Sequence of Compact Spaces

  • Thread starter sammycaps
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  • #1
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I was just googling around and I came across this problem.

Let (X,d) be a metric space.

Let (An)n [itex]\in[/itex] N be a sequence of closed subsets of X with the property An [itex]\supseteq[/itex] An+1 for all n [itex]\in[/itex] N. Suppose it exists an m [itex]\in[/itex] N such that Am is compact. Prove that [itex]\bigcap[/itex]n[itex]\in N[/itex]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.
 

Answers and Replies

  • #2
jgens
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The result is true with the added restriction that An ≠ ∅ for each n in N.
 
  • #3
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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?
 
  • #4
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If this were the case, then why would we need compactness?

Just because each [itex]A_n[/itex] is nonempty, doesn't mean that the intersection is.

For example, take [itex]A_n=[n,+\infty[[/itex], then the intersection [itex]\bigcap A_n = \emptyset
[/itex]. We need a compactness hypothesis somewhere.
 
  • #5
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Just because each [itex]A_n[/itex] is nonempty, doesn't mean that the intersection is.

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

Oh I see, that was stupid. Thanks!
 
  • #6
Bacle2
Science Advisor
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I was just googling around and I came across this problem.

Let (X,d) be a metric space.

Let (An)n [itex]\in[/itex] N be a sequence of closed subsets of X with the property An [itex]\supseteq[/itex] An+1 for all n [itex]\in[/itex] N. Suppose it exists an m [itex]\in[/itex] N such that Am is compact. Prove that [itex]\bigcap[/itex]n[itex]\in N[/itex]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 [itex]\supseteq[/itex] 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.
 
  • #7
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Notice that since Am is assumed compact, X is metric ( so Hausdorff ), and

Am [itex]\supseteq[/itex] 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.
 
  • #8
Bacle2
Science Advisor
1,089
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Yeah, I just had a momentarily lapse in brain function .....

Aah,.., welcome to the club :) .
 

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