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I need to find an isomorphism between [0,1] to (0,1) can you help me with this please?

thanks. benny

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Possible duplicate: math.stackexchange.com/questions/418/… –  Nuno Nov 26 '10 at 15:28
    
Also let me know if the tag elementary-set-theory here is inappropriate. –  Nuno Nov 26 '10 at 15:36

6 Answers 6

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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.

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This duplicates part of the answer by Martin Brandenburg posted 8 hours ago: math.stackexchange.com/questions/9274/… –  Jonas Meyer Nov 8 '10 at 0:40

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.

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sorry, I meant sets –  benjamin Nov 7 '10 at 15:29
    
@user3224: an Isomorphism is a special Homomorphism, which is a structure-preserving map between two algebraic structures. so, what is the structure on those sets? or do you only want any bijective map? –  comonad Nov 7 '10 at 15:42
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isomorphism in the category of sets = bijection, and this is meant here. sets are also algebraic structures (with empty signature). –  Martin Brandenburg Nov 7 '10 at 16:28

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.

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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.

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Isn't this one of the tricks behind the Banach-Tarski decomposition? I seem to remember some discussion of this in a Mathematical Intelligencer article... –  Steven Stadnicki Nov 7 '10 at 17:53
    
It is used. In Wagon's excellent book, The Banach-Tarski Paradox, one of the first things proved is equivalent to this, a bijection between the unit circle and the unit circle missing a point. He uses rotations that are multiples of 1 radian –  Ross Millikan Nov 7 '10 at 18:00

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. ;)

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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.

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"It's also completely useless and nonconstructive. " Indeed ;) –  Martin Brandenburg Nov 7 '10 at 16:51

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