# 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.

• what do you mean by compact open sets? Compact sets are always closed – hunch Oct 22 '16 at 9:11
• Compact sets that are also open. – iago Oct 22 '16 at 9:12
• Are you assuming $X$ is a subset of $\mathbb{R}^n$ or at least Hausdorff? Otherwise statement (1) may not be true. – Eric Wofsey Oct 22 '16 at 9:16
• $X$ is Hausdorff. – iago Oct 22 '16 at 9:21

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).
• Which are the open sets containing $0$? – iago Oct 22 '16 at 9:34
• I require $X$ to be Hausdorff – iago Oct 22 '16 at 9:35
• A set containing $0$ is open iff it's cofinite. This is just the subspace topology on $X$ as a subset of $\mathbb{R}$. – Eric Wofsey Oct 22 '16 at 9:35