Isomorphism between [0,1] to (0,1) I need to find an isomorphism between [0,1] to (0,1) can you help me with this please?
thanks. benny
 A: The maps $(0,1) \to [0,1], x \mapsto x$ and $[0,1] \to (0,1), x \mapsto 1/4 + x/2$ are injective. The theorem of Cantor-Schröder-Bernstein implies that there is a bijection. It is actually possible to work out the proof of the theorem in these simple examples and get an explicit bijection, namely
$g : [0,1] \to (0,1), x \mapsto 1/2 \pm 1/2^{n+2}$ if $x = 1/2 \pm 1/2^{n+1}$ for some $n$, and $x \mapsto x$ else.
See also this article if you know german. ;)
A: You can identify a countably infinite collection of points starting with 0 and converging to some limit less than 1/2.  Just shift each point one down the line.  Now repeat from the top end.  All points not involved go to themselves.
A: What do you mean by "isomorphism"? Are you considering these sets with a given group or ring structure (if then, which one?), or as topological spaces (in which case you'd be looking for a "homeomorphism"), or just as sets?
I don't know of a canonical group structure on either set, but if you want your map between the two to be continuous you might have some luck by looking at the notion of compactness and how it behaves under continuous functions.
A: If you just want a bijection there are many ways to do it. Here's a rather contrived one:
Let $\alpha$ be a limit ordinal of the same cardinality as $(0,1)$. By the Cantor-Bernstein-Schröder theorem there exists a bijective map $f: (0,1) \to \alpha$. Let $S$ denote the successor "function" (I know it's not a function - hence the quotes). Note that $\alpha$ does not have a maximal element, so it makes sense to define $\phi: [0,1] \to (0,1)$ by
$$\phi(x) = \begin{cases}
(f^{-1} \circ S \circ S \circ f)(x) & \text{ for $x \in (0,1)$}, \\
f^{-1}(1) & \text{ for $x = 1$}, \\
f^{-1}(0) & \text{ for $x = 0$}.
\end{cases}$$
Showing that $\phi$ is a bijection is an easy exercise. It's also completely useless and nonconstructive. 
A: The bijection 1/x turns (0,1) into (1,$\infty$) which you can chop up into countably many (]'s. Also you can split [] into [)(] and these can be divide countably many times to bisect with the (]'s but it flips the ordering around.
A: The easiest way (if you can apply Cantor-Bernstein-Schroder theorem) is to find a bijection from $[0,1]$ onto a subset, say $[1/4,3/4]$; that is easy to do. Since the identity map from $(0,1)$ to $[0,1]$ is a bijection, it follows from the theorem that the two sets are isomorphic.
