According to Brris & Sankappanavar's "A course in universal algebra," the set $L_C$ of closed subsets of a set $A$ forms an algebraic lattice under $\subseteq$, given that the closure oprator is an algebraic closure operator on $A$.
Here, a subset $X$ of $A$ is said to be closed if $C(X) = X$, and the closure operator $C$ on $A$ is said to be algebraic if $C$ satisfies C4, on top of C1 - C3, below:(For any $X, Y \subseteq A$)
C1: $X \subseteq C(X)$
C2: $C^2(X) = C(X)$
C3: $X \subseteq Y \Rightarrow C(X) \subseteq C(Y)$
C4: $C(X) = \bigcup \{C(Y): Y$ is a finite subset of $X \}$A lattice $\langle L, \le \rangle$ is called algebraic if it is complete and compactly generated; A lattice $\langle L, \le \rangle$ is said to be compactly generated if each of its elements is the supremum of a set of compact elements of $\langle L, \le \rangle$; and an element $a$ of a lattice $\langle L, \le \rangle$ is called a compact element of $\langle L, \le \rangle$ if for each subset $\Delta \subseteq L$ s.t. $a \le sup \Delta$, there is a finite subset $\Gamma \subseteq \Delta$ s.t. $a \le sup \Gamma$.
B&S' proof that $\langle L_C, \subseteq \rangle$ forms a compactly generated lattice seems to assume the following [1] without explanation (probably because it's obvious for most people):
[1] If $C$ is an algebraic operator on a set $A$ and each of $X$ and $Y$ is a finit subset of $A$, then $C(X) \cup C(Y) = C(X \cup Y)$.
But, I cannot so far show [1]. I will appreciate any help.
Here is how and where I think [1] is assumed:
Let $X$ be an arbitrary subset of $A$, $\Delta = \{ C(Y): Y$ is a finite subset of $ X \}$. (And $\Delta ' = \{Y \subseteq A: Y$ is a finite subset of$ X \}$.) Then, $C(X) = \bigcup \Delta$ by C4. If I can assume [1], then $\bigcup \Delta = C( \bigcup \Delta ')$, so, $C(X) = C(\bigcup \Delta ')$. (Or, if I can assume [3] below, $C(X) = C(\bigcup \Delta ')$ immeiately.) Meanwhile, $C( \bigcup \Delta ') = sup \Delta$ (in $L_C$) as a property of the complete lattice $L_C$ due to C1-C3. Thus $C(X) = sup \Delta$. Since each $C(Y) \in \Delta$ is a compact element of $L_C$ (which is explained in detail in B&S), $C(X)$ is the supremum of a set of compact elements of $L_C$.
I have two thoughts.
- If [2] below holds for $X, Y$ of [1], then [1] follows from [2] and C4.
[2] $\{ Z \subseteq X \cup Y: Z$ is finite$ \} = \{ Z \subseteq X: Z$ is finite$ \} \cup \{ Z \subseteq Y: Z$ is finite$ \}$.
But, [2] seems false.
- If [3] below holds in general, then the need of [1] is obviated.
[3] $X = \bigcup \{ Y \subseteq X: Y$ is finite$ \}$ for any subset $X$ of any set $A$.
But, assuming [3] somehow seems epistemologically illicit or something, when $X$ is not finite...