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My understanding is that the more general case of this (the countable union of countable sets) requires choice to prove. However in the case of ordinals, it should be provable without choice since there is an explicit well ordering on ordinals. Based on this, how does one go about constructing the explicit surjection needed to prove this?

MMR
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    The short answer is that just because the ordinals are well-ordered does not mean that the set of the enumeration is well-ordered. And the choices you make using the axiom of choice are from the sets of enumerations of each ordinal. Feferman and Levy constructed a model in which $\omega_1$ is indeed the countable union of countable ordinals. – Asaf Karagila Nov 28 '18 at 13:05
  • I'm unsure if I understand what you mean by the set of enumerations. – MMR Nov 28 '18 at 13:13
  • The way you prove a countable union of countable sets is countable is that you first enumerate each set, namely you find an injection into $\omega$, and then you make this into an injection from the union into $\omega\times\omega$, which we can prove is countable "by hand". Choice kicks into the game when you need to choose the enumeration for each set. But even if each set is in fact a countable ordinal, there is no canonical way to enumerate all the countable ordinals. – Asaf Karagila Nov 28 '18 at 13:26
  • I'm not sure if I see why there is no way to enumerate the union, since the union is itself an ordinal and has well ordering defined on its elements. Or is this only true of a finite union? – MMR Nov 28 '18 at 13:40
  • Yes, the union is an ordinal. But why is it countable? Enumerate in the broad sense means "put in bijection with an ordinal", yes. But here we use it to specifically talk about countable sets. – Asaf Karagila Nov 28 '18 at 13:40
  • I see, thank you. – MMR Nov 28 '18 at 13:42
  • @AsafKaragila Why is this different from this proof, where there is an injection onto $\omega_1$? Surely an injection onto this set is functionally the same as being countable? – MMR Nov 28 '18 at 15:32
  • (1) Note the difference between "injection into" and "injection onto", the latter means a bijection. (2) Well, $\omega_1$ is not countable. Moreover the supremum of any set of countable ordinals is at most $\omega_1$ by definition. So there is always an injection into $\omega_1$. But it is not enough to conclude the injection is never a bijection. – Asaf Karagila Nov 28 '18 at 15:34

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