Elegant way to make a bijection from the set of the complex numbers to the set of the real numbers [duplicate]

Make a bijection that shows $|\mathbb C| = |\mathbb R|$

First I thought of dividing the complex numbers in the real parts and the complex parts and then define a formula that maps those parts to the real numbers. But I don't get anywhere with that. By this I mean that it's not a good enough defined bijection.

Can someone give me a hint on how to do this?

marked as duplicate by Ross Millikan calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 17 '17 at 4:37

• Are you sure that you need to show that $|\mathbb R| = \aleph_1$? – Chris Culter Oct 17 '17 at 4:18
• It was an extra problem to make in the class. Maybe it was just to find a bijection from $\mathbb R$ to $\mathbb C$ to show that those 2 sets have the same cardinality number. – Anonymous196 Oct 17 '17 at 4:20
• Is it wrong to say that $|\mathbb R| = \aleph_1$ – Anonymous196 Oct 17 '17 at 4:21
• That statement, better known as the continuum hypothesis, is independent of the axioms you're likely to study in elementary set theory. In particular, it is not provable from those axioms. – Chris Culter Oct 17 '17 at 4:23
• math.stackexchange.com/questions/245141/… – d.k.o. Oct 17 '17 at 4:34

You can represent every complex number as $z=a+ib$, so let us denote this complex number as $(a,b) , ~ a,b \in \mathbb R$. Hence we have cardinality of complex numbers equal to $\mathbb R^2$.

So finally, we need a bijection in $\mathbb R$ and $\mathbb R^2$.

This can be shown using the argument used here.

Note that since there is a bijection from $[0,1]\to\Bbb R$ (see appendix), it is enough to find a bijection from the unit square $[0,1]^2$ to the unit interval $[0,1]$. By constructions in the appendix, it does not really matter whether we consider $[0,1]$, $(0,1]$, or $(0,1)$, since there are easy bijections between all of these.

Mapping the unit square to the unit interval

There are a number of ways to proceed in finding a bijection from the unit square to the unit interval. One approach is to fix up the "interleaving" technique I mentioned in the comments, writing $\langle 0.a_1a_2a_3\ldots, 0.b_1b_2b_3\ldots\rangle$ to $0.a_1b_2a_2b_2a_3b_3\ldots$. This doesn't quite work, as I noted in the comments, because there is a question of whether to represent $\frac12$ as $0.5000\ldots$ or as $0.4999\ldots$. We can't use both, since then $\left\langle\frac12,0\right\rangle$ goes to both $\frac12 = 0.5000\ldots$ and to $\frac9{22} = 0.40909\ldots$ and we don't even have a function, much less a bijection. But if we arbitrarily choose to the second representation, then there is no element of $[0,1]^2$ that is mapped to $\frac12$, and if we choose the first there is no element that is mapped to $\frac9{22}$, so either way we fail to have a bijection.

This problem can be fixed.

First, we will deal with $(0,1]$ rather than with $[0,1]$; bijections between these two sets are well-known, or see the appendix. For real numbers with two decimal expansions, such as $\frac12$, we will agree to choose the one that ends with nines rather than with zeroes. So for example we represent $\frac12$ as $0.4999\ldots$.

Now instead of interleaving single digits, we will break each input number into chunks, where each chunk consists of some number of zeroes (possibly none) followed by a single non-zero digit. For example, $\frac1{200} = 0.00499\ldots$ is broken up as $004\ 9\ 9\ 9\ldots$, and $0.01003430901111\ldots$ is broken up as $01\ 003\ 4\ 3\ 09\ 01\ 1\ 1\ldots$.

This is well-defined since we are ignoring representations that contain infinite sequences of zeroes.

Now instead of interleaving digits, we interleave chunks. To interleave $0.004999\ldots$ and $0.01003430901111\ldots$, we get $0.004\ 01\ 9\ 003\ 9\ 4\ 9\ldots$. This is obviously reversible. It can never produce a result that ends with an infinite sequence of zeroes, and similarly the reverse mapping can never produce a number with an infinite sequence of trailing zeroes, so we win. A problem example similar to the one from a few paragraphs ago is resolved as follows: $\frac12 = 0.4999\ldots$ is the unique image of $\langle 0.4999\ldots, 0.999\ldots\rangle$ and $\frac9{22} = 0.40909\ldots$ is the unique image of $\langle 0.40909\ldots, 0.0909\ldots\rangle$.