Can we represent $\mathbb{R}$ to $\mathbb{C}$, such that: $\forall c \in \mathbb{C}: \exists x \in \mathbb{R}: f(x)=c$?

Question

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!

Possible duplicate of 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
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

score 1 · Answer 1 · answered Apr 15 '19 at 14:11

1

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 $\Bbb C$.

answered Apr 15 '19 at 14:11

Arthur

199,419

I am not sure how to craft a bijection out of this proof, which I understand... – TVSuchty Apr 15 '19 at 14:26
My list would be for Reals:
01010101010... 10001010100... ... and then there would be a new element 110...
– TVSuchty Apr 15 '19 at 14:29

Can we represent $\mathbb{R}$ to $\mathbb{C}$, such that: $\forall c \in \mathbb{C}: \exists x \in \mathbb{R}: f(x)=c$?

1 Answers1