Using homeomorphism to induce an equivalent metric.

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Suppose $(X,d_1)$ and $(Y,d_2)$ are homeomorphic.Then $\exists f:X\to Y$ homeomorphism.Define $d$ on $X$ by,

$d(x,y)=d_2(f(x),f(y))$ which is a metric on $X$ as $f$ is injective.Now,$f$ is an isometry between $(X,d)$ and $(Y,d_2)$ and hence the topology $\tau_d=\{f^{-1}(U):U\in \tau_2\}$.But $(X,d_1)$ and $(X,d_2)$ have homeomorphism $f$ between them.So,$ \{f^{-1}(U):U\in \tau_2\}=\tau_1$.So,$(X,d_1)$ and $(X,d)$ are equivalent metrics.So,using a homeomorphism between two metric spaces,we can construct an equivalent metric.

Can someone tell me some nice applications of this result?

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It shows e.g. that we can give $\Bbb R$ and equivalent non-complete metric and similarly, we can give $(0,1)$ an equivalent complete metric. It shows how we cannot expect metric notions to be preserved between homeomorphic metric spaces, but that studying isometries and uniform equivalences has some merit in that context. Completeness is "non-topological" in that sense. (Though there is a more general notion of topological completeness, which is more appropriate for the purely topological setting).

The "transport of metric" idea also shows that being metrisable is a topological property.