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This question already has an answer here:

Is there any bijection from $[0,1]^3$ to $[0,1]$? How can I construct it?

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marked as duplicate by Mark McClure, tilper, Willie Wong, carmichael561, Community May 8 '17 at 18:46

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    $\begingroup$ Do you know how to construct a bijection $[0,1] \to [0,1]^2$? $\endgroup$ – MooS May 8 '17 at 11:52
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    $\begingroup$ Then I do not get your problem. There is a trivial (and constructive) way to get a bijection $X \to X^3$ if you have already established a bijection $X \to X^2$. $\endgroup$ – MooS May 8 '17 at 11:57
  • $\begingroup$ Could you explain me this method? $\endgroup$ – STF May 8 '17 at 12:01
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    $\begingroup$ Thanks, but I am still confused: for $t\in [0,1]$ the injection that I have gives me $f(t)=(x,y)\in [0,1]^2$. How can I proceed from here? $\endgroup$ – STF May 8 '17 at 12:15
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    $\begingroup$ Let $g(t,u)=(x,y,u)$ to get an iso $[0,1]^2 \to [0,1]^3$. Then compose it. $\endgroup$ – MooS May 8 '17 at 12:25
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Hint:

If there exists a surjection between $A$ to $B$ and a surjection between $B$ to $A$, then there exists a bijection between $A$ to $B$. In your case, space filling curves are surjections from $[0,1]$ to $[0,1]^3$. It should be easy to find a surjection going the other way.

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  • $\begingroup$ Is this at all related to the Schröder-Bernstein Theorem? I only heard of this result with injective functions. $\endgroup$ – aras May 8 '17 at 11:48
  • $\begingroup$ Using the Axiom of Choice it is obvious that the formulation with injective functions is equivalent to the statement with surjective functions. But still this does not answer the question, since the OP asked for some constructive method. $\endgroup$ – MooS May 8 '17 at 11:51
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    $\begingroup$ @STF You do know about surjections from $[0,1]$ to $[0,1]^3$. You even said so in your question. The problem with my hint is that it can't be used to construct a bijection, just to prove it exists. $\endgroup$ – 5xum May 8 '17 at 11:52
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I'm not sure if this is correct, but I like the idea and would like to see how it is broken if it is.

Given any real number you can express it uniquely using its canonical continued fraction. So from three numbers you can produce three sequences. You can interleave these three sequences ($s_1, t_1, u_1, s_2, t_2, u_2, \ldots)$ and evaluate it to a real number.

Perk: This is constructive for three rational numbers.

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