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I know that the set of all rational numbers is countable, and can be enumerated by a sequence, say $\{a_n\}$. But can we construct a monotonic $\{a_n\}_{n=1}^{\infty}$, e.g. with $a_k<a_{k+1}$? It doesn't seem plausible to me, because then $a_1$ would be the smallest rational number, which clearly can't be any finite number. Am I mistaken?

wchargin
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syeh_106
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    @CarstenS You're countably right! $\mathbb Q$ inherits an order from any bijection $\mathbb N\to\mathbb Q$. This is NEVER the order we are used to (which is equivalent to answering the post). Tending to delete my previous comment. – Hanno Sep 05 '17 at 08:19
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    As an aside, a better phrase for what you are asking about is "increasing sequence". The phrase "ordered sequence" usually just means an ordinary sequence except that it emphasizes we care about the order relation on the indices –  Sep 05 '17 at 08:31
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    Not possible because in the set $\mathbb Q$ there is no successor for any element. – Piquito Sep 05 '17 at 19:24
  • @Hurkyl Thanks a lot for pointing this out. I'll change the title to a "monotone sequence" to prevent confusion. – syeh_106 Sep 06 '17 at 03:34
  • What about the Well-Ordering principle? – Geoff Pointer Sep 06 '17 at 08:48
  • @GeoffPointer It is simply not needed in the 'countable context' $\ddot\smile$ – Hanno Sep 06 '17 at 08:55
  • @Hanno I just mentioned it because the OP speaks of a smallest rational number and the Well-Ordering principle states that for any set, an ordering can be found so that all subsets have a first element. – Geoff Pointer Sep 06 '17 at 15:01
  • @Hanno: I believe you're thinking of the well-orderability of $\mathbb{R}$. An explicit well-ordering on $\mathbb{Q}$ can be given in terms of any of the various bijections between the rationals and the naturals. – jwodder Sep 06 '17 at 21:20
  • Yep, true. Cf also my first comment ... $\Rightarrow$ Deleted my last comment. – Hanno Sep 07 '17 at 04:41

5 Answers5

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if $a_k< a_{k+1}$ then $x := \frac{a_{k+1}+a_k}{2}$ is a rational number between those two, so no.

Thomas
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    Yours is the best and simplest method, but I prefer if $a_k = \frac{b_k}{c_k}$ then $\frac{b_k+b_{k+1}}{c_k+c_{k+1}}$ is between the two. – Mitch Sep 05 '17 at 21:42
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    Implication: There is a countable infinity that can't be counted as 1,2,3,... – Joshua Sep 06 '17 at 02:24
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    @Mitch care to elaborate? That seems way more difficult and unintuitive, at a glance – MichaelChirico Sep 06 '17 at 03:00
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    @MichaelChirico: What Mitch wrote leads to Farey sequences and the Stern-Brocot tree. IMHO, I don't think Mitch's idea makes an easier solution to this specific question in isolation; but it gives an answer that connects up to further interesting topics. – Zach Teitler Sep 06 '17 at 03:46
  • @Joshua Depends what you mean. There are countable ordinals which are not $\omega$, but the rationals are not wellfounded under the usual order, so this isn't evidence for that assertion. There are no countable cardinals which can't be counted, because "countable" precisely means "can be counted". – Patrick Stevens Sep 06 '17 at 11:00
  • @MichaelChirico What Zach said. The median (my number) is, as you say, surely not intuitive, but it is elementary to check. Both Thomas's and my intermediate number (called the 'mediant' show that '<' does not provide a countable ordering for rationals. But mine (via Stern-Brocot) covers all and only the rationals via the Farey sequence, which does provide a countable ordering . But the arithmetic mean only covers rationals with denominator a power of 2. – Mitch Sep 06 '17 at 12:22
  • @PatrickStevens: http://www.homeschoolmath.net/teaching/rational-numbers-countable.php https://math.stackexchange.com/questions/659302/how-to-prove-that-mathbbq-the-rationals-is-a-countable-set ; Rationals are countable. – Joshua Sep 06 '17 at 15:16
  • @Joshua I am fully aware that the rationals are countable. However, the rationals can be counted. They are not evidence in favour of your assertion that "there is a countable infinity that can't be counted". – Patrick Stevens Sep 06 '17 at 15:35
  • @PatrickStevens: The rationals do not count in their correct order. The correct-order rationals must inherently also be countable as the number of elements did not change, but trying to project to 1,2,3,... won't work. – Joshua Sep 06 '17 at 15:39
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    @Joshua So you mean "there is a countable total order which is not $\omega$"? I mean, $\mathbb{Z}$ is a simpler one. – Patrick Stevens Sep 06 '17 at 16:05
  • @Mitch thanks! I figured there must be something deeper going on behind that choice. Perhaps worth adding as a separate answer! – MichaelChirico Sep 06 '17 at 18:37
  • @MichaelChirico haha. that's math. There's always something deeper. $a+b=b+a$? Really pretty boring. But ... what if...? – Mitch Sep 06 '17 at 19:03
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You are not mistaken. Even assuming you want just the non-negative numbers, so that $a_0=0$, you cannot pick $a_1$ correctly because you will have skipped $a_1/2$.

Alternatively, you could allow negative indices (and have $\ldots, a_{-1}, a_0, a_1,\ldots$), which would also solve your "there is no smallest rational number" problem, but it still has the problem that there are rational numbers between any two numbers. Specifically, if the list is complete, we must have some $n$ such that $a_n=0$. But then we necessarily miss $a_{n+1}/2$.

Arthur
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As Arthur already stated, such a numbering cannot start with a finite index, i.e. either your sequence will count from $-\infty$ to $\infty$ or you only count the non-negative rational numbers (i.e. $\mathbb Q^+_0$). And as Thomas showed, a "normal" sequence also won't work since you'll always find a number in between.

However, you can define a sequence of sequences $\{\{a_{nm}\}_{n=-\infty}^\infty\}_{m=1}^\infty$ such that its $\lim_{m\to\infty}$ yields a sequence counting all rational numbers. As an example, consider the typical enumeration sequence of $\mathbb Q$ (see e.g. here) and let $\{a_{nm}\}_n$ be the ordered sequence of the first m rational numbers obtained that way. The thing is, though, you will just end up with $\mathbb R$...

Tobias Kienzler
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  • I'm pretty sure there's no such thing as the $\lim_{m\rightarrow \infty}$ you describe - and if there is, it's not a sequence. – MartianInvader Sep 05 '17 at 22:53
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    @MartianInvader For a mathematically sound proof one would definitely need more elaboration, and yes, calling this thing a "sequence" may be handwaving, it's rather a meta-sequence (or something analogous to calling a function of functions a functional...) – Tobias Kienzler Sep 06 '17 at 07:39
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To parallel Thomas's answer with a more obscure one:

If $a_k < a_{k+1}$ and $a_k = \frac{b_k}{c_k}$ then the mediant $x:= \frac{b_k+b_{k+1}}{c_k+c_{k+1}}$ is a rational number between those two, so no.

($b_k$ and $c_k$ are relatively prime/are in lowest terms)

It is elementary but not obvious that the mediant $x$ is between the two (exercise for the reader). The reason this answer is even a thing (Thomas gave a much simpler and easier to understand answer) is because using the mediant we can construct an ordering on the rationals that is countable, and in addition gives all and only the rationals.

The usual proof that the (positive) rationals have equal cardinality to the naturals is by 'dovetailing'. counting along antidiagonals of pairs and then ignoring rationals that have already been seen (ignoring if gcd $\neq 1$). This is somewhat unsatisfying because it doesn't give an explicit bijection with the naturals. To get the implied rational-natural bijection from gcd, see the Stern-Brocot tree for details.

Mitch
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As stated, the answer is no, because the question uses the symbol $<$ which has the implied meaning: The usual ordering of $\mathbb{Q}$ where $\frac{a}{b}<\frac{c}{d}$ iff $ad < bc$ in $\mathbb{Z}$.

But.

As mentioned in another answer, $\mathbb{Q}$ can be well-ordered, i.e. one can define a different order $\prec$ with the property that every nonempty subset of $\mathbb{Q}$ contains a least element with respect to $\prec$. For this ordering, a monotone sequence containing all of the rationals is easy to construct: let $x_1$ be the smallest rational, let $x_2$ be the smallest element of $\mathbb{Q} \setminus \{x_1\}$, etc.

Sort of Damocles
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  • And here is how you can build such an ordering. Take a bijection $f\colon\mathbb Q\to\mathbb N$, and, well, that is your ordering: $p\prec q$ iff $f(p)<f(q)$. – Joker_vD Sep 06 '17 at 17:54