The identity map in two metric spaces is a homeomorphism

Question

We have given two metric spaces (M,$\tau_{d}$) and (M,$\tau_h$) whereby the metric $h$ is given as:

$h(x,y)=\frac{d(x,y)}{1+d(x,y)}$.

Now I have to show that the function

$id_{M}$ : $(M, \tau_{d}) \rightarrow (M,\tau_h)$ which sends $x \rightarrow x$ is a homeomorphism.

I don't know if it would suffice to say that the identity function is a homeomorphism following the fact that it is a bijection and continuous and would we therefore be done or if there might be a more specific, elegant way.

Thanks in advance

Edit: I realised that my problem is the fact that I would only know how to show that the identity is contnuous in the case that we have the same metric spaces. But how can you show it if we have two metric spaces with two different induced topologies?

Just show that $d(x_n,x) \to 0$ if and only if $h(x_n,x) \to 0$. — Kavi Rama Murthy, Nov 17 '20 at 11:38
Are you assuming that every continuous bijection is a homeomorphism? — José Carlos Santos, Nov 17 '20 at 11:38
@JoséCarlosSantos that and the inverse function also has to be continuous, so 3 conditions — Annalisa, Nov 17 '20 at 11:55
So, yes, the way to go is to prove that $\operatorname{id}_M$ is continuous in both directions. — José Carlos Santos, Nov 17 '20 at 11:57
@JoséCarlosSantos but how do we do it in the case that the metrics are not the same — Annalisa, Nov 17 '20 at 12:27
Do you know the definition of continuity of a function $f : X \to Y$ between two metric spaces $(X,\tau_g)$ and $(Y,\tau_h)$, where the metrics $g : X \times X \to [0,\infty)$ and $h : Y \times Y \to [0,\infty)$ are given? If you do, apply that to the situation where $X=Y$ and $f = \text{Id}$. — Lee Mosher, Nov 17 '20 at 13:34
$ x/(1+x)$ is a monotone continuous function on the positive reals — lhf, Jul 20 '22 at 00:08

José Carlos Santos · Answer 1 · 2022-07-19T23:21:50.113

2

Those two metrics induce the same topology. In other words, those are equivalent metrics. You will find a proof of that assertion here. So, since the continuity of a map depends only upon the topologies that you are working with, $\operatorname{id}_M$ is continuous in both directions.

edited Jul 19 '22 at 23:21

answered Nov 17 '20 at 13:27

José Carlos Santos

427,504

i thank you for your answer, however i am still not fully clear how from the definition of the continuity in topological spaces we concluded that idM is continuous in both directions.. – Annalisa Nov 17 '20 at 20:47
This is done in two steps. First: those two distances induce the same topology. Second: on any topological space $(M,\tau)$, $\operatorname{id}_M$ is a homeomorphism from $(M,\tau)$ onto itself. Is it clear now? – José Carlos Santos Nov 17 '20 at 20:55

The identity map in two metric spaces is a homeomorphism

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