Q:

The set of compact open subsets of a compact set is a $\sigma$-algebra?

Let X be a compact set such that its compact open subsets form a basis for the topology. I ask if they form a $\sigma$-algebra. Let's denote this set by $\tau_c(X)$.
The first property is easy:
1)$U\in\tau_c(X)\Longrightarrow X\setminus U\in\tau_c(X)$
2) $\{U_i\}_{i\in\mathbb{N}}\subset\tau_c(X)\Longrightarrow\bigcup_{i\in\mathbb{N}}{U_i}\in\tau_c(X)$.
For the second property it is clear that the union is open. Therefore, my question reduces to ask if this set is compact.
If the answer is yes, I think it is key the fact that $X$ is compact.

A:

A counterexample is $X=\{1,1/2,1/3,1/4,\dots\}\cup\{0\}$ as a subspace of $\mathbb{R}$: every singleton except $\{0\}$ is open, so $\{0\}$ is a countable intersection of compact open sets but is not open.  In fact, virtually any example is a counterexample: it is possible to show that the compact open sets are a $\sigma$-algebra iff $X$ is finite (or iff the $T_0$ quotient of $X$ is finite, if you don't require $X$ to be Hausdorff).

