Intersection of a family of compact sets in Hausdorff topological space is compact

Question

Let $\{G_i\}_{i \in \Lambda}$ be a family of compact sets in a Hausdorff topological space. Show that the intersection $\bigcap_{i \in \Lambda} G_i$ is compact.

My attempt was:

Since $G_j$ for some $j \in \Lambda$ is compact, there exists an open, finite covering. And $\bigcap_{i \in \Lambda} G_i \subset G_j$. So I can take the same covering for the intersection as for $G_j.$

Is that correct?

No. You need to show that for any open cover of the intersection that there is a finite subcover. Hint, the intersection is closed. — copper.hat, Jan 06 '21 at 22:00
You need that $\Lambda$ is not empty. Incidentally, this correction is also a hint — Hagen von Eitzen, Jan 06 '21 at 22:26
I've corrected my proof. May I ask you to have a look on it? — , Jan 08 '21 at 10:06

score 1 · Answer 1 · answered Jan 06 '21 at 22:06

It is definitely not correct. First, you seem to have misunderstood compactness. It does not say that a compact set has a finite open cover: in any space $X$ the family $\{X\}$ of open sets is a finite open cover of every subset of $X$, compact or not. It says that every open cover of a compact set has a finite subcover.

Secondly, you have not used the hypothesis that the space is Hausdorff, which is essential: the result is not true in general for non-Hausdorff spaces. Let $X$ be the line with two origins as described here; then the sets $\{p\}\cup(0,1]$ and $\{q\}\cup(0,1]$ are compact subsets of $X$ whose intersection $(0,1]$ is not compact.

HINT: Show (if you’ve not already done so) that every compact subset of a Hausdorff space is closed. What do you know about the intersection of closed sets? What do you know about closed subsets of compact sets?

I've corrected my proof in another answer. Does that work? – Jan 08 '21 at 10:05 — , Jan 08 '21 at 10:05

score 1 · Accepted Answer · answered Jan 08 '21 at 10:03

I see my mistakes and realised that I misunderstood the definition of compactness. I would correct the proof as following:

Since a compact set in a Hausdorff topological space is closed, it follows that $G_i \forall i \in \Lambda$ is closed. Hence, $\bigcap_{i \in \Lambda} G_i$ is closed. Now define $F:=(\bigcap_{i \in \Lambda} G_i)^{c}$ which is the complement. Note that $F$ is open.

Let $\bigcup_{i \in I} U_i$ be an open covering of $\bigcap_{i \in \Lambda} G_i$. My goal is to show that this covering is finite or has a finite sub-covering.

Note that $\bigcup_{i \in I} U_i \cup F$ is the whole space. In particular it is a covering of $G_{i_0}$ for $i_0 \in \Lambda$. By compactness of $G_{i_0}$ there exists finite $I' \subset I$ s.t. $G_{i_0} \subset $ $\bigcup_{i \in I'} U_i \cup F$.

Note that:

$\bigcap_{i \in \Lambda} G_i \subset G_{i_0} \subset \bigcup_{i \in I'} U_i \cup F$. But since $\bigcap_{i \in \Lambda} G_i \cap F = \emptyset$ we have that:

$\bigcap_{i \in \Lambda} G_i \subset \bigcup_{i \in I'} U_i$. So we have found a finite open covering and hence $\bigcap_{i \in \Lambda} G_i$ is compact.

Is that correct?

Much better! Yes, this works. – Brian M. Scott Jan 08 '21 at 21:05 — Brian M. Scott, Jan 08 '21 at 21:05

Intersection of a family of compact sets in Hausdorff topological space is compact

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