I constructed a bijection by using decimal expansions of two real numbers and taking numbers 1 by 1 consecutively. (It took me hours to come up with this). I remember someone saying appealing to some sort of expansion is the only way to do this, and I think this type of method can never be continuous. Is there anyway to prove that no such function exists?

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    $\begingroup$ I don't know, but continuous is probably too weak as a condition. In fact there is the Peano curve, which is an example of continuous surjection $[0,1] \to [0,1] \times [0,1]$. $\endgroup$
    – Andrea
    Apr 10 '13 at 12:23
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    $\begingroup$ en.wikipedia.org/wiki/Space-filling_curve $\endgroup$ Apr 10 '13 at 12:23
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    $\begingroup$ This doesn't answer your question, but... "I constructed a bijection by using decimal expansions of two real numbers and taking numbers 1 by 1 consecutively." -- this is reminiscent of a trap I fell down when trying to solve this problem for the first time. Make sure you didn't assume that $\mathbb{R}$ is countable or anything like that! $\endgroup$ Apr 10 '13 at 12:23
  • $\begingroup$ To expand on Willie Wong's link, the Peano curve $[0,1]\rightarrow [0,1]\times[0,1]$ can easily be extended to a continuous surjection $(0,1)\rightarrow\mathbb{R}^2$. Or in your case, by a composition by homeomorphism on the domain, a continuous surjection $\mathbb{R}\rightarrow\mathbb{R}^2$. $\endgroup$
    – Dan Rust
    Apr 10 '13 at 13:11
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    $\begingroup$ See "No differentiable space-filling curve can exist." and this proof $\endgroup$ Apr 10 '13 at 13:16

Yes, there exists a continuous surjection from $\mathbb R$ to $\mathbb R^2$. The following is a simple way to construct one, although there should be more elegant constructions. Let $f:[0,1]\to[-1,1]^2$ be a spacefilling curve which starts and ends at the origin, for example the Sierpinski curve (although one could of course start with any space-filling curve $[0,1]\to[0,1]^2$, obtain a new space-filling curve $[0,1]\to[-1,1]^2$ by translation and scaling, and then append paths to the beginning and end of the curve to make it start and end at the origin). By scaling, we can define a space-filling curve $g_n:[0,1]\to[-n,n]^2$ by $g_n(x)=nf(x)$ for any integer $n>0$, which starts and ends at the origin. We can now define a space-filling curve from $[0,\infty)$ to $\mathbb R^2$ by first walking along the curve $g_1$, then walking along the curve $g_2$, then walking along the curve $g_3$, and so on. This is a continuous curve and it passes through every point of $(x,y)\in \mathbb R^2$ since for every $(x,y)$ there exists an $n>0$ such that $(x,y)\in[-n,n]^2$.

To be explicit, the curve $F:\mathbb R\to\mathbb R^2$ defined by $F(x)=0$ if $x<1$, and $F(x)=g_n(x-n)$ if $x\in[n,n+1)$, is a continuous surjection.

  • $\begingroup$ Hey Samuel! You say one could start with a space-filling curve $[0,1]\to[0,1]^2$ and obtain a new space-filling curve $[0,1]\to[-1,1]^2$. I don't understand how exactly this is done. Could you please elaborate? $\endgroup$
    – rehband
    Oct 21 '14 at 14:13
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    $\begingroup$ @rehband: Consider the square $S=[0,1]^2$. Move it a distance $1/2$ to the left and another distance $1/2$ down, so that it is centered about the origin, and then scale it by a factor $2$ so that it becomes twice as large. Then you have obtained the square $2(S-(1/2,1/2))=[-1,1]^2$. Thus you should be able to convince yourself that if $f:[0,1]\to[0,1]^2$ is space-filling, then $g:[0,1]\to[-1,1]^2$ is spacefilling, where $g(t)=2(f(t)-(1/2,1/2))$. $\endgroup$
    – Samuel
    Oct 21 '14 at 18:13
  • $\begingroup$ Great, I see. Thank you. Say I start with the Peano curve and obtain a space-filling curve through the procedure in your comment. Explicitly, how do I ensure that it begins and ends in the origin? $\endgroup$
    – rehband
    Oct 22 '14 at 13:09
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    $\begingroup$ @rehband: Append a path from the origin to the starting point of the Peano curve and append a path from the ending point of the Peano curve to the origin. So if $f:[0,1]\to[-1,1]^2$ is a Peano curve such that $f(0)=(-1,-1)$ and $f(1)=(1,1)$, then define $g:[0,1]\to[-1,1]^2$ by $g(t)=(-3t,-3t)$ if $t\in[0,1/3)$, $g(t)=f( 3(t-1/3) )$ if $t\in[1/3,2/3)$ and $g(t)=(3(t-2/3),3(t-2/3))$ if $t\in[2/3,1]$. Then $g$ will be spacefilling and start and end at the origin. $\endgroup$
    – Samuel
    Oct 22 '14 at 14:38
  • $\begingroup$ Could you expalin why you new defined function on $[0,\infty)$ is continuous? Could you give detailed answer, please? $\endgroup$
    – ZFR
    Nov 6 '19 at 4:23

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