Is there something like a "countable Zorn's lemma" which is equivalent to the axiom of countable choice?


Not exactly. But there is one equivalent to Dependent Choice, the big brother of countable choice, and the principle most mathematicians think of when they think of countable choice.

Theorem. The Principle of Dependent Choice is equivalent to the statement that every partial order has a maximal element or a countable chain.

The proof of this ranges from almost trivial to simple to "there's a small trick to remember", depending on how you define Dependent Choice.

Note, however, that the statement is not exactly the Zorn's Lemma condition, but it's not far from it: it omits the assumption "every chain which is finite has an upper bound". That is due to the fact that this is a theorem of $\sf ZF$, so it can be omitted. But if we want to generalize the equivalence, we need to bring it back. Which leads us to the following,

Theorem. The Principle of Dependent Choice is equivalent to the statement: If $P$ is a partial order where every finite chain has an upper bound, then $P$ contains a countable chain or a maximal element.

Which can now be generalized to arbitrary cardinals.

Theorem. The principle $\sf DC_\kappa$ is equivalent to the statement: If $P$ is a partial order that every chain of order-type $<\kappa$ has an upper bound, then $P$ has a maximal element or a chain of type $\kappa$.

These were proved by Wolk in the following paper

Wolk, Elliot S., On the principle of dependent choices and some forms of Zorn’s lemma, Can. Math. Bull. 26, 365-367 (1983). ZBL0491.04005.

  • $\begingroup$ If by "chain" you mean a linearly ordered subset, doesn't a finite chain always have an upper bound (the maximal element in the chain)? $\endgroup$ – Ur Ya'ar Feb 28 '18 at 7:52
  • $\begingroup$ Yes. Indeed. You can rephrase the first one as "no maximal element". But it doesn't lend itself to the generalization. $\endgroup$ – Asaf Karagila Feb 28 '18 at 7:53
  • $\begingroup$ I don't understand, you need DC to conclude that a finite linear order has a maximal element? Are you sure your statment of the first theorem is correct? Or am I missing something (maybe about what definition of finite you are using?) $\endgroup$ – Ur Ya'ar Feb 28 '18 at 7:56
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    $\begingroup$ Okay. Let's go Socratic. How about you reformulate this in a way that makes sense to you, and then we can see how well it generalizes? (Because the point of the redundancy of the countable case is necessary for the generalization.) $\endgroup$ – Asaf Karagila Feb 28 '18 at 8:00
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    $\begingroup$ Is everyone happy now? (I know I am, it's snowing like crazy in Norwich.) $\endgroup$ – Asaf Karagila Feb 28 '18 at 8:12

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