Let $\mathbb P = (P, ≤, 1)$ be a partial order.
We call an antichain $A$ of $P$ maximal in case every $p ∈ P$ is compatible to some $a ∈ A$.
We say that a set $D ⊆ P$ is predense in $P$ in case every $p ∈ P$ is compatible to some $d ∈ D$
(so - trivially - an antichain of $P$ is maximal if and only if it is predense.)
Can someone give me an example of a partial order $P$, and a predense subset $D$ of $P$, such that no subset of $D$ is a maximal antichain of $P$.
Let $P=\{1\}\sqcup(\omega\times\{0,1\})$, where $\omega\times\{0,1\}$ is partially ordered by saying $(m,i)\leq (n,0)$ iff $m\leq n$. The set $D=\omega\times\{0\}$ is predense, since for any $(m,i)$, $(m,1)$ is a common extension of $(m,i)$ and the element $(m,0)\in D$. But no subset of $D$ is a maximal antichain. Indeed, $D$ is a chain, so any subset which is an antichain has at most one element. But for any $(m,0)\in D$, $(n,1)$ is incompatible with $(m,0)$ for any $n>m$, so $\{(m,0)\}$ is not a maximal antichain.