So it's easy to show that the rationals and the integers have the same size, using everyone's favorite spiral-around-the-grid.

Can the approach be extended to say that the set of complex numbers has the same cardinality as the reals?

  • 1
    $\begingroup$ One can show that $|\mathbb R| = |\mathbb R^2| = |\mathbb C|$ $\endgroup$
    – Stefan
    Nov 26, 2012 at 19:05
  • 10
    $\begingroup$ It's quite sad, but it's easier to write an answer than finding the duplicate. And I am sure this question has been asked before. $\endgroup$
    – Asaf Karagila
    Nov 26, 2012 at 19:08
  • 3
    $\begingroup$ The best treatment of this in an existing answer is probably here. $\endgroup$ Nov 26, 2012 at 19:31

4 Answers 4



$$|\mathbb R|=2^{\aleph_0}; |\mathbb C|=|\mathbb{R\times R}|=|\mathbb R|^2.$$

We have if so:

$$|\mathbb C|=|\mathbb R|^2 =(2^{\aleph_0})^2 = 2^{\aleph_0\cdot 2}=2^{\aleph_0}=|\mathbb R|$$

If one wishes to write down an explicit function, one can use a function of $\mathbb{N\times 2\to N}$, and combine it with a bijection between $2^\mathbb N$ and $\mathbb R$.

  • $\begingroup$ What is $2^\mathbb{N}$? $\endgroup$
    – Filippo
    Jun 10 at 18:14

Of course. I will show it on numbers in $[0,1)$ and $[0,1)\times[0,1)$. Consider $z=x+iy$ with $x=0.x_1x_2x_3\ldots$ and $y=0.y_1y_2y_3\ldots$ their decimal expansions (the standard, greedy ones with no $9^\omega$ as a suffix). Then the number $f(z)=0.x_1y_1x_2y_2x_3y_3\ldots$ is real and this map is clearly injective on the above mentioned sets. Generalization to the whole $\mathbb C$ is straightforward. This gives $\#\mathbb C\leq\#\mathbb R$. the other way around is obvious.

  • 8
    $\begingroup$ This requires a bit more work. The map isn’t well-defined until you deal with the $0.4999\dots=0.5000\dots$ issue; if you deal with that straightforwardly, it’a nor surjective. $\endgroup$ Nov 26, 2012 at 19:14
  • $\begingroup$ Yes, you are right. However, they all all (complex) rational hence of no interest for the sets of continuum cardinality. I'll add a comment. $\endgroup$
    – yo'
    Nov 26, 2012 at 19:16
  • $\begingroup$ And btw, usually a string with suffix $9^\omega$ is not considered to be an expansion (it is only a representation), in the usual greedy expansions as defined by Rényi in 1957. $\endgroup$
    – yo'
    Nov 26, 2012 at 19:22
  • $\begingroup$ I’ve never seen anyone make a distinction between representation and expansion, and I very much doubt that the distinction can be considered standard; certainly it does not qualify as well-known, so if you use it, you need to explain it. $\endgroup$ Nov 26, 2012 at 19:29
  • 1
    $\begingroup$ And yes, I know that only countably many numbers are affected and that this does not affect the result, but I don’t know that the OP knows this. $\endgroup$ Nov 26, 2012 at 19:33

Consult #4b in http://faculty.lasierra.edu/~jvanderw/classes/m415a03/hw8ans.pdf.

A straightforward bijection $B : \mathbb{R}^2 \rightarrow \mathbb{C}$ is: $B(a,b) = a + bi$. I omit the verification of injectivity and surjectivity. Then $|C| = |\mathbb{R}^2|$. The separate result that $|\mathbb{R}^k| = |\mathbb{R}| \; \forall \; k \in \mathbb{N}$ implies $|\mathbb{R}^2| = |\mathbb{R}|$. Altogether, $|\mathbb{C}| = |\mathbb{R}^2| = |\mathbb{R}|$.


One particularly nice class of bijections from $\Bbb R$ to $\Bbb C = \Bbb R^2$, which is in my opinion a little bit similar to the spiral around the grid, is given by the space-filling curves.

  • 8
    $\begingroup$ This is incorrect. Space filling curves are not injective. $\endgroup$
    – Dan Rust
    Aug 8, 2014 at 15:01
  • $\begingroup$ @DanRust: Can you explain why? $\endgroup$
    – DaBler
    Oct 15, 2018 at 14:58
  • 5
    $\begingroup$ If an injective and surjective curve in the square existed, then this would imply that such a curve is in fact a homeomorphism, as the interval is compact, and the square is Hausdorff. We know they are not homeomorphic as the interval has a cut point and the square does not. $\endgroup$
    – Dan Rust
    Oct 15, 2018 at 15:05

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