every set of non compact elements has a maximum element

Question

Let $P$ be a poset. An element $x$ of $P$ is compact if for every directed set $D\subseteq D$, $x\leq \sup P$ implies $x\in {\downarrow\!\! D}$.

Is it true if the set of compact elements of $P$, denoted by $K(P)$, then $P-K(P)$ has a maximum element?

my attempt: Since the supremum of a chain of noncompact elements is noncompact, by Zorn’s Lemma, $P-K(P)$ has a maximum element.

Answer 1

Let $P=\Bbb R^2$ with the product partial order: $\langle x_0,y_0\rangle\preceq\langle x_1,y_1\rangle$ iff $x_0\le x_1$ and $y_0\le y_1$. Let $\langle x_0,y_0\rangle\in P$ be arbitrary, and let $D=\{\langle x,y\rangle\in P:x<x_0\text{ and }y<y_0\}$; then $D$ is directed, and $\langle x_0,y_0\rangle=\sup D$, but $\langle x_0,y_0\rangle\notin{\downarrow\!\!D}$. Thus, $K(P)=\varnothing$, so $P\setminus K(P)=P$, which has no maximum element.