Where in Zorn's lemma do we use that P is a poset?

Question

In wikipedia there is a proof sketch of Zorn's lemma:

A sketch of the proof of Zorn's lemma follows, assuming the axiom of choice. Suppose the lemma is false. Then there exists a partially ordered set, or poset, P such that every totally ordered subset has an upper bound, and every element has a bigger one. For every totally ordered subset T we may then define a bigger element b(T), because T has an upper bound, and that upper bound has a bigger element. To actually define the function b, we need to employ the axiom of choice.

Using the function b, we are going to define elements a0 < a1 < a2 < a3 < ... in P. This sequence is really long: the indices are not just the natural numbers, but all ordinals. In fact, the sequence is too long for the set P; there are too many ordinals (a proper class), more than there are elements in any set, and the set P will be exhausted before long and then we will run into the desired contradiction.

The ai are defined by transfinite recursion: we pick a0 in P arbitrary (this is possible, since P contains an upper bound for the empty set and is thus not empty) and for any other ordinal w we set aw = b({av: v < w}). Because the av are totally ordered, this is a well-founded definition.

This proof shows that actually a slightly stronger version of Zorn's lemma is true:

If P is a poset in which every well-ordered subset has an upper bound, and if x is any element of P, then P has a maximal element greater than or equal to x. That is, there is a maximal element which is comparable to x.

Where in this proof do we use the hypotheses, namely that P is a poset:

specifically antisymmtry, transitivity, and reflexivity?

Answer 1

The important property the relation needs to have is transitivity.

This is used in the "successor" step of the construction: Suppose that the chain we're looking at has a largest element $x$, but that $x$ is not itself a maximal element of the entire poset. Then we need to extend the chain upwards in order to make the transfinite induction continue. From the premises we know that it has an upper bound, but that upper bound could be $x$ itself so that doesn't help us. But since $x$ is not maximal in the entire $P$ we know at least that there is some $y$ that is larger than $x$. Transitivity is what guarantees that $y$ will also be larger than every other element of the chain, such that the chain remains a chain when we add $y$ to it.

Reflexivity and antisymmety are only minor players. Having them allows us to state the lemma more succinctly, but even if we didn't have them we could still muddle through, at the cost of a slightly less natural (but still "morally equivalent") definition of exactly what "maximal element" that the lemma guarantees us means.

Answer 2

You only need transitivity and that you need for the concept of an upper bound to even make sense.

So why does transitivity suffice? If $(P_0; R_0)$ is transitive, we can define an equivalence relation $$ x \sim y \iff x R_0 y \wedge y R_0 x. $$ Since $R_0$ is transitive, we now obtain a well-defined antisymmetric, reflexive and transitive relation on the equivalence classes $[x]_\sim := \{ y \in P \mid x \sim y \}$ by setting $$ [x]_\sim R_1 [y]_\sim \iff x R_0 y \vee x = y $$ Let $P_1 = \{[x]_\sim \mid x \in P_0 \}$. If $[x]_\sim$ is a maximal element of $(P_1, R_1)$, it's easy to see that there is no $y \in P_1$ such that $x R_0 y$ and $y \not R_0 x$. In this sense $x$ is a maximal element of $(P_0, R_0)$. Note, however, that $x$ may not be maximal in the usual sense as there may be many different $y$ such that $x R_0 y$ and $y R_0 x$, i.e. $x$ is only maximal up to the above equivalence.