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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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    Space filling curve2011-05-14
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    Are you looking for a continuous function, or just a function?2011-05-14
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    I guess one can consider the vector projection: http://en.wikipedia.org/wiki/Vector_projection2011-05-14
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    Continuous sets?2011-05-14
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    @PEV. But that's not bijective...2011-05-14
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    @Chandra: space-filling curves are continuous, so they aren't bijections, as noted here: http://en.wikipedia.org/wiki/Space-filling_curve#Properties2011-05-14
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    @Chandru1: But that's not bijective...2011-05-14
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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.2011-05-14
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    @mac: a curve being continuous doesn't mean it's not bijective. ("space filling curves are continuous, **so** they aren't..." What I'm saying is the use of "so" in this way is conveying that continuity implies non-bijection. But you're right: space-filling curves are continuous. Also, they are surjective (hence the "filling" portion of space-filling). But not injective, hence not bijective.2011-05-14
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    @Amy J.M.: sorry, what I wrote wasn't that clear. I meant: "Space-filling curves are continuous; by the argument here there is no continuous bijection from the interval to the square, so space-filling curves aren't bijections."2011-05-14
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    A space-filling curve from a side of a square could fill a square, but perhaps what's being asked for here is whether a bijective, continuous function exists from the side of a square to an square (not filled/solid)?2011-05-14
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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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    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.2011-05-14
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    True, but this doesn't really address the issues being raised by pi_is_good's question.2011-05-14
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    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.2011-05-14
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    What is delta IxI ?2011-05-14
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    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.2011-05-14
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    @Theo: Sure. I was taking "side" to mean "boundary".2011-05-14