# Intersection of family of closed compact sets is compact

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.

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

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.)

• Nov 27 at 4:56

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 $$C_{j_0}$$.