Example of an infinite compact measurable space

Question

Let $X$ be a nonempty set with a $\sigma$-algebra $\mathcal{A}$. The notion of $\sigma$-algebra strictly lies between Boolean algebras and complete Boolean algebras. Clearly, $\mathcal{A}$ is a distributive lattice (moreover a $\sigma$-frame).

A measurable space $(X,\mathcal{A})$ is said to be compact if $\mathcal{A}$ is a compact lattice.

Any set $X$ with a finite $\sigma$-algebra $\mathcal{A}$ is a compact measurable space. I want the following example.

Does there exist an infinite compact measurable space?

By infinite I mean that the $\sigma$-algebra $\mathcal{A}$ has infinite cardinality.

Answer 1

Let $(X,\mathcal{A})$ be a measurable space. Let us prove:

If $\mathcal{A}$ has infinite cardinality then $\mathcal{A}$ is not a compact lattice (a complete lattice where all elements are compact).

Proof : Suppose $\mathcal{A}$ has infinite cardinality and it is a compact lattice. Let us derive a contradiction.

For each $x \in X$ let us define $f(x) = \bigcap \{A \in \mathcal{A} : x\in A\}$. Since $\mathcal{A}$ is a complete lattice, we have that for all $x \in X$, $f(x) \in \mathcal{A}$. (Intuitively $f(x)$ is the smallest element of $\mathcal{A}$ to which $x$ belong)

It is easy to see that if $x,y \in X $ and $x \neq y$ then $f(x) \cap f(y) = \emptyset$ or $f(x) =f(y)$.

Let $\mathcal{F} = \{ f(x) : x \in X\}$. If $\mathcal{F}$ has finite cardinality, then $\mathcal{A}$ has finite cardinality. In fact, all set $B \in \mathcal{A}$ can be written as unions of elements of $\mathcal{F}$. Since we are assuming that $\mathcal{A}$ has infinite cardinality, we must have that $\mathcal{F}$ has infinite cardinality.

Let $\{F_1, F_2, ... \}$ be any countable infinite subset of $\mathcal{F}$. Let $F = \bigcup_i F_i$. Clearly $F \in \mathcal{A}$, $F \subseteq \bigcup_i F_i$. Since we are assuming that all elements of $\mathcal{A}$ are compact, there must exist a finite sub-collection $\{F_{i_j}\}_j$ of $\{F_i\}_i$ such that $F = \bigcup_j F_{i_j}$. Contradiction, because $F = \bigcup_i F_i$ and all the members of $\{F_i\}_i$ are disjoint.