Let $\{C_j:j \in J\}$ be a family of closed compact subsets of a topological space $(X,\tau)$. Prove that $\bigcap_{j \in J}C_j$ is compact.

I realized this is not a metric space, so compactness in general topology does not imply closed or boundedness. But if we use the subcover definition of compactness, it should always by possible find a finite open subcover right since each $C_j$ is compact and each has a finite subcover, the intersection of finite open subcover is still open and is still a subcover for the intersection of all the compact sets.

One concern I have: even though each $C_j$ has a finite subcover, if there are an uncountable number of $C_j$, then we are taking an infinite intersection of open subcovers which is not open.

  • 2
    $\begingroup$ 1 Any intersection of closed sets is closed. 2 A closed subset of a compact set is compact. This is all you need. $\endgroup$ Jan 30, 2021 at 8:19
  • $\begingroup$ You also need to assume that $J\ne\varnothing$. If $J=\varnothing$ then $\bigcap_{j\in J}C_j=X$ may not be compact. $\endgroup$
    – user14111
    Nov 27 at 2:04

2 Answers 2


Perhaps worth adding this example for why "closed" is a necessary assumption here.

Let $X=\mathbb Z\cup\{-\infty,\infty\}$ have the topology $\{U\subseteq X:U\cap\{-\infty,\infty\}\not=\emptyset\Rightarrow \mathbb Z\subseteq U\}$. Then $\mathbb Z\cup\{\infty\}$ and $\mathbb Z\cup\{-\infty\}$ are compact, not closed, and their intersection is the infinite discrete (and thus not compact) subspace $\mathbb Z$.

(Such a space is $T_0$, but not $T_1$. However, it can be modified to be $T_1$: use $\{U\subseteq X:U\cap\{-\infty,\infty\}\not=\emptyset\Rightarrow \mathbb Z\setminus U \text{ finite}\}$ instead.)

Since this was marked as the accepted answer, here's a writeup answering the question.

Let $0\in J$ and consider the compact subset $C_0\subseteq X$. Then for $j\in J$, note that $C_0\cap C_j$ is closed in the subspace topology for $C_0$. Then $\bigcap_{j\in J} C_j=\bigcap_{j\in J}(C_0\cap C_j)$ is the intersection of closed subsets of the space $C_0$, and is therefore closed. Since a closed subset of a compact space is compact, we have $\bigcap_{j\in J} C_j$ compact.

(Note that I do assume $J$ is non-empty by letting $0\in J$ - see user14111's comment.)


HINT: Fix $j_0\in J$, and let $\mathscr{C}=\{C_j\cap C_{j_0}:j\in J\}$. Then

$$\bigcap\{C_j:j\in J\}=\bigcap\mathscr{C}\,,$$

and $\bigcap\mathscr{C}$ is a closed subset of $C_{j_0}$.


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