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With the axiom of choice, we can select a ration number $r_k$ in each interval $[a_k,b_k]$ and we can easily prove $Q$ is countable,so this complete the proof. However,without the axiom of choice, does each interval has a rational number which we can select?

Andrés E. Caicedo
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J.Guo
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1 Answers1

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You don't need the axiom of choice to pick a rational number in each interval. Since one can define an explicit bijection between $\mathbb Q$ and $\mathbb N$, you can decide once and for all to pick the rational in each interval that maps to the smallest natural. Then there's no choice (in the sense of the axiom of choice) going on.

hmakholm left over Monica
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    Nice answer. Sorry about my earlier comment. – Kavi Rama Murthy Sep 24 '18 at 06:20
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    How exactly does this go through, formally? Is it just that you can write a formula in $ZF$ expressing the choice function? – Jack M Sep 24 '18 at 06:21
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    Henning.Nice answer.Perhaps one could mention that since Q is dense there are rationals in the interval, and hence this subset of Q is not empty?Thanks. – Peter Szilas Sep 24 '18 at 06:30
  • Perhaps using dyadic rational makes picking points even simpler. – Kavi Rama Murthy Sep 24 '18 at 09:01
  • @JackM: Yes, since you can write a formula in ZF that expresses the bijection with $\mathbb N$, and you can write a formula in ZF that expresses a choice function on $\mathbb N$. – hmakholm left over Monica Sep 24 '18 at 10:45
  • @KaviRamaMurthy: In isolation it would probably be simpler to say something like "choose the smallest denominator that is represented in the interval, and if there are more than one rational with that denominator use the smallest of those" -- but I wanted to suggest the point that we don't need to use choice for anything that is known to have an $\aleph$ number as cardinal. – hmakholm left over Monica Sep 24 '18 at 10:47
  • @HenningMakholm And so in particular, it's possible to justify the existence of the choice function (as a set) using only the axioms of specification, pairing, union, etc (whatever else you end up needing) of ZF? Really the existence of a formula is not the important point, but maybe formulae in ZF are in some sense isomorphic to proofs of existence using ZF axioms? – Jack M Sep 24 '18 at 10:48
  • @JackM: What you actually need for the proof to go through to have is the formula, not the choice function. (If you had a choice function, what you would use it for is to construct a formula that mentions it). See this answer for a longer explanation. – hmakholm left over Monica Sep 24 '18 at 10:52
  • @PeterSzilas: That part of the argument appears to be implicit in what the OP already had in the question. – hmakholm left over Monica Sep 24 '18 at 10:52
  • Henning.Thanks for your comment. – Peter Szilas Sep 24 '18 at 11:10