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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.

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

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

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