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I understand that the set of all 2x2 matrices over the reals is of the same cardinality as the set of the reals themselves. However, I am curious if a specific, explicit bijection is known to exist, and, if so, what that bijection is. This will be interesting and useful to me in a project that I am working on.

user1551
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Jebruho
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    $M_2(\Bbb R)$ is really just $\Bbb R^4$. Given a bijection $f\colon\Bbb R\to\Bbb R^2$ simply set $F\colon\Bbb R\to\Bbb R^4$ by $F(t)=\langle f(f(t)_x),f(f(t)_y)\rangle$ (where $\bullet_x$ is the projection on the $x$-coordinate, and similarly with $y$), or some other composition. See http://math.stackexchange.com/questions/183361/bijective-map-from-mathbbr3-rightarrow-mathbbr for a bijection between $\Bbb R$ and $\Bbb R^2$. – Asaf Karagila Apr 17 '13 at 15:38
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    The difficulty when we’re new to these concepts is to expect such functions to be continuous, or nearly so. But topology tells us that such a map has to be highly discontinuous. – Lubin Apr 17 '13 at 15:45
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    I would suggest closing this as a duplicate of the question Asaf Karagila linked to. It needs his comment to be a proper duplicate, but the answers there are very good. – Ross Millikan Apr 17 '13 at 15:51
  • One, I don't expect the answers to be continuous. I just expect there to be some explicit bijection. Two, I read the answers to the question Asaf linked to and they didn't do much in the way of providing a bijection as merely showing that a bijection exists. Even the interleaving argument he uses specifically applies to (0,1). So, that answer doesn't really answer my question. – Jebruho Apr 17 '13 at 15:55
  • You do know that $$x\mapsto\frac{x-\frac12}{x-x^2}$$ is a bijection from $(0,1)$ to $\Bbb R$? It's pretty explicit, and if you compose two explicit functions you should have an explicit function (which is a somewhat of an ambiguous term, in my opinion). – Asaf Karagila Apr 17 '13 at 17:45

1 Answers1

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$\newcommand{\bbR}{\mathbb{R}}$ $\newcommand{\bbZ}{\mathbb{Z}}$

The fundamental reason why the bijection that you are working for exists is that the cardinalities agree. There are, however, a lot of sets with this cardinality, so you should not expect any especially nice or explicit or distinguished bijection in general (note that the proof that such bijection exists is not quite constructive). We will construct an example, but bear in mind that this bijection will respect pretty much no structure that there is on $\bbR$ (or on the matrices).

Note that $2 \times 2$ matrices are the same as quadruples of reals, i.e. $\bbR^4$, so it will suffice to find the bijection between $\bbR$ and $\bbR^4$. To make life a little easier, let us consider something a little simpler: a bijection between $\bbR$ and $\bbR^2$ (it will be obvious how to generalise). It is convenient to think of reals as their decimal expansion. Given $x \in \bbR$, you can write it as $x = x_n x_{n-1} \dots x_1x_0.x_{-1}x_{-2} \dots$, where $x_i \in \{0,\dots,9\}$ is the digit corresponding to $10^{i}$. You can think of these digits as of a sequence indexed by $\bbZ$ with $x_i = 0$ for $i$ large enough. Let us denote the set of these sequences by $S$.

It is almost the case that $\bbR$ can be identified with $S$, but there is one twist: some numbers have two different expansions, like $0.99999... = 1$. However, there are only countably many of these, so let us not bother with them for the time being.

Now, it is fairly easy to identify $S^2$ with $S$: given two sequences $x = x_n x_{n-1} \dots x_1x_0.x_{-1}x_{-2} \dots$ and $y = y_n y_{n-1} \dots y_1y_0.y_{-1}y_{-2} \dots$ (wlog, they have the same number of significant digits, else fill in some extra $0$'s) assign to them $$f(x,y) := y_n x_n y_{n-1} x_{n-1} \dots y_1x_1y_0x_0.y_{-1}x_{-1}y_{-2}x_{-2} \dots$$ The reverse map is given by: $$ g(z_{2m+1}z_{2m-1}\dots z_1z_0,z_{-1}\dots) = (z_{2m}z_{2m-2}\dots z_0,z_{-2}\dots, z_{2m+1}z_{2m-1}\dots z_1,z_{-1}\dots) $$ Combining this with identification of $S$ with $\bbR$, you find your bijection (modulo some technicalities for $S$ not being quite the same as $\bbR$)

Jakub Konieczny
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