Let $(L,\leq)$ be a complete lattice. We say an $a\in L$ is compact in $L$ iff for $\forall A\subset L$ such that $a\leq \bigvee A$, there exists a finite subset $A'\subset A$, sucth that $a\leq\bigvee A'$.
Show that the following 2 conditions are equivalent:
(i) Every element of $L$ is compact in $(L,\leq)$, i.e. $L=\text{Comp}(L)$ where $\text{Comp}(L)$ denotes the set of all compact elements of $L$
(ii) Every chain in $(L,\leq)$ has a greatest element.
I can show (i) implies (ii), But am totally stuck on the reverse implication. Any help and assistance will be greatly appreciated. My workings for (i) implies (ii) are as follows:
Let $S$ be a chain in $(L,\leq)$. Then as $(L,\leq)$ is complete, $\bigvee S\in L$, and so by (i) we have a finite $S'\subset S$, such that $\bigvee S\leq\bigvee S'$, thus $\bigvee S'=\bigvee S$ and as $S'$ is finte, $\bigvee S'\in S'\subset S$, thus $\bigvee S\in S$, i.e. every chain has a greatest element. I think this is fine but if you see any problems with it please let me know, and any help on (ii) implies (i) is really needed as I'm totally stuck there. Thanks in advance
Let $y\in L$ and $X\subset L$ such that $y\leq\bigvee X$, and let $P=\{\bigvee X'| X'\subset X\space\text{&} \space X'\space\text{is finite}\}$. Take a chain $S$ in $P$, then as $P\neq\emptyset$ (since $(L,\leq)$ is complete) and by (ii) we can then apply Zorn's Lemma, thus $P$ has a maximal element, say $b=\bigvee X'$ for some finte $X'\subset X$. It is then easy to see that $b=\bigvee X$; as if not then for some $x\in X$ we have $x\nleq b$, and so $\bigvee (X'\cup\{x\})>b$, contradicting the maximality of $b$. Thus $y\leq\bigvee X'$