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Is there a simple geometric proof that there exists a continuous bijective function between a square and its side?

And is there some explicit continuous function or formula $f^1(z)\mapsto (x,y)$ and $f(x,y)\mapsto z$, with $(x,y) \in [0,1]\times[0,1]$ and $z \in [0,1]$?

And is there a constructive proof that two sets of the same cardinality have a bijective function?

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marked as duplicate by Lord_Farin, Vedran Šego, Martin Sleziak, user1729, Tobias Kildetoft Oct 1 '13 at 12:07

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Space filling curve –  user9413 May 14 '11 at 17:28
    
Are you looking for a continuous function, or just a function? –  wckronholm May 14 '11 at 17:30
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Continuous sets? –  Asaf Karagila May 14 '11 at 17:31
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@Chandra: space-filling curves are continuous, so they aren't bijections, as noted here: en.wikipedia.org/wiki/Space-filling_curve#Properties –  mac May 14 '11 at 17:33
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If you want a bijection, that's easy. If you want a continuous surjection that's doable. A continuous bijection is impossible though. –  JSchlather May 14 '11 at 17:33
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As Chandru1 commented, there are several space filling curves. These on the other hand are usually only surjective. With Cantor-Schröder-Bernstein's Theorem we find a bijection. This is not geometric though and I couldn't imagine to have a geometric proof for this, because $[0,1]$ and $[0,1]^2$ are not homeomorphic. As for the last question, having a bijection between two sets is the definition of having the same cardinality.

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As edited, the answer to the first question is that such a continuous bijection doesn't exist.

The fact that sets of the same cardinality are in bijective correspondence is my definition of "same cardinality", so I'm not sure what the second question is really asking.

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A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Since $I\times I$ is not homeomorphic to $\partial(I\times I)$, there are no maps like the ones you are looking for.

To address your last question, it is usually taken as a definition that two sets have the same cardinality iff there is a bijection between them.

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Easier: If you remove a point from the interior of $I$ then it becomes disconnected. If you remove a point from the interior if $I \times I$ then it remains connected. –  t.b. May 14 '11 at 17:42
    
True, but this doesn't really address the issues being raised by pi_is_good's question. –  wckronholm May 14 '11 at 17:43
    
Ah, I interpreted "its side" as a side not "its boundary" in view of the second paragraph. If the latter is meant, I agree with you. –  t.b. May 14 '11 at 17:46
    
What is delta IxI ? –  user10903 May 14 '11 at 17:52
    
The $\partial$ symbol is used in topology (and elsewhere) to refer to the boundary of a space. So here, $\partial(I\times I)$ is the sides of a square. –  wckronholm May 14 '11 at 18:00
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