# Do the real numbers and the complex numbers have the same cardinality?

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?

• One can show that $|\mathbb R| = |\mathbb R^2| = |\mathbb C|$ Nov 26, 2012 at 19:05
• 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. Nov 26, 2012 at 19:08
• The best treatment of this in an existing answer is probably here. Nov 26, 2012 at 19:31

Yes.

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

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 $\mathbb R$.

• What is $2^\mathbb{N}$? 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.

• 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. Nov 26, 2012 at 19:14
• 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.
– yo'
Nov 26, 2012 at 19:16
• 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.
– yo'
Nov 26, 2012 at 19:22
• 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. Nov 26, 2012 at 19:29
• 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. Nov 26, 2012 at 19:33

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.

• This is incorrect. Space filling curves are not injective. Aug 8, 2014 at 15:01
• @DanRust: Can you explain why? Oct 15, 2018 at 14:58
• 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. Oct 15, 2018 at 15:05