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

Star
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    Do you know how to construct a bijection $[0,1] \to [0,1]^2$? – MooS May 08 '17 at 11:52
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    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 $X \to X^2$. – MooS May 08 '17 at 11:57
  • Could you explain me this method? – Star May 08 '17 at 12:01
  • Or you just think about it yourself for a few minutes: $X \to X^2$ yields $X^2=X \times X \to X \times X^2=X^3$ and then just compose... – MooS May 08 '17 at 12:03
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    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? – Star May 08 '17 at 12:15
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    Let $g(t,u)=(x,y,u)$ to get an iso $[0,1]^2 \to [0,1]^3$. Then compose it. – MooS May 08 '17 at 12:25

2 Answers2

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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.

yberman
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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.

5xum
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  • Is this at all related to the Schröder-Bernstein Theorem? I only heard of this result with injective functions. – aras May 08 '17 at 11:48
  • 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. – MooS May 08 '17 at 11:51
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    @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. – 5xum May 08 '17 at 11:52