X$, $B(x_0,\epsilon)$ can be identified with $(\{x_0\}\times X)\cap d^{-1}([0,\epsilon))$. Note that $[0,\epsilon)$ is relatively open in the subspace $\Bbb R_{\ge0}$.
By hypothesis $(\{x_0\}\times X')\cap d^{-1}([0,\epsilon))$ is relatively open, hence $B(x_0,\epsilon)=\pi_2((\{x_0\}\times X')\cap d^{-1}([0,\epsilon)))$ is an open set of $X'$.
I'm fairly sure this argument works, but not entirely confident. Is this right? Is there an easier way?
Best Answer
We have the family of maps indexed by $x\in X$
$$ X \stackrel {d(x,-)} \longrightarrow\mathbb R.$$
The preimages of open sets here forms a basis for the topology $\tau$ of the metric space $X$. If this map is continous wrt to another topology $\tau'$, that means that all open sets of $\tau$ are contained already in $\tau'$.