Suppose that $(L,\leq)$ is a lattice satisfying the ascending chain condition on elements of $L$. Let $S \subseteq L$ be a subset. An element $c \in L$ is called compact if $c \leq \bigvee S$ implies $c \leq \bigvee S'$ for some finite subset $S' \subseteq S$. How can one prove that the ACC implies that every element of $L$ is compact?
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1You've misquoted the definition of "compact". – Andreas Blass Feb 03 '20 at 17:20
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My bad, I corrected the error. – Feb 03 '20 at 17:29
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Do you know a proof in this case? – Feb 04 '20 at 19:29
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Let us start by stating (without proof) a lemma (a formal proof of which would require de Axiom of Choice, or some equivalent, but maybe you already know this).
Lemma. A poset satisfies (ACC) iff every nonempty subset of this poset has a maximal element.
Now we will prove that if $L$ satisfies (ACC) then for every non-empty subset $S$ of $L$, there exist a finite subset $S'$ of $S$ such that $\bigvee S = \bigvee S'$.
So let $\varnothing \neq S \subseteq L$, where $L$ is a lattice satisfying (ACC), and let
$$T=\{ \bigvee F : F \Subset S \}.$$
(Here, $A\Subset B$ means that $A \subseteq B$ and $A$ is finite.)
Clearly $\varnothing \neq T \subseteq L$, and so by the Lemma, there exist $S' \Subset S$ such that $m = \bigvee S'$ is a maximal element of $T$.
Let $s \in S$. Then $S' \subseteq S'\cup\{s\}$, whence $m = \bigvee S' \leq \bigvee(S'\cup\{s\})$, and therefore $m = \bigvee(S' \cup \{s\})$ (since $\bigvee(S'\cup\{s\})\in T$).
Thus $s \leq m$, yielding that $m$ is an upper bound of $S$.
Now let $r$ be any upper bound of $S$.
It follows that $r$ is an upper bound of $S'$, whence $m \leq r$.
Hence $m = \bigvee S = \bigvee S'$.
amrsa
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Notice that a proof of the lemma stated above is more or less what you asked in this question, where you have an answer. – amrsa Feb 04 '20 at 21:50