I am struggling through Abbott's Understanding Analysis and have been asked if there is a sequence that contains subsequences that converge, severally, to each term of the harmonic sequence. The rational numbers between zero and one would seem to fit the requirement; but are they, if supposed to be in order of size (and could they be put in order of size? There is no rational number that is the first after zero) a sequence?

They could not be listed in order of size, but there is a natural number for every rational number; one could map rationals to naturals; but is that enough for the rational numbers to be a function of the real numbers? (And hence be a sequence.)

I have looked through all the previous questions on this topic but none seem to answer my question. Or perhaps they do but they are too advanced for me to understand them.

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    $\begingroup$ The rationals are dense in themselves so they cannot be sequenced and respect the order. Consider the sequence $(1/2,1/3,1/2,1/3,1/4,1/2,1/3,1/4,1/5,....)$ to solve your problem. $\endgroup$ Nov 25, 2021 at 1:17
  • $\begingroup$ @CyclotomicField I was saving that one for the next exercise which requires a sequence that does not converge to any number not in the harmonic sequence! So you have answered my question: 'no, the rational numbers cannot be a sequence' ? $\endgroup$
    – Kang
    Nov 25, 2021 at 16:05
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    $\begingroup$ They can be sequenced, but they can't be sequenced in a way that respects the order. You can probably prove this yourself by contradiction and using the density of the rationals. $\endgroup$ Nov 25, 2021 at 17:14

2 Answers 2


To list the rationals: sort them by the sum of numerator + denominator.

2/1, 1/2.
3/1, 1/3.
4/1, 3/2, 2/3, 1/4.
6/1, 5/2, 4/3, 3/4, 2/5, 1/6.

and so on.

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    $\begingroup$ Each denominator in the harmonic sequence would appear, and they would be roughly sequential... $\endgroup$
    – abiessu
    Nov 24, 2021 at 18:42

I doubt an infinite set of rationals can be ordered but a finite set can be ordered as follows.

  1. Generate an arbitrary number of rationals using the pairing function

  2. Find the least common multiple of these denominators

  3. Express each rational with this LCM as the denominator

  4. Order the resulting rationals by numerator.

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    $\begingroup$ Of course you can order the rationals, just use cantor's diagonilization (as in the other answer) to create a bijection between the rationals and the naatural numbers, then use the induced order from natural number < $\endgroup$
    – Alan
    Nov 25, 2021 at 3:51

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