A criterion for compactness in complete lattices

Question

Let $(X,\le)$ be a complete lattice, $a\in X$ and for each chain $C\subseteq X$, $$a\le\sup C\to (\exists c\in C)(a\le c)$$

Is $a$ compact?

Answer 1

Yes, $a$ is compact, here is the proof.

For a given $A \subseteq X \colon a \leqslant \sup A$ we need to show that there exists finite $K \subseteq A \colon a \leqslant \sup K$. We will use induction on the cardinality of $A$. For finite $A$ take $K = A$. Now assume that $A$ is infinite, $|A| = \alpha$ and the proposition holds for all $B \subseteq X \colon |B| < \alpha$. Write $A$ as $\{a_{\gamma}\ \colon \gamma < \alpha\}$ and let $A_{\beta} = \{a_{\gamma}\ \colon \gamma < \beta\}$ for $\beta < \alpha$. Then $|A_{\beta}| < \alpha$ and $\{\sup A_{\beta}\ \colon \beta < \alpha\}$ is a chain and $a \leqslant \sup A = \sup\{\sup A_{\beta}\ \colon \beta < \alpha\}$, hence (by the condition about chains) $a \leqslant \sup A_{\beta}$ for some $\beta < \alpha$ and by the induction hypothesis we have $a \leqslant K$ for some finite $K \subseteq A_{\beta} \subset A$.