My Claim:
Let (D,$\le$) be a poset and f a function from the set of linearly ordered subsets of D to D. Assume that for every linearly ordered subset U of D we have $\forall$x $\in$ U, x $\lt$ f(U). Then there is no maximal element in D.
My proof:
Suppose for a contradiction x is a maximal element in D. Clearly, {x} $\subset$ U is linearly ordered. Then x $\lt$ f({x}) $\in$ U. This contradicts with the assumption x being a maximal element in D.
On the other Zorn's Lemma implies that D has a maximal element because every linearly ordered subset U of D is upper bounded by f(U) $\in$ D.
Clearly, there is something wrong here. There are three possibilities that might explain these contradicting results.
- My proof of the claim is incorrect.
- The use of Zorn's Lemma is incorrect.
- There exist no such function f and thus there is no contradiction.
What do you think about this?
Working in $\sf ZFC$, every partial order has a maximal chain. Your requirement about $f$ is that $f$ is a strict upper bound, but a maximal chain cannot have a strict upper bound, since adding a strict upper bound to a chain will result in a strictly larger chain.
So assuming $\sf ZFC$, there can be no such $f$. But the same can be said in $\sf ZF$, actually. What $f$ is doing is to choose an upper for every chain, and it is easy to see that it is exactly the choice function needed for the proof of Zorn's lemma, so by assuming it exists we circumvent to appeal to the Axiom of Choice. Consequently, it means that by applying $f$ recursively, we can create a maximal well-ordered chain, which again by maximality cannot have a strict upper bound.
So in fact such $f$ cannot exist, even without assuming the Axiom of Choice. So in either case, your proof is completely vacuous.