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I was just browsing through the Puzzle section on Noam Elkies website. The puzzle can be found here.

The solution to the puzzle proves that any well-ordered subset of $\mathbb{R}$ is countable. In the solution, Elkies defines $s(x)$ to be the immediately following element of $x$ for any $x\in S$ for a well ordered subset $S\subset \mathbb{R}$.

I want to understand the added note at the bottom. It says to consider any interval $(x,s(x))$. I take it he means to view this interval as an subset of $\mathbb{R}$, for it is empty in $S$? Then I know there always exists a rational between any two reals since $\mathbb{Q}$ is dense in $\mathbb{R}$, so one could map $(x,s(x))$ to some $r(x)\in (x,s(x))$, which would show $S$ is in bijection with a subset of $\mathbb{Q}$, and hence countable.

What I don't get is why this mapping doesn't require choice. For any interval $(x,s(x))$, how could we distinquish some $r(x)$ to pick? Are we possibly talking about taking the least rational in $(x,s(x))$ for any $x$? Would that be a way to distinguish which rational to choose?

Martin Sleziak
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Waldott
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    There are explicit well-orderings of the rationals; pick one. Then define $r(x)$ to be the smallest rational, under the chosen well-order, that is in the given interval. This does not require the Axiom of Choice. – Arturo Magidin Apr 10 '11 at 01:23
  • Thanks Arturo, do you have any references to a common explicit well ordering of $Q$ that I could look up? I googled but didn't find any. – Waldott Apr 10 '11 at 01:28
  • You can also take the smallest n such that there exists a k such that $\frac{k}{2^n}$ is in your interval, and then choose the smallest xu h $\frac{k}{2^n}$ – Thomas Andrews Apr 10 '11 at 01:30
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    @Waldott: One example is to sort p/q (in reduced terms) first by denominator, then by absolute value of the numerator, then minus before plus. The well order does not need to (and cannot) correspond to the usual order on $\mathbb{Q}$. – Ross Millikan Apr 10 '11 at 01:32
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    In addition to Ross' comment a good way to think about his ordering is to think of the well-ordering induced by any bijection from $\mathbb{N}$ to $\mathbb{Q}$. What Ross is describing is the ordering induced by the "typical" bijection one is taught from the naturals to the rationals. – JSchlather Apr 10 '11 at 01:45
  • @Waldott: In fact, you don't need an explicit well-ordering; all you know is to know there are well-orderings, because you don't need AC to make a single selection (in this case, a well-ordering of $\mathbb{Q}$). But the fact that there are explicit such orderings emphasizes the fact that we don't need AC for this. – Arturo Magidin Apr 10 '11 at 02:25
  • BTW there are several questions on this site about this problem. For example http://math.stackexchange.com/questions/408300/countable-ordinals-are-embeddable-in-the-rationals-bbb-q-proofs-and-their, http://math.stackexchange.com/questions/496488/universality-of-mathbb-q, http://math.stackexchange.com/questions/44559/formal-proof-for-a-subset-of-the-real-numbers-well-ordered-with-the-normal-orde, http://math.stackexchange.com/questions/37151/showing-any-countable-dense-linear-ordering-is-isomorphic-to-a-subset-of-mat and several questions which are shown among linked questions there. – Martin Sleziak Apr 12 '14 at 10:15

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The fact that there are explicit well-orderings of $\mathbb{Q}$ makes it clear that AC is not needed. But in fact, if you have any countable dense subset $A$ of $\mathbb{R}$, the argument can be used to give a one-to-one function from the well-ordered subset $S$ of $\mathbb{R}$ into $A$ that does not require the Axiom of Choice, even if you cannot produce an explicit well-ordering for $A$ (or even if you don't know what $A$ is, beyond knowing that it is both dense in $\mathbb{R}$ and countable).

Since we are assuming that $A$ is countable, that means that the set of bijections $f\colon A\to\mathbb{N}$ is nonempty. Let $\mathcal{F}$ be any such bijection. This does not require AC, because we are only choosing one element from a single nonempty set.

Then we define $r\colon S\to A$ as described: given $x\in S$, we know that $(x,s(x))\cap A\neq\emptyset$. Let $m\in\mathbb{N}$ be the smallest element of the set $\{\mathcal{F}(a)\mid a\in A\cap (x,s(x))\}$. This does not require choice, since we are only using the well-ordering of $\mathbb{N}$. Then define $r(x) = \mathcal{F}^{-1}(m)$. By construction, $r(x)\in (x,s(x))\cap A$.

(In short: a bijection from $A$ to $\mathbb{N}$ induces a well-ordering $\preceq$ on $A$; pick a bijection, and define $r(x)$ to be the $\preceq$-smallest element of $A$ in $(x,s(x))$).

The Axiom of Choice is not needed, because the only "choice" was the selection of $\mathcal{F}$, and this single choice from a nonempty set does not require the Axiom of Choice.

Arturo Magidin
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A variant of Thomas Andrews' comment: choose the dyadic rational (terminating binary decimal) with the smallest number of $1$s in the interval. If there are ties, choose the smallest one in absolute value, then the positive one (if there's two with the same absolute value).

Qiaochu Yuan
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