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I was learning Zorn's lemma yesterday and I couldn't find any example, where Zorn's lemma fails when we only require that every countable chain in P has a maximal element in P. Does anyone know an example?

Long
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3 Answers3

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Consider the collection of countable subsets of $\Bbb R$, ordered by inclusion.

Every countable chain has an upper bound, since the countable union of countable sets is countable; but there is no "maximal countable subset".

(It is consistent for the axiom of choice to fail and $\Bbb R$ to be the countable union of countable sets, in which case this example is not a counterexample; and the same goes for Brian's example. But if the axiom of choice fails that bad, then there are other counterexamples to your question.)

Asaf Karagila
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  • +1, but: I think you should explain better that your example needs AC(-omega), because otherwise a countable union of countable sets need not be countable. Also, I think that Brian's example does not need AC. Afaik you can prove the existence of uncountable well-ordered sets without AC. – Carsten S Oct 25 '13 at 07:40
  • @Carsten: If $\omega_1$ is the countable union of countable sets, then it does. And if $\Bbb R$ is the countable union of countable sets, then so is $\omega_1$. While choice is not needed to prove that there are uncountable well-ordered sets; it is needed to prove that there are well-ordered sets whose cofinality is uncountable. – Asaf Karagila Oct 25 '13 at 07:41
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    Maybe it's over my head, but I thought Zorn's lemma and AC are equivalent? – Long Oct 25 '13 at 07:41
  • I see, I was not aware of that. Thanks for the clarification! Still I think it would be good if you pointed out more precisely in your answer at which point you need AC. It is somewhat surprising to first read a proof and then that the proven fact need not hold ;) – Carsten S Oct 25 '13 at 07:47
  • @user1412 Yes, they are. – Carsten S Oct 25 '13 at 07:49
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The first uncountable ordinal, $\omega_1$, is an example: every countable subset of $\omega_1$ has a least upper bound in $\omega_1$, but $\omega_1$ itself has no maximal element.

Brian M. Scott
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Here's an example from topology.

Given some complete topological space (like $\mathbb R$), we can consider the set of open dense subsets of $\mathbb R$. It's clear that there isn't a minimal element (order by inclusion) in this set (remove any element from an open dense subset and its still open and dense), but the Baire Category theorem implies that every countable chain has a bound.

Rushabh Mehta
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    Interestingly enough, in terms of how much AC is needed to make the counterexamples brought up on this page actual counterexamples we go: your general counterexample needs more than Brian's, which needs more than mine, which needs more than your specific example of $\Bbb R$ which holds if ZF. – Asaf Karagila Apr 21 '21 at 07:24