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Let $(X,\mathcal T)$ be a (not necessarily Hausdorff) topological space. It is well-known that $X$ is compact (in the sense of every open cover has a finite subcover) if and only if for every family of closed subsets $(C_i)_{i\in I}$ of $X$ satisfying the finite intersection property, $\bigcap _{i\in I}C_i$ is nonempty. Now does the following assertion hold:

If $(X,\mathcal T)$ is compact and $(C_i)_{i\in I}$ is a family of $\textbf{compact}$ subsets satisfying the finite intersection property, then $\bigcap_{i\in I}C_i$ is compact.

Since compact subsets in non-Hausdorff spaces need not be closed, I am not sure if this holds. Is there a counterexample?

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2 Answers 2

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This is not true in general. For instance, let $X=[0,1]$ with the topology that a set is open iff it is downward closed (i.e., $x\in U$ and $y\leq x$ implies $y\in U$). Then $X$ is compact since any open set containing $1$ is the whole space. Now for $i\in (1/2,1]$, let $C_i=[0,1/2)\cup(1/2,i]$. Each $C_i$ is similarly compact, and they have the finite intersection property. But their intersection $[0,1/2)$ is not compact, since it is covered by the open sets $[0,r)$ for $r<1/2$.

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Take the set $X=\mathbb{N}\cup\{a_1,a_2\}$ where $a_1,a_2$ are some points outside of $\mathbb{N}$. We define the following topology: a set is open if it is $\mathbb{N}\cup\{a_1\},\mathbb{N}\cup\{a_2\}$, $\mathbb{N}\cup\{a_1,a_2\}$ or any subset of $\mathbb{N}$. It is easy to see that this is a compact topological space and $\mathbb{N}\cup\{a_1\},\mathbb{N}\cup\{a_2\}$ are compact subsets with non empty intersection. However, their intersection is $\mathbb{N}$ which is not compact.

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