# Are there as many real numbers as there are imaginary numbers?

On the one hand, I know that $\mathbb{R}$ and $\mathbb{I}=\{xi:x\in\mathbb{R}\setminus\{0\}\}$ are both uncountable sets, so they have the same number of elements (i.e. the same cardinality)

On the other hand, there's no bijection between $\Bbb{R}$ and $\Bbb{I}$: $0$ is not mapped to anything in $\Bbb{I}$, so, by definition, $\require{enclose} \enclose{horizontalstrike}{\mathbb{R}}$ and $\enclose{horizontalstrike}{\mathbb{I}}$ have different sizes (cardinalities) .

These two statements seem to contradict each other, so which one is correct?

Please excuse my ignorance and/or lack of correct terminology; I'm a rookie when it comes to set theory.

Edit: the statements with strikeouts are erroneous and have later been shown to be nonsense, but I've included them for completeness.

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Whatever you want to get mapped to $0$. Where's the problem? – blue Jul 10 '14 at 0:27
Sets have the same cardinality if there exists SOME bijection between them. – StrangerLoop Jul 10 '14 at 0:29
@alexqwx Why would they have to be mapped to their imaginary counterparts like that? Who died and made you king and gave you the authority to stipulate where a bijection must send everything? – blue Jul 10 '14 at 0:30
Also, it is not correct to write $\mathbb{I}=\mathbb{C}$ \ $\mathbb{R}$, if by $\mathbb{I}$ you mean the pure imaginary numbers. – StrangerLoop Jul 10 '14 at 0:31
$\{2,3,\cdots\}$ has one fewer element than $\{1,2,3,\cdots\}$, but they have the same size. Cardinality is not measured by where two subsets sit in relation to each other in the lattice of subsets of a bigger set. Instead, cardinality is an equivalence class defined by bijections. – blue Jul 10 '14 at 0:38

I can give you an explicit bijection from $\mathbb{R} \mapsto \mathbb{I^+}$. Map $x \mapsto i e^x$.

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And $0$ goes to ...? – blue Jul 10 '14 at 0:36
OP did not ask for a bijection with $\Bbb I^+$ – blue Jul 10 '14 at 0:51

There is a bijection between $\Bbb R$ and $\Bbb C$. Therefore there is an injection from $\Bbb I$ into $\Bbb R$, and of course there is an injection from $\Bbb R$ into $\Bbb I$. By the Cantor-Bernstein theorem, there is a bijection between the two sets.

To see why there is a bijection between $\Bbb R$ and $\Bbb C$ it's very easy to note that there is a bijection between $\Bbb R$ and $\Bbb{N^N}$ (the set of infinite sequences of natural numbers), and then observe that: $$(\Bbb{N^N})^2\approx\Bbb{N^{2\times N}}\approx\Bbb{N^N}.$$

Therefore $\Bbb C$, which naturally has a bijection with $\Bbb R^2$, has the same cardinality as $\Bbb R$.

The question has been edited, and now it redefines $\Bbb I$ as the set $\{xi\mid x\in\Bbb R\setminus\{0\}\}$.

Here a bijection is easily definable. It is true that $x\mapsto xi$ is not a bijection since $0$ is causing us problems. There are two easy ways to solve this problem:

1. $\Bbb I$ maps injectively into $\Bbb R$ by mapping $xi$ to $x$, obviously; and in the other direction $x\mapsto e^xi$ is an injection as well. Therefore $\Bbb R$ and $\Bbb I$ have the same cardinality. But we can do better, we can write down an explicit bijection.

2. Note that only one element is causing us problems, so we just need to "shift" some elements around. For example: $$x\mapsto\begin{cases} xi & x\notin\Bbb N\\(x+1)i & x\in\Bbb N\end{cases}$$ and in this context, $0\in\Bbb N$.

The key issue is that not every bijection needs to be "very simple" or even continuous. Or even definable by nice means.

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I was going to biject $[ 0,\infty)$ with $[ 0,1)$ with $(0,1)$ with $(0,\infty)$, and keep $(-\infty,0)$ the same, using the argument of Did over here that $[ 0,1)\cong(0,1)$. Apparently though Did's shifting argument is the same exact idea. – blue Jul 10 '14 at 0:46
About your comment (1.)--what would be mapped to the negative imaginary numbers, since $e^x>0$ for all $x \in \mathbb{R}$? – alexqwx Jul 10 '14 at 0:47
@alexqwx: We don't worry about that. The Cantor-Bernstein theorem tells us that if there is an injection from $A$ into $B$, and an injection from $B$ into $A$, then there is a bijection between $A$ and $B$. So I don't have to sit and write one by hand. I can just show that there are two injections. – Asaf Karagila Jul 10 '14 at 0:48
@blue: And also with a bi-injection argument which used the map $x\mapsto e^xi$. – Asaf Karagila Jul 10 '14 at 0:54
@alexqwx: That is [almost] correct. If $f\colon\Bbb R\to\Bbb R$ is injective and $0$ is not in the range of $f$, then $x\mapsto f(x)i$ is an injection from $\Bbb R$ into $\Bbb I$. – Asaf Karagila Jul 10 '14 at 0:57