1. Given a chain-complete poset $P$, every $x\in P$ lies below some maximal element.

  2. Every inductive poset has enough maximal elements a maximal element.

Chain-complete means every chain has a least upper bound, inductive means every chain has an upper bound.

Are the above statements both correct and equivalent? What exactly does "enough maximal elements" mean? Can one proof $1\implies 2$ without any further assumptions?

/edit: The "enough" thing is from handwritten notes maybe it's a mistake. "All chains have upperbounds, then there is a maximal element" seems to be the most common definition. If we take this as $2$, are $1$ and $2$ equivalent? If I'm not mistaken, $1$ has both a stronger condition and a stronger conclusion?

  • 2
    $\begingroup$ Where have you found the second statement? Any finite chain has exactly one maximal element (while being trivially inductive). The most common version of Zorn's lemma states that there is at least one maximal element. $\endgroup$ Jan 23 '16 at 22:29
  • 1
    $\begingroup$ A note on (1): If $\mathbb{P}$ is any poset, let $\mathbb{P}'$ be the set of downwards-closed chains in $\mathbb{P}$, ordered by inclusion. Then $\mathbb{P}'$ is chain-complete, and if every chain in $\mathbb{P}$ has an upper bound (which is weaker than $\mathbb{P}$ being chain complete) then a maximal element of $\mathbb{P}'$ corresponds to a maximal element of $\mathbb{P}$. $\endgroup$ Jan 23 '16 at 22:31
  • $\begingroup$ Whether 1 $\to$ 2 or not really depends upon what "enough" means. After reading 2. I was expecting a definition :) Perhaps someone will know. $\endgroup$
    – BrianO
    Jan 23 '16 at 22:34
  • $\begingroup$ Maybe the "enough" is just a mistake, see my edit above. If we replace it by the version Random Jack pointed to, is this equivalent to the "stronger" statement in 1.? $\endgroup$
    – akkarin
    Jan 23 '16 at 23:05
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    $\begingroup$ "Enough maximal elements" reminds me of the statement that a category has "enough injective objects" or "enough projective objects". I'm guessing that whoever stated this comes from algebra and category theory? $\endgroup$
    – Asaf Karagila
    Jan 23 '16 at 23:14

Yes, one can easily prove they are equivalent.

The reason is that in order to prove the Axiom of Choice, for example, we resort to a partial order where each chain has a least upper bound. Similarly, to prove Hausdorff's maximality principle from Zorn's lemma we use only partial orders in which a chain has a least upper bound.

And this can be used to prove Zorn's lemma in its inductive version from the chain-completeness version. Given an inductive partial order $P$ and $p\in P$, first show there is a maximal chain $D$ such that $p\in D$, and you can do that by considering all the chains which have $p$ as a member, and showing this is a chain-complete partial order. Then conclude that an upper bound of $D$ must be a maximal element which lies above $p$.


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