22
$\begingroup$

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?

  • 2
    The best treatment of this in an existing answer is probably [here](http://math.stackexchange.com/a/183383/12042).2012-11-26

4 Answers 4

19

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 $\mathbb R$.

10

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.

  • 0
    This argument seems strange to me. Let me show why. Consider$a$real number $a$ with $\ldots x_2 x_1 x_0 . y_0 y_1 y_2 \ldots$ its decimal expansion. Then the number $f(a) = \ldots x_2 y_2 x_1 y_1 x_0 y_0$ is an integer, right? Thus $\# \mathbb R \le \# \mathbb Z$. Do you see the issue?2018-10-15
1

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, $|C| = |\mathbb{R^2}| = |\mathbb{R}|$.

-3

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.

  • 0
    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.2018-10-15