Homeomorphic metric spaces where the identity map is not a homeomorphism

Question

In our Topology lectures, we were told that there are examples where two metric spaces $(M, d)$ and $(M, d')$ are homomeomorphic (i.e. there exists some homeomorphism between them, which is not necessarily the identity map) but the identity map $\operatorname{Id}\colon(M, d) \rightarrow (M, d')$ is not a homeomorphism. I was hoping someone could provide an example of this.

Answer 1

Take any metric space $(X,d)$ for which there is a bijective discontinuous function $f\colon M\longrightarrow M$ and define $d'(x,y)=d\bigl(f(x),f(y)\bigr)$. Then $f$ is a homeomorphism (actually, it's an isometry) from $(M,d')$ into $(M,d)$. However, the identity is not a homeomorphism from $(M,d')$ into $(M,d)$.

Answer 2

Let $M=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, and let $d$ be the usual metric on $M$. Define a new metric $d'$ on $M$ as follows:

$$d'(x,y)=\begin{cases} |x-y|,&\text{if }0<x,y<1\text{ or }x=y\text{ or }\{x,y\}=\{0,1\}\\ x,&\text{if }0<x<y=1\\ y,&\text{if }0<y<x=1\\ 1-x,&\text{if }0=y<x<1\\ 1-y,&\text{if }0=x<y<1 \end{cases}$$

This simply interchanges the rôles of $0$ and $1$: the map

$$h:M\to M:x\mapsto\begin{cases} x,&\text{if }0<x<1\\ 1,&\text{if }x=0\\ 0,&\text{if }x=1 \end{cases}$$

is a homeomorphism between $\langle M,d\rangle$ and $\langle M,d'\rangle$ in either direction, but the identity map is not: $0$ is the only non-isolated point with respect to $d$, while $1$ is the only non-isolated point with respect to $d'$.

Answer 3

$X = [0,1)$.

$d =$ usual distance.

$d'(x,y) = |x - y|$ for $x,y\in(0,1)$;

$d(0,x) = |x - 1|$ for $x\in(0,1)$.

Idea for $d'$: take $(0,1]$ and rename $1$ as $0$.