# 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.

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For $(0,1)$ and $\mathbb{R}$ or $(0,\infty)$ – think of possible distances. For $\mathbb{ℝ}$ and $(0,\infty)$, remove an unbounded, connected set that isn't the whole space. – k.stm Jan 22 '13 at 17:51
By the way, I think you mean to prove that $(0,1)$ and $\mathbb{ℝ}$ are homeomorphic with tangens. Then it'd be correct. But more elementary, you can compose $(0,1) \to (-1,1),\; x \mapsto 2x - 1$ and $(-1,1) \to \mathbb{R},\; y \mapsto \tfrac{y}{1-\lvert y \rvert}$ as the former is obviously a homeomorphism and the latter is a (well-defined) continuous map with continuous inverse $\mathbb{R} \to (-1,1),\; z \mapsto \tfrac{z}{1+\lvert z \rvert}$. – k.stm Jan 22 '13 at 18:05
Markdown comment: you can create a header using #, or html codes <h1>, <h2>, etc. Blockquote doesn't look right for this purpose. – user53153 Jan 22 '13 at 19:07
@5PM: Thanks for the advice, I'll be sure to use that for future posts. – Mathmo Jan 22 '13 at 22:10

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 $(0,\infty)$.
Hint: To show that the two spaces are not isometric note that in $(0,1)$ no two points have distance $1$, whereas in $\mathbb R$ a lot of them have an even greater distance. To show that $(0,\infty)$ and $\mathbb R$ are not isometric you can show that adding a new point to the "left" of the space we have $[0,\infty)$ which is a metric space, and $[-\infty,+\infty)$ which is not a metric space.
You could also do it by showing that if $f$ is an isometry then it is continuous and we can always choose $\varepsilon=\delta$ in the proof; whereas in any homeomorphism between the given spaces this would be impossible.
Discussion: This is because one topology may be compatible with many different metrics, and homeomorphism is a topological property, whereas isometry deals with a particular metric and is a much stricter requirement. This is similar to how $\mathbb N$ and $\mathbb Z$ are both isomorphic as sets, but not as ordered sets (with their natural orders, of course).