Let $A$ be a set. Let $C:Su(A)\longrightarrow Su(A)$ be a function, where $Su(A)$ denotes the set of all subsets of $A$. Suppose that
1) $X\subseteq C(X)$
2) $X\subseteq Y\rightarrow C(X)\subseteq C(Y)$
3) $C^2(X)=C(X)$
for every $X,Y\subseteq A$.
Let $L_C$ be the set of all $C(X)$, $X$ varying in $SU(A)$. Then $L_C$ is a poset, with $\subseteq$ as partial order, $L_C$ is also a lattice, with the two operations:
$$\bigvee C(A_i):=C(\bigcup_iA_i)$$
$$\bigwedge C(A_i):=\bigcap_iC(A_i)$$
Suppose now that $C$ satisfies the further property:
4) $C(X)=\bigcup \{C(Y): Y\subseteq X \textrm{ and } Y \textrm{ is finite }\}$, for every $X\subseteq A$.
I want to prove that $C(X)$ is compact if and only if $X$ is finite. I recall the definiton of compactness in this context: an element $a$ in a lattice $L$ is compact iff, whenever $a\leq\bigvee A$ for some $A\subseteq L$ then $a\leq\bigvee B$ for some finite $B\subseteq A$. I am able to prove that $X$ finite implies $C(X)$ compact, but I can't prove the converse. My textbook says what follows: suppose $C(Y)$ is not equal to $C(X)$ for any finite $X$. From
$$C(Y)\subseteq\bigcup\{C(X):X\subseteq Y\textrm{ and } X \textrm{ is finite }\}$$
it is easy to see that $C(Y)$ cannot be contained in any finite union of the $C(X)$'s, hence $C(Y)$ is not compact.
I can't undersand the last sentence: $C(Y)$ cannot be contained in any finite union of the $C(X)$'s. Why?
Assume that $C(Y) \subseteq \bigcup_{i = 1}^n C(X_i)$ for some family of finite subsets $X_i \subseteq Y, i = \overline{1, n}$.
It follows that $C(Y) = \bigcup_{i = 1}^n C(X_i)$. (Why we have the inverse inclusion?)
Hence $C(Y) = C(C(Y)) = C(\bigcup_{i = 1}^n C(X_i)) = \bigvee_{i = 1}^n C(C(X_i)) = \bigvee_{i = 1}^n C(X_i) = C(X)$, where $X = \bigcup_{i = 1}^n X_i$ is a finite subset of $Y$. So we get a contradiction with ''$C(Y)$ is not equal to $C(X)$ for any finite $X$''.