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This question already has an answer here:

I know that $|\mathbb R|=|\mathbb R\times\mathbb R|$, and that $|[0,1]|=|\mathbb R|$, which suggests that $|[0,1]|=|[0,1] \times [0,1]|$ but I would like to know a bijection between the interval and square.

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marked as duplicate by Asaf Karagila, user7530, user61527, azimut, Stefan Hansen Oct 18 '13 at 8:24

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  • $\begingroup$ The first thing that popped into my head was $t\rightarrow (t,\lim_{\omega\rightarrow\infty} \cos(2\pi\omega t))$, does this fit the bill and is it even well defined? $\endgroup$ – Dan Oct 18 '13 at 7:41
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    $\begingroup$ In the suggested duplicate MJD gave an answer which includes a bijection from the unit square to the unit interval, hence answering your question. $\endgroup$ – Asaf Karagila Oct 18 '13 at 7:42
  • $\begingroup$ @user7530: this is not a bijection $\endgroup$ – Thomas Rot Oct 18 '13 at 7:48
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This is surprisingly subtle. Lets do this for the open interval $(0,1)\cong (0,1)\times (0,1)$ The obvious map $(0.a_1a_2\ldots,0.b_1b_2\ldots)=(0,a_1b_1a_2b_2\ldots)$ doesn't work, this is not a bijection. This is because $0,899\ldots=0.900\ldots$. It is not hard to make an injection $(0,1)\times(0,1)\rightarrow (0,1)$ by modifying this example, by choosing a representation. There is also an injection $(0,1)\rightarrow (0,1)\times (0,1)$, sending $x\mapsto (x,x)$. The theorem Cantor-Bernstein-Schroeder now gives a bijection, but this is not very explicit.

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  • $\begingroup$ If you can find a subset $C\subset[0,1]$ so that $C'=[0,1]\setminus C$ is countable, and using your obvious map from $C\times C\to C$, then you can map countable $C'\times C'$ to countable $C'$. $\endgroup$ – Carlos Eugenio Thompson Pinzón Oct 18 '13 at 8:03

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