Convergence of Compact Sets in Metric Spaces

[sammycaps]

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

Let (X,d) be a metric space.

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


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

 

[jgens]

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

 

[sammycaps]

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

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

 

[micromass]

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.

 

[sammycaps]

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


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

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

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

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

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

 

[sammycaps]

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

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

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

A_m. 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]

Yeah, I just had a momentarily lapse in brain function ...

Aah,.., welcome to the club :) .