(It's likely that this entire question is answered with a "yes, you're right" comment. I'm just proceeding with extreme caution after discovering an error in my notes.)

I was reading my years-old university notes (Part II, Logic and Set Theory, lectured by Imre Leader at Cambridge), and I discovered a note which is by now cryptic to me.

We did not use the axiom of choice in proving Hartogs's lemma. The only time we used AC in the Ordinals chapter was in the remark that $\omega_1$ is uncountable.

This presumably was sloppy note-taking and is false, because unless I'm horribly wrong, $\omega_1$ is defined (without Choice) to be the first uncountable ordinal: it's the Hartogs ordinal of $\omega$.

Someone else's notes phrased it thus:

… apart from the remark that $\omega_1$ is not a countable supremum — which used the fact that a countable union of countable sets is countable.

I agree that "a countable union of countable sets is countable" is a countable-choice statement, but a well-ordered union of well-ordered sets (such as ordinals) is certainly well-orderable without any choice principle, right?

My best guess is that this note is referring to the following remark, which was lectured very near to the definition of $\omega_1$:

Every countable sequence of countable ordinals is bounded by its sup, which is a countable ordinal. Hence every countable sequence of ordinals "does not get close to" $\omega_1$.

Is the choicy part the italicised "which is a countable ordinal", and in fact if Choice fails, $\omega_1$ could be a countable sup (i.e. $\aleph_1$ could be singular)? This would match up with the answer of Existence of a regular uncountable $\aleph_{\alpha}$ without $\mathsf{AC}$ .


1 Answer 1


Yes, you're right, it should have said that choice is necessary for proving that $\omega_1$ is regular. And indeed it is consistent with $\sf ZF$ that $\omega_1$ is singular.

However, you're wrong about a well ordered union of well ordered sets. For example, the real numbers could also be a countable union of countable sets (this is true in the first model in which $\omega_1$ is singular, due to Feferman and Levy). Or, the classic example of a set of socks which is a countable union of pairs which is Dedekind finite (so there is no infinite set of these pairs that admits a choice function).

Of course, if we understand "well ordered set" as a set with a specified well order, then it is indeed the case that a well ordered union of well ordered sets is well ordered. The key problem in the above examples is exactly that we cannot choose a well ordering in a uniform manner.

  • $\begingroup$ I'm not entirely convinced by the socks example here. The pairs aren't individually well-ordered, and as I understand it, that's the whole point: we need a choice principle to define a "least" sock from every pair simultaneously. So in my understanding, you have a countable (maybe even well-ordered) union of countable sets (pairs), but not of well-ordered sets. But I'll go and think about it. Thanks! $\endgroup$ Aug 18, 2021 at 10:33
  • $\begingroup$ Okay, then maybe you need to clarify what you mean exactly by well-ordered. If you mean that there is a well-order attached to each set, then yes, you're right. I'll clarify this in the answer. $\endgroup$
    – Asaf Karagila
    Aug 18, 2021 at 10:34
  • $\begingroup$ As Bertrand Russell said, AC is needed for socks but not for shoes. $\endgroup$ Aug 18, 2021 at 10:34
  • $\begingroup$ @PatrickStevens . The problem with the socks is not whether a pair can be well-ordered. As you said, if $\{P_n:n\in\omega\}$ is a set of pairs of socks, the Q is whether there exists a function $S=\{(n, S(n)):n\in \omega\}$ such that $S(n)\in P_n$ for all $n.$ Equivalently, whether there exists $\{W_n:n\in\omega\}$ where each $W_n$ is a well-order of $P_n.$ $\endgroup$ Aug 18, 2021 at 10:48
  • 1
    $\begingroup$ @Patrick: Drop a penny, lose a penny. $\endgroup$
    – Asaf Karagila
    Aug 18, 2021 at 17:37

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