# Metrizable Topological Space In Many Ways

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??

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.

• A singleton set can have many metrics Oct 27, 2020 at 7:27
• @user728159: No, it can’t: the only metric on $\{x\}$ is the function $$d:\{x\}\times\{x\}\to\Bbb R:\langle x,x\rangle\mapsto 0\,.$$ Oct 27, 2020 at 7:30

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.

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 $$s\in(0,r)$$.