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Prove that if a topological space $(X, T)$ is metrizable then it is metrizable in infinitely many ways. $$$$As the given topological space is metrizable so there exists a metric $d$ on the set $X$ such that the it can generate a class of open subsets which is the topology $T$. Now as we know that whenever $d$ is a metric on $X$, then the function satisfying $$d'(x, y)=\frac{d(x, y)}{1+d(x, y)}$$, is also a metric on $X$. Now suppose $A$ be an open subset of $X$ under the metric $d$. Now chose a $x \in A$, then there exists some $\epsilon$ such that for all $y$ satisfying $$d(x, y) < \epsilon$$ lie in $A$. Now for the metric $d'$ we see that $\frac{\epsilon}{1+\epsilon}$ works and for all $y$ satisfying $$d'(x, y) < \frac{\epsilon}{1+\epsilon}$$ satisfy the above equation and hence lie in the set $A$ and hence $A$ is also open under the metric $d'$. So the class of open sets generated by the metric $d$ can also be generated by the metric $d'$ and hence $d'$ can also induce the topology $T$. Similarly we can find infinitely many metric like $d''$ satisfying $$d''(x, y)=\frac{d'(x, y)}{1+d'(x, y)}$$. And hence the topological space $(X, T)$ is metrizable in many ways. $$$$Is the proof Correct??

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3 Answers 3

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The statement is false: a set with a single element admits one and only one metric.

If we assume that $X$ has more than one element, then, although your proof works, I think that it is simpler to say that, if $d$ is a metric on $X$, then, for each $k>0$, $kd$ is another metric on $X$ which induces the same topology.

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  • $\begingroup$ A singleton set can have many metrics $\endgroup$
    – user728159
    Oct 27, 2020 at 7:27
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    $\begingroup$ @user728159: No, it can’t: the only metric on $\{x\}$ is the function $$d:\{x\}\times\{x\}\to\Bbb R:\langle x,x\rangle\mapsto 0\,.$$ $\endgroup$ Oct 27, 2020 at 7:30
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You proved that open sets in $d$ are open in $d'$. But you also have to prove the converse. For this you just have to replace $\frac {\epsilon} {1+\epsilon}$ by $\frac {\epsilon} {1-\epsilon}$ in your argument (taking $\epsilon <1)$. Except for this your construction of the metrics is is fine.

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The idea is fine, but you really should include an actual demonstration that if $d'(x,y)<\frac{\epsilon}{1+\epsilon}$, then $d(x,y)<\epsilon$ and hence $y\in A$. If

$$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}<\frac{\epsilon}{1+\epsilon}\,,$$

then

$$(1+\epsilon)d(x,y)<\epsilon\big(1+d(x,y)\big)\,,$$

so

$$d(x,y)+\epsilon d(x,y)<\epsilon+\epsilon d(x,y)\,,$$

and hence $d(x,y)<\epsilon$.

More important, you also need to show that $d$-open sets are $d'$-open. If we let $\epsilon'=\frac{\epsilon}{1+\epsilon}$, we can solve for $\epsilon$ to find that $\epsilon=\frac{\epsilon'}{1-\epsilon'}$, a fact that should suggest how to do this.

There is, however, an easier way to get infinitely many different equivalent metrics. (It does require that $X$ have at least two points, but so does any approach.) Let $x$ and $y$ be two distinct points of $X$, and let $r=d(x,y)$. For each $s\in(0,r)$ define a metric $d_s$ on $X$ by setting $$d_s(u,v)=\min\{d(u,v),s\}$$ for all $u,v\in X$. It’s easy to verify that $d_s$ and $d$ generate the same topology, since they have the same open balls of all radii less than $s$, and they are clearly distinct, because $d_s(x,y)=s$ for each $s\in(0,r)$.

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