Proving that a dense set in a poset by Zorn's lemma have a maximal antichain as a subset

Question

So I am little confused here; Zorn's lemma says that if every chain has a supremum in a poset one defines, then the poset must have a maximal element. My question is, how does this lead to the proof that a dense set in a poset by Zorn's lemma have a maximal antichain as a subset of the dense set?

Answer 1

Suppose $D$ is a dense subset of a poset $P$. Define $Q$ to be the family of all antichains of $P$ made up of elements of $D$, ordered by inclusion. By Zorn's Lemma $Q$ has a maximal element, call it $A$. In particular this means that for every $y \in D$ either $y \in A$ or $A \cup \{ y \}$ is not an antichain, which implies that $y$ is compatible with some element of $A$; in either case $y$ is compatible with some element of $A$.

To show that $A$ is actually a maximal antichain in $P$ it suffices to show that every $x \in P$ is compatible with an element of $A$. Given $p \in P$a s $D$ is dense in $P$ there is a $y \in D$ with $y \leq x$. By the above there is an $a \in A$ such that $a$ and $y$ are compatible. It follows that $x$ and $a$ are compatible.