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I'm doing a introductory course on commutative algebra, and have just been introduced to the chain conditions. I know there's a lot of questions probably similar, but I haven't really understood this. In an online lecture on chain conditions, the following Proposition was proved:

Prop 1: TFAE

  1. Every chain $a_1\leq a_2\leq a_3\dots $ in a poset $\Sigma$ terminates, i.e. $\exists n_0$ s.t. $a_n=a_{n+1}, \quad \forall n\geq n_0$.
  2. Every non-empty subset of $\Sigma$ has a maximal element.

Then we have Zorn's:

Zorn's Lemma: Suppose a partially ordered set P has the property that every chain in P has an upper bound in P. Then the set P contains at least one maximal element.

Question: To me it then seems as if Prop 1 $\Rightarrow$ Zorn's Lemma, since the first part of Zorn's seems imply 1 of Prop 1. And the whole of $P$ is certainly a non-empty subset of $P$, so it has a maximal element by Prop 1. What have I misunderstood?

I did read this thread, but I didn't get any wiser. I think my problem understanding this is even more fundamental/trivial; But perhaps I missed something.

EDIT: Or is it just that perhaps it isn't true that part 1 of zorn's imply 1 of Prop 1? Or something else/more?

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No. The assumption of Zorn do not imply part 1 of your proposition.

For instance, consider the ordinal $\omega+1$ as an ordered set. It has a greater element, $\omega$. However, it has an infinite increasig sequence given by $a_n=n$.

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  • $\begingroup$ I suspected this a short moment after I posted, as I edited, but I wanted to see if it was perhaps something more; But this is essentially where it breaks down I guess. $\endgroup$ – Christopher.L Feb 11 '19 at 21:59

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