I am doing exercises from Universal Algebra by Clifford Bergman.
Question (Bergman 2.34.5) Let $a,b$ be compact elements in algebraic lattice $L$. (2) Must $a\wedge b$ be compact?
My thoughts I think I found a counterexample on web (see below) but it is rather involved and it seems to me that the point of the exercise is, that there are simpler counter-examples.
I would like to find some more natural counter-example to $(2)$.
Definitions Let $L$ be a complete lattice
(a) Element $a$ in $L$ is called compact if for every $X\subset L$:
$a\leq \bigvee X \rightarrow (\exists \text{ finite } Z\subset X) ~a\leq \bigvee Z $
(b) $L$ is called algebraic if every element is the join of compact elements.
"My" counter-example
Wikipedia lists an example of two compact sets $U,V$ in Hausdorf space $X$ whose intersection is not compact https://en.wikipedia.org/wiki/Compact_space#cite_note-17.
I came to the conclusion, that unlike for general closure operator from topology, for this particular topology the closer operator is indeed Agebraic. So the lattice of closed subsets of $X$ is an algebraic lattice and $U,V$ are the desired counterexample.
Best Answer
So essentially you have the following lattice:
where $a$ and $b$ are compact but $a \wedge b$ is not compact.
Notice that you can define a topological space from this lattice as the one whose open sets are, up to isomorphism, the elements of the lattice.
Make $X = \{ u_0, u_1, u_2, \ldots, a, b \}$ and the open sets are $$\varnothing, U_0, U_1, U_2, \ldots, U, V, W, X,$$ where $U_i = \{ u_0, \ldots, u_i \}$, $U = \bigcup_i U_i, V = U \cup\{a\}$ and $W = U \cup \{b\}$