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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?

Asaf Karagila
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Ethan
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4 Answers4

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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 $\mathbb R$.

Asaf Karagila
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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.

J. W. Tanner
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yo'
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    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. – Brian M. Scott Nov 26 '12 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 '12 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 '12 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. – Brian M. Scott Nov 26 '12 at 19:29
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    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. – Brian M. Scott Nov 26 '12 at 19:33
  • The sequence $a_1a_2a_3\ldots$ represents $x$ if $x=\sum a_i\beta^{-i}$ (let's say it is a property of the string), you can speak about a greedy one (maximal in the lexicographical order) etc., but then, you only add more properties. The Rényi expansion of $x$ is the one given by $x_0=x$, $a_i=\lfloor \beta x_{i-1}\rfloor$, $x_i=\beta x_{i-1}-a_i$, which is an algorithm for obtaining the representations, and then you can speak about expanding the number. The fact that such representation is greedy (in the sense of maximality w.r.t. some order) has to be proved. – yo' Nov 26 '12 at 19:35
  • Yes, I inferred all that from your earlier comments. It doesn’t affect mine. – Brian M. Scott Nov 26 '12 at 19:37
  • your statement "the other way around is obvious" might be lost on some beginners. can you show how $f$ is surjective? – chharvey Sep 27 '17 at 02:01
  • 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? – DaBler Oct 15 '18 at 14:55
  • @DaBler That doesn't work in the case you mention; integers have to have finite length as strings, so your f(a) isn't actually an integer, or even a number of any sort. But decimal expansions have countably infinite length, so this argument (modulo the issues mentioned earlier around non-unique representation) works fine for interweaving two decimal expansions. Or any number of expansions, for that matter. – Idran Aug 12 '19 at 18:21
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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, $|\mathbb{C}| = |\mathbb{R}^2| = |\mathbb{R}|$.

joseville
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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.

Dominik
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    This is incorrect. Space filling curves are not injective. – Dan Rust Aug 08 '14 at 15:01
  • @DanRust: Can you explain why? – DaBler Oct 15 '18 at 14:58
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    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. – Dan Rust Oct 15 '18 at 15:05