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I know that there are many counterintuitive situations that can't be disproved in $ZF$, some enumerated in this answer (I would add the existence of a finite undetermined game, from this answer [by the same author]). However, some of these can be solved with a weaker choice principle than the full Axiom of Choice. I'm looking for ones that can't. What are some counterintuitive statements one can deduce from $ZF + \lnot AC$?

Examples of other statements that can be proven in $ZFC$ but not in $ZF+DC$ (dependent choice) would also be interesting.

acupoftea
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  • Given that dependent choice implies countable choice, there's no need to specify both. – Mark Saving Sep 09 '21 at 19:25
  • @MarkSaving oh I forgot about that, thanks and I'll edit it – acupoftea Sep 09 '21 at 19:26
  • One example is: there are sets $A, B$ such that neither $|A| \leq |B|$ nor $|B| \leq |A|$. – Mark Saving Sep 09 '21 at 19:32
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    The question about counterintuitive statements that can be deduced from $ZF + \neg AC$ is basically asking for propositions $P$ such that $ZF \vdash P \iff AC$ and taking the negation of $P$. So you're really asking for intuitive statements that are equivalent to the axiom of choice in the first part of your question. – Mark Saving Sep 09 '21 at 19:33
  • @MarkSaving Strictly speaking, counterintuitive consequences of $\mathsf{ZF}$ would also be counterintuitive consequences of $\mathsf{ZF+\neg AC}$ :P. – Noah Schweber Sep 09 '21 at 19:41
  • @NoahSchweber I guess you got me – Mark Saving Sep 09 '21 at 19:43
  • @MarkSaving I guess that would be a formally valid answer, but I don't see how that's what the question is "basically asking for" (unless it's in the sense of "asking for a beating" ;) ). Couldn't the $P$ be strictly stronger than $AC$ ? Or perhaps of unknown implication status – acupoftea Sep 09 '21 at 19:59
  • Related: https://math.stackexchange.com/questions/2378211/what-are-some-unintuitive-consequences-if-we-assume-that-axiom-of-choice-is-wron https://math.stackexchange.com/questions/1017361/what-is-the-opposite-of-the-axiom-of-choice https://math.stackexchange.com/questions/199087/which-set-is-unwell-orderable https://math.stackexchange.com/questions/1777364/what-is-an-example-of-two-sets-which-cannot-be-compared (If people feel that this should be closed as a duplicate, I'm happy to delete my answer, although I feel that it adds nicely to the already existing discussion as a whole). – Asaf Karagila Sep 09 '21 at 20:11
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    @acupoftea ZF + $\lnot$AC proves exactly 1) the consequences of ZF 2) the negation of every sentence that implies AC over ZF. How could something be of unknown implication status and still qualify? – spaceisdarkgreen Sep 10 '21 at 00:40
  • @spaceisdarkgreen oops, I should've thought the basic logic through before commenting. Thank you for the correction. – acupoftea Sep 10 '21 at 14:29

1 Answers1

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The thing is that $\lnot\sf AC$ is just as non-constructive as $\sf AC$. Just like $\sf AC$ tells you that every set can be well-ordered, but it doesn't specify the well-ordering; $\lnot\sf AC$ simply tells you that some sets cannot be well-ordered, but it does not specify which sets.

So in general, all you can prove is that statements that are equivalent to $\sf AC$ fail. This may end up as counterintuitive in some cases, but that really depends on your intuition. Some examples:

  1. Some family of non-empty sets does not have any choice functions.
  2. Some two sets cannot be compared in cardinality.
  3. Some vector space does not have a basis.
  4. Some tree (a partial order where every downwards cone is well-ordered) is closed, in the sense that every increasing sequence has an upper bound, but there are no chain which meets every level of the tree.
  5. Some forcing notion $\Bbb P$ and some formula $\varphi(x)$ satisfy that $1\Vdash\exists x\varphi(x)$, but there is no $\Bbb P$-name $\dot x$ such that $1\Vdash\varphi(\dot x)$.
  6. Some surjection does not admit an inverse.
  7. Some set does not admit a group structure.
  8. Some commutative ring with a unit does not have any maximal ideals.

Again, we can just enumerate all the equivalences of the axiom of choice and consider their negation. That's pretty much all we can do here.

We cannot prove that $\sf BPI$ holds, but we cannot prove that it fails. We cannot prove that $\sf DC$ holds, but we cannot prove that it fails either. We cannot prove that $\mathcal{PPPPPPPPPPPPPP}\Bbb R$ can be well-ordered, but we cannot prove that it cannot be well-ordered either. And so on.

But, here are a few things that are at the very least consistent with $\sf ZF+DC+\lnot AC$.

  1. $\Bbb R$ can be partitioned into strictly more parts than elements. Or, just as well, it can be partitioned so that the partition is incomparable with $\Bbb R$ itself.
  2. Some infinite set $A$ satisfies that $|A|<|A\times 2|$.
  3. $\mathcal P(\Bbb R)$ cannot be linearly ordered.
  4. $\Bbb R$ does not have a Hamel basis over $\Bbb Q$.
  5. Some vector space over some field satisfies that every proper subspace has a countable basis, but the space itself does not.
  6. For every set $x$, there is some $A_x$ such that $A_x$ is the union of $\aleph_1$ sets of size $\aleph_1$, and $\mathcal P(A_x)$ can be mapped onto $x$.

And, again, there are a lot more.

Asaf Karagila
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  • What is a good reference for proves of the results you mention as consistent with $\sf ZF+DC+\lnot AC$? – mathcounterexamples.net Sep 09 '21 at 20:25
  • One reference: Alan Taylor and Stan Wagon, A paradox arising from the elimination of a paradox, Amer. Math. Monthly vol. 126 (2019), pp. 306-318. – Dan Velleman Sep 09 '21 at 20:45
  • @DanVelleman: I helped with article! (And as far as I understand, actually helped inspire it!) – Asaf Karagila Sep 09 '21 at 20:47
  • @mathcounterexamples.net: 1, 3, 4 follow from $\sf ZF+DC+BP/LM$; 2 follow from the existence of an $\aleph_1$-amorphous set, which is consistent with $\sf ZF+DC$; 5 appears in my M.Sc. thesis; and 6 follows from my results in "The Morris model" and "Iterated failures of choice". – Asaf Karagila Sep 09 '21 at 20:49
  • I don't understand why it wouldn't be possible that $ZF + \lnot AC$ proves the negation of something stronger than $AC$, or even something like the other implication direction is unprovable, or at least unkown. I'm not expecting it but I don't see the connection with your first paragraph. – acupoftea Sep 09 '21 at 21:05
  • @acupoftea: Sure, it proves everything that is stronger than AC. But isn't that obvious? And if that's what you're after, wouldn't the right question be "what axioms are stronger than AC" instead? – Asaf Karagila Sep 09 '21 at 21:09
  • @AsafKaragila the question was about counterintuitive statements $S$ provable from $\lnot AC$, regardless of the status of the implication $\lnot S \implies AC$. In principle it doesn't (I think) reduce to any single one of the questions where we also assert that the implication must be true, false, independent of $ZF$ or of unknown status, only to all of them at once. Perhaps it does reduce to a single one in practice, then I think that a perfect answer should address that. I don't see how your first paragraph relates to it, even though from the second one it looks like it's supposed to. – acupoftea Sep 09 '21 at 21:38
  • @acupoftea: Statements stronger than AC are not usually considered to be counterintuitive, nor do their negations. As both are usually consistent with ZFC. – Asaf Karagila Sep 09 '21 at 21:56
  • @acupoftea: The point is that "intuitive" usually refers to a subtheory of ZFC (not closed under implications or logical equivalence, usually, obviously). GCH is not "intuitive", nor its negation is "counterintuitive". So saying that it should be mentioned because it's a consequence of the negation of AC seems to be a bit redundant, as far as answers go. – Asaf Karagila Sep 09 '21 at 23:21