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I am working on how the Axiom of choice implies Zorn's Lemma. I am using Devlin's Joy of Sets, and it is on page. 60. I have a small question concerning the end of the proof. I have taken photos of the entire proof, but if the photos are too hard to see [they look good to me though] I can write out the entire proof so people know what came before.

Devlin writes:

Let Let $\eta$ be least such that $f(\eta) = \lambda$. If $\eta$ is a limit ordinal, then the sequence $\langle p_{f(\xi)}\mid\xi < \eta\langle$ is a chain in $P$ with no upper bound, which is impossible [i.e. because it violates the assumption of Zorn's Lemma: "every chain in P has an upper bound]. Hence $\eta = \nu + 1$ [i.e. a successor ordinal]. Clear $p_{f(\nu)}$ is a maximal element of $P$.

Question: Why does $\nu$ not run into the same problem as $\eta$, namely, being a limit ordinal such that the new sequence would again have no upper bound?

Recall: a maximal element $a$ of $P$ is maximal if and only if there is no $b$ such that $a < b$.

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  • $\begingroup$ Why would it be a problem if $\nu$ is a limit ordinal? All that matters is that $p_{f(\nu)}$ is a maximal element of $P$. $\endgroup$ – Mees de Vries Nov 3 '17 at 14:45
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Because a successor ordinal is an ordinal that as a linear order has a maximal element.

The point is that $\nu$ is the last ordinal before $\nu+1=\eta$, and therefore $p_{f(\nu)}$ is the last member of $P$ which we defined in the first $\eta$ steps. Therefore, $p_{f(\nu)}$ is an upper bound of the chain, and by choice of $\eta$ also a maximal element.

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