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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?

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  • $\begingroup$ When you say $X\times X\to \Bbb R$, I assume you mean the metric $d$. It helps to be clear about that. Same for $X'\times X'\to \Bbb R$. Tell us that $d$ is the function you're referring to, so we don't have to guess. $\endgroup$ – Arthur May 14 '15 at 22:11
  • $\begingroup$ Yes, I'll fix that. $\endgroup$ – Tim Raczkowski May 14 '15 at 22:11
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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 $\tau'$.

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  • $\begingroup$ Thanks Dan. I this is much cleaner. I knew I was missing something simple. $\endgroup$ – Tim Raczkowski May 16 '15 at 15:19

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