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This is a question on chap. 9 of Introduction to Set Theory of Hrbacek and Jech. The book define that the Zorn's Lemma is: If every chain in a partially ordered set has an upper bound, then the partially ordered set has a maximal element.

This is very confused to me. I tried find a upper bound for each chain (is the $\subseteq$-maximal element ?) but I don't know how to try this. Furthermore how I will find the maximal element for the partially ordered set?

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    $\begingroup$ You can't necessarily find the max element. Zorn only tells you that such an element exists. $\endgroup$
    – user4894
    Dec 1, 2014 at 19:09
  • $\begingroup$ You didn't really ask about any equivalences of Zorn's lemma. So I took the liberty of changing the title. $\endgroup$
    – Asaf Karagila
    Dec 1, 2014 at 19:16

2 Answers 2

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The point of the lemma is that it guarantees the existence of a maximal element. It doesn't tell you which element it is.

Just like the axiom of choice guarantees that there is a function $f\colon\mathcal P(\Bbb R)\to\Bbb R$ such that for every non-empty set $A$, $f(A)\in A$. It doesn't tell us what is $f(\Bbb R)$ or what is $f(\Bbb{R\setminus Q})$ or anything like that. It exists, move along.

Zorn's lemma, the axiom of choice, the well-ordering principle, and so on, only claim that certain objects exist. Since modern mathematics is often non-constructive, we don't have to actually define and construct the objects we use. It suffices to prove their existence, and Zorn's lemma is a wonderful tool for just that.

So Zorn's lemma tells us, essentially, that if whenever we pick a chain in $P$, then we can find an element which is larger (or equal) to all the elements in the chain; then we can be sure that there is a maximal element in $P$.

This upper bound doesn't have to be unique, or "tight" either. It just has to be an upper bound.

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You apply Axiom of Choice every time to pick a larger element for your incomplete chain. By every time, you are constructing a function on all ordinal numbers to your poset, since the function is increasing, so 1-1, eventually it stops at some ordinal, then the corresponding one is your maximal element.

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