Compact elements in a lattice satisfying ACC I am trying to prove that in a complete lattice $L$  satisfying ACC (ascending chain condition), all elements are compact. 
To do so, I start taking $x\in L$ and $S\subseteq L$ such that $x\leq \bigvee S$. I have to prove that there exists a finite subset $T$ os $S$ satisfying that $x\leq \bigvee T$, but I don't know how and where to use my hypothesis...
 A: Actually, in a lattice satisfying (ACC), you can find a finite subset $T$ of $S$ such that $\bigvee T = \bigvee S$.
Assume this is false, ie for any finite $T \subseteq S$, $\bigvee T < \bigvee S$. The game is to build a suitable infinite chain that will contradict (ACC).
First pick any $s_1 \in S$. By hypothesis, $\bigvee \{ s_1 \} = s_1 < \bigvee S$. Hence you must have $s_2 \in S$ such that $s_2 \not \leq s_1$ (since $s_1$ is not a majorant). For such a $s_2$, $\bigvee \{s_1,s_2\} > s_1$.
Once again, because $\bigvee \{ s_1, s_2 \} < \bigvee S$, we can pick $S \ni s_3 \not \leq \bigvee \{ s_1, s_2 \}$. This means that $\bigvee \{ s_1, s_2, s_3 \} > \bigvee \{ s_1, s_2 \}$.
You can go on and build a sequence $(s_i)_{i \geq 1}$ of elements of $S$ such that if for $n \geq 1$, $t_n = \bigvee \{ s_i \mid i \leq n \}$, then $t_{n+1} > t_n$. But now $\{ t_n \mid n \geq 1 \}$ contradict (ACC).

You can even prove it directly, if you already know that in a latice $L$ satisfying (ACC), then any non empty subset must have a maximal element (the proof is very similar to the above one). Then you can apply it to the set $B = \{ \bigvee T \mid T \subseteq S, T \text{ finite } \}$ and find a maximal element $m_0 = \bigvee T_0$. For any $s \in S$, $T_0 \subseteq T_0 \cup \{ s \}$ so $\bigvee T_0 = m_0 \leq \bigvee(T_0 \cup \{ s \})$ but $T_0 \cup \{s\}$ is finite thus $\bigvee (T_0 \cup \{ s \}) \leq m_0$ hence $\bigvee(T_0 \cup \{s\}) = m_0$. Since $s \leq \bigvee(T_0 \cup \{s \}) = m_0$ it follows easily that $\bigvee T_0 = m_0 = \bigvee S$.
