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I'm searching for an easy example of a set that cannot be totally ordered without axiom of choice.

I know that amorphous sets are a possible candidate, but it's quite difficult to prove that amorphous sets exist (i.e. it's difficult to prove that ZF+"amorphous sets exist" is consistent relative to ZF).

Are there other examples of sets that cannot be totally ordered without axiom of choice, such that it's really easy to prove the existence?

effezeta
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  • A slightly "simpler" argument for the fact that $\mathcal P(\Bbb R)$ is not provable linearly orderedable in ZF is simply that it may contain an amorphous set. – Asaf Karagila Jun 13 '23 at 13:28
  • But irrespective of anything, in order to prove anything like "some sets might not be linearly orderable" you will, at the end, have to either rely on some difficult to prove fact, or prove it yourself. There's no escape from the conservation of work here. – Asaf Karagila Jun 13 '23 at 13:29
  • @AsafKaragila: in this way I can prove that ZF+"amorphous sets exist" is consistent relative to ZF. But can I prove that in a generic metatheory (ZF or others) without axiom of choice this result holds? – effezeta Jun 13 '23 at 13:40
  • @AsafKaragila Your comments are on point as always. I just wanted to point out that the way I read the question, the OP is not asking for a set $X$ such that it is easy to prove that $X$ may not be linearly orderable, but rather a set $X$ such that (a) it is easy to prove that $X$ exists (maybe it's better to say: it's easy to describe/construct $X$), and (b) it is consistent that $X$ is not linearly orderable (but it is not necessarily easy to prove this). – Alex Kruckman Jun 13 '23 at 13:43
  • Ok, but now I'm cofused by the OP's latest comment, so I'm going to bow out of this conversation! – Alex Kruckman Jun 13 '23 at 13:45
  • Much like @Alex, I don't understand your comment at all, effezeta. – Asaf Karagila Jun 13 '23 at 15:22
  • In the case of ZF everything is clear: ZF + "amorphous sets exist" is consistent relative to ZF, so ZF cannot prove that every set can be totally ordered. Now consider the case of a generic metatheory T (instead of ZF) without axiom of choice: can we say that T cannot prove that every set can be totally ordered? – effezeta Jun 13 '23 at 16:00
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    But these are consistency statements. These are statements about the natural numbers. If T is even remotely capable formalising first-order logic (e.g. PRA), it will be strong enough to have the consistency proofs done in it. But what is a "generic metatheory"? And what does the metatheory have to do with this? The metatheory is where our proofs live, not where the objects are. You seem to be asking "will a generic foundation of mathematics prove that all sets can be linearly ordered", to which there is no answer since "generic foundation" is too vague of a term. – Asaf Karagila Jun 13 '23 at 16:44

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