Given a chain-complete poset $P$, every $x\in P$ lies below some maximal element.
Every inductive poset has
enough maximal elementsa 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?
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 $p$.