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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.

beep-boop
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  • What makes you say there is no bijection between $\Bbb R$ and $\Bbb I$? – anon Jul 10 '14 at 00:26
  • @blue As I've said, what gets mapped to 0? – beep-boop Jul 10 '14 at 00:27
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    Whatever you want to get mapped to $0$. Where's the problem? – anon Jul 10 '14 at 00:27
  • @blue Well, isn't every real number mapped to its imaginary counterpart? e.g. $a \mapsto ai$, but, then, $0$ can't be mapped to $0i=0$ since $0 \notin \mathbb{I}$. – beep-boop Jul 10 '14 at 00:29
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    Sets have the same cardinality if there exists SOME bijection between them. – StrangerLoop Jul 10 '14 at 00:29
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    @alexqwx - Just because you can't think of a bijection doesn't prove that no such bijection exists... – Hao Ye Jul 10 '14 at 00:29
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    @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? – anon Jul 10 '14 at 00:30
  • So what would be an example of a bijection from $\mathbb{R}$ to $\mathbb{I}$, then? – beep-boop Jul 10 '14 at 00:30
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    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 00:31
  • @StrangerLoop Why not? The set of imaginary numbers is all the real numbers removed from the set of complex numbers. – beep-boop Jul 10 '14 at 00:32
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    Pure imaginary means $xi$ for some $x\in\Bbb R$. But $1+i$ is neither purely imaginary nor is it real. You're basically saying the $y$-axis is the whole plane minus the $x$-axis. Nonsense! – anon Jul 10 '14 at 00:33
  • @blue Ah, yes. My mistake. – beep-boop Jul 10 '14 at 00:34
  • @blue Surely, if there is one fewer element of $\mathbb{I}$ than $\mathbb{R}$, they can't have the same size? – beep-boop Jul 10 '14 at 00:38
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    ${2,3,\cdots}$ 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. – anon Jul 10 '14 at 00:38
  • @alexqwx: There are many discussions on this site about how infinite sets behave differently than finite sets when it comes to cardinality. I strongly suggest that you find them, and read them. – Asaf Karagila Jul 10 '14 at 00:40
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    @alexqwx Next you should be wondering how $(0, 1)$ and $[0, 1]$ can have the same cardinality! It's essentially the same as what you tried to ask, without any confusion about the definition (you wanted to exclude $0$ from $\mathbb I$, but that goes against the definition). – M. Vinay Jul 10 '14 at 02:34
  • "so they have the same number of elements" That's a dangerously imprecise way of speaking, and it can lead to well know paradoxes/contradictions. To have the same cardinality is a precise concept. "the number of elements" of an infinite set is not. – leonbloy Jul 10 '14 at 11:19
  • @leonbloy So is the 'number of elements' in an infinite set not a well-defined concept, then? – beep-boop Jul 10 '14 at 12:18
  • @alexqwx: No, it is a well-defined concept. It just doesn't agree with the notion "Strict subset, smaller cardinality" that we learn intuitively with finite cardinals. I will also point out that you have a grave mistake in your first sentence of the current revision. The fact two sets are uncountable doesn't imply they have the same cardinality. Not even remotely. – Asaf Karagila Jul 10 '14 at 16:40
  • @AsafKaragila Just for my intuition's sake, could you give me an example of two uncountably-infinite sets whose cardinalities are different? – beep-boop Jul 10 '14 at 16:45
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    Of course. $\Bbb R$ and $\mathcal P(\Bbb R)$. – Asaf Karagila Jul 10 '14 at 16:52

2 Answers2

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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.

Asaf Karagila
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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. – anon Jul 10 '14 at 00:46
  • About your comment (1.)--what would be mapped to the negative imaginary numbers, since $e^x>0$ for all $x \in \mathbb{R}$? – beep-boop Jul 10 '14 at 00:47
  • @blue: Yes, it is the same idea, because both use the same facts about the real numbers, and about any other set which has a countably infinite subset. – Asaf Karagila Jul 10 '14 at 00:48
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    @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 00:48
  • @AsafKaragila But what is (an example of) the injection from $\mathbb{R}$ to $\mathbb{I}$? – beep-boop Jul 10 '14 at 00:52
  • @alexqwx: I gave one $x\mapsto e^xi$. Injection doesn't mean a bijection. It just means injection. The fact that the range is not the entire set is irrelevant. – Asaf Karagila Jul 10 '14 at 00:53
  • @alexqwx Asaf edited his answer thirteen minutes ago with an explicit bijection. – anon Jul 10 '14 at 00:54
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    @blue: And also with a bi-injection argument which used the map $x\mapsto e^xi$. – Asaf Karagila Jul 10 '14 at 00:54
  • @AsafKaragila So, for an injection from $\mathbb{R}$ to $\mathbb{I}$, could you use $x \mapsto f(x)i$ where $f$ is any injective function? Does that equate to the same thing? – beep-boop Jul 10 '14 at 00:56
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    @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 00:57
  • @blue I'm aware of that. I just wasn't sure how this was a bijection. – beep-boop Jul 10 '14 at 00:58
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I can give you an explicit bijection from $\mathbb{R} \mapsto \mathbb{I^+}$. Map $x \mapsto i e^x$.

StrangerLoop
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