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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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    @Amy To an filled square2011-05-14

3 Answers 3

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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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    @Theo: Sure.$I$was taking "side" to mean "boundary".2011-05-14