- #1

- 587

- 247

- Homework Statement
- If ##X## is a topological space and ##S_i \subset X## is an arbitrary collection of closed subspaces, at least one of which is compact, then ##\bigcap_i S_i## is also closed and compact.

- Relevant Equations
- (o.o)_)~

Given that one of the ##S_i## (let's name it ##S_{compact}##), is compact. Assume there is an open cover ##\mathcal V## of ##S_{compact}##. By definition of a compact subspace, there is a subcover ##\mathcal U## with ##n<\infty## open sets. Notice that ##\forall x\in (\bigcap_i S_i)##, ##x\in S_i##, which implies that the intersection of all the ##S_i##'s is contained in anyone of the ##S_i##'s. Since the intersection of the ##S_i##'s is contained in ##S_{compact}##, the subcover for ##S_{compact}## is also a subcover of the intersection of the ##S_i##'s. Also, given that every ##S_i## is closed, their arbitrary intersection is also closed. It follows that ##\bigcap S_i## is closed and compact.

Last edited: