TITLE: Homeomorphic and Isometric Spaces
QUESTION [5 upvotes]: Problem
I'm currently studying metric spaces, and the lectruer's notes make the remark:
Clearly $(0,1)$, $(0,\infty)$ and $\mathbb{R}$ are homeomorphic under the standard metrics, but no two of them are isometric
I want to be able to show this for myself (it isn't a "clear" to me as it apparently should be!)
Progress
To show $(0,1)$, $(0,\infty)$ are homeomorphic, I think we can just take the map $\frac{1-x}{x}$, which is a bijection with continuous inverse $\frac{1}{1+x}$.
To show $(0,\infty)$, $\mathbb{R}$ are homeomorphic, we can take the map $\tan(\pi (x-\frac{1}{2}))$, with continuous inverse $\frac{1}{\pi}\arctan(x)+\frac{1}{2}$.
Now, to show $(0,1)$, $\mathbb{R}$ are homeomorphic, can we just take a composition of the previous two maps? I think this works.
I am not yet able to show that the spaces are not isometric however. I would be very grateful if someone could show me how this is done.
Thanks.

REPLY [5 votes]: In $(0,\infty)$ we have that $d(1,3)=2$ this is not possible in $(0,1)$. In $\mathbb R$ for a fixed $x \in \mathbb R$ the equation $d(x,y)=2$ has two unique solutions. But $d(1,y)=2$ has only one solution in $(0,\infty)$.