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I am wondering why the complex plane is not defined to be of higher cardinality than the reals. Since there should not be a function:

$$f: \mathbb{R} \rightarrow \mathbb{C}$$ such that, $$\forall c \in C: \exists x \in \mathbb{R} : f(x)=c$$

Can we prove this hypothesis? Or am I wrong? In this question, there is proven that $|\mathbb{R}| = |\mathbb{C}|$, but what would be an bijection? I do not see the bijection, although the proof is given.

The prove that $2^{\mathbb{|N|}}=\mathbb{|R|}$ as far as I know works like this:

Give me any list where each "packet" contains infinitely many digits, there is one the packet you missed. (Now you show that the List is incomplete).
But I have no idea how I make the $2^{\mathbb{|N|}^2}$ work. Doubling the list will not let me represent all numbers.

Thanks!

Jyrki Lahtonen
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TVSuchty
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    Possible duplicate of [Do the real numbers and the complex numbers have the same cardinality?](https://math.stackexchange.com/questions/245141/do-the-real-numbers-and-the-complex-numbers-have-the-same-cardinality). The top-voted answer there explains how to obtain a bijection. – Lee Mosher Apr 15 '19 at 14:11
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    I tried to explain, why this is not a duplicate. – TVSuchty Apr 15 '19 at 14:49
  • Have you seen arguments for the fact that $\Bbb{R}$ and $\Bbb{R}^2$ have the same cardinality? For example the one where the general idea is to interlace two decimal expansions in $[0,1)$ (and picking every other decimals to form a kind of an inverse)? That argument is a bit unclean, because you need to worry about the non-uniqueness of decimal expansions, which is why the approach based on $2^{|\Bbb{N}|}$ is often better. Oh, you do know that $|\Bbb{N}|=|\Bbb{N}^2|$? – Jyrki Lahtonen Apr 16 '19 at 03:13

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Most proofs of $|\Bbb C|=|\Bbb R|$ are done by constructing a bijection. It may be hidden behind other notation, but it is always there. For instance, in the top answer in the post you linked, it is written$$|\mathbb C|=|\mathbb R|^2 =(2^{\aleph_0})^2 = 2^{\aleph_0\cdot 2}=2^{\aleph_0}=|\mathbb R| $$Each one of those $=$ has a relatively simple bijection behind it, and if you compose all those bijections, you get a single bijection between $\Bbb R$ and $\Bbb C$.

Arthur
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