# Sequence of Compact Spaces

I was just googling around and I came across this problem.

Let (X,d) be a metric space.

Let (An)n $\in$ N be a sequence of closed subsets of X with the property An $\supseteq$ An+1 for all n $\in$ N. Suppose it exists an m $\in$ N such that Am is compact. Prove that $\bigcap$n$\in N$An is not empty.

I'm wondering if there is a typo here. Take some metric space. The we can set Am = ∅, and this is compact and closed, so it satisfies the conditions but the intersection is empty. What am I missing, thanks.

jgens
Gold Member
The result is true with the added restriction that An ≠ ∅ for each n in N.

The result is true with the added restriction that An ≠ ∅ for each n in N.

If this were the case, then why would we need compactness?

If this were the case, then why would we need compactness?

Just because each $A_n$ is nonempty, doesn't mean that the intersection is.

For example, take $A_n=[n,+\infty[$, then the intersection $\bigcap A_n = \emptyset$. We need a compactness hypothesis somewhere.

Just because each $A_n$ is nonempty, doesn't mean that the intersection is.

For example, take $A_n=[n,+\infty[$, then the intersection $\bigcap A_n = \emptyset$. We need a compactness hypothesis somewhere.

Oh I see, that was stupid. Thanks!

Bacle2
I was just googling around and I came across this problem.

Let (X,d) be a metric space.

Let (An)n $\in$ N be a sequence of closed subsets of X with the property An $\supseteq$ An+1 for all n $\in$ N. Suppose it exists an m $\in$ N such that Am is compact. Prove that $\bigcap$n$\in N$An is not empty.

I'm wondering if there is a typo here. Take some metric space. The we can set Am = ∅, and this is compact and closed, so it satisfies the conditions but the intersection is empty. What am I missing, thanks.

Notice that since Am is assumed compact, X is metric ( so Hausdorff ), and

Am $\supseteq$ Am+1 , all of which are closed, then

the Am+i ;i=1,2,... , can be seen as compact subspaces of

Am. This is a standard theorem in Analysis/Topology.

Notice that since Am is assumed compact, X is metric ( so Hausdorff ), and

Am $\supseteq$ Am+1 , all of which are closed, then

the Am+i ;i=1,2,... , can be seen as compact subspaces of

Am. This is a standard theorem in Analysis/Topology.

Yeah, I just had a momentarily lapse in brain function where I didn't realize closed and nested doesn't imply a non-empty intersection.

Bacle2