A problem in my homework is to prove the cardinality of $\mathbb R$ and the cardinality of $\mathbb R^2$ is the same. I'm in my first semester and I have no clue how to do this, do I need the axiom of choice to prove this? Can I find a bijection, or must I use Cantor-Schroeder-Bernstein? Apparently if we assume choice we can go further and prove the cardinality of any set $A$ is is equal to the cardinality of $A^2$ thank you very much, I need help.


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    $\begingroup$ Here's the answer. math.stackexchange.com/questions/243590/… $\endgroup$ – Math.StackExchange Oct 30 '14 at 1:07
  • $\begingroup$ Do you know already that $|\mathbb R|=|\mathcal P(\mathbb N)|$? $\endgroup$ – hmakholm left over Monica Oct 30 '14 at 1:07
  • $\begingroup$ no, that question literally came out of the blue, the rest of the homework is full of really simple questions. Thanks $\endgroup$ – Yorch Oct 30 '14 at 1:09
  • $\begingroup$ If you're using Bernstein, one direction is trivially easy. For the other direction, a hint: Cantor's proof (or one of them) was by interlacing decimal digits ... $\endgroup$ – hmakholm left over Monica Oct 30 '14 at 1:12
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    $\begingroup$ Yes, it's very independent of choice. The Cantor-Bernstein theorem is independent of choice as well. $\endgroup$ – Asaf Karagila Oct 30 '14 at 13:39

I just wanted to get the selfie badge. The solution I found at the end is the following:

We want an injection from $\mathbb R^2 $ to $\mathbb R$ we take a surjection from $\mathbb R$ to $\mathbb R^2$ and what we want is the right inverse.

As an example of a surjection I took the following: let the number in $\mathbb R$ be $(-1)^n(k+.d_1d_2d_3\dots)$ if $k$ is not of the form $2^a3^b5^c7^d$ send it to zero. if it is send it it to $((-1)^ad_1d_3d_5\dots,((-1)^bd_2d_4d_6\dots)$. And leave $c$ and $d$ digits before adding the decimal point.


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