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As for example stated in the answer to this question: Why does Cantor's diagonal argument not work for rational numbers?

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HemnesMirrorball
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  • We don't know, and we don't care. It might be $\pi$, if that's what the procedure produces from a given countable list of reals. – BrianO Sep 15 '16 at 14:26

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There's a common misconception that Cantor's diagonal argument produces a "special real." That's really not what's going on. Let's look at exactly what Cantor's argument shows.

The argument shows that, if $L$ is any list of real numbers (formally: $L$ is a function from natural numbers to reals - think of the $i$th entry in the list as being $L(i)$), then there is some real number $r$ not in $L$ (formally: not in the range of $L$). Call the real built in this way "$Diag(L)$."

So "the real we get from Cantor's argument" isn't unique: it depends on what list we feed it. There are some lists $L$ such that $Diag(L)$ is rational (a good exercise), but these are lists that don't contain every rational number. If $L$ is a list that does contain every rational number, then since $Diag(L)\not\in L$, we know $Diag(L)$ is not rational.

Note that Cantor's argument is usually presented as a proof by contradiction. I think this obscures what's actually going on: the diagonal method gives a perfectly constructive procedure for, given a list of reals $L$, outputting a real not in $L$. This is really the "meat" of the argument. EDIT: As Mitchell points out below, there is no constructive method to go from a countable set of reals (that is, unlisted) to a real not in that set. (Precisely: ZF ( = set theory without choice) doesn't even prove that there is any function $d$ from $\{$countable sets of reals$\}$ to $\mathbb{R}$ with $d(X)\not\in X$.) The ordering is a crucial ingredient here.

One way to phrase the conclusion which I've found helpful is

"Any countable set of real numbers, isn't all of $\mathbb{R}$."

Of course, this just means "the reals are uncountable," but I think it makes things less mysterious.

Noah Schweber
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  • Ah thank you! That is a great answer! – HemnesMirrorball Sep 15 '16 at 14:31
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    A small point here: What Cantor's diagonal argument gives is a constructive procedure for going from a specific counting of a countable set $L$ to a real number not in $L.$ The existence of a function (constructive or not) which maps each countable set $L$ of reals to a real not in $L$ requires some use of the axiom of choice -- in fact, something stronger than $AC_\omega.$ (Working in $ZF+AC_\omega,$ if you assume such a function exists, then you can apply transfinite induction to get an injection from $\aleph_1$ to the reals, which is known not to be provable from $ZF+AC_\omega.)$ – Mitchell Spector Sep 15 '16 at 14:55
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    @MitchellSpector This is quite right; I was being careless. Fixed. – Noah Schweber Sep 15 '16 at 14:56
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    Yes! I cannot for the life of me understand why it is so often presented as a proof by contradiction; that’s an unnecessary complication that seems to cause some of the problems that students have with the argument. Getting them to see it as a ‘machine’ for producing a number not in the list is often all that’s needed to get them sorted. – Brian M. Scott Sep 15 '16 at 19:18