Nested interval theorem

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The Nested interval theorem: If An = [an, bn] is a sequence of closed intervals such that An+1 [itex]\subseteq[/itex] An for all n [itex]\in[/itex] N, then [itex]_{n \in n}\bigcap[/itex]A = ∅.
I think of the case where a1=a2=...=an and b1=b2=...=bn for all n, hence every set A(n+1) will be the "subset" of A(n) and the intersection is the original closed interval. So I think the theorem in my textbook have some problem. Any correction for this ?
 

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  • #2
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The Nested interval theorem: If An = [an, bn] is a sequence of closed intervals such that An+1 [itex]\subseteq[/itex] An for all n [itex]\in[/itex] N, then [itex]_{n \in n}\bigcap[/itex]A = ∅.
I think of the case where a1=a2=...=an and b1=b2=...=bn for all n, hence every set A(n+1) will be the "subset" of A(n) and the intersection is the original closed interval. So I think the theorem in my textbook have some problem. Any correction for this ?
It should be:
If [itex]A_n=[a_n,b_n][/itex] is a sequence of closed intervals such that [itex]A_{n+1}\subseteq A_n[/itex] for all [itex]n\in\mathbb{N}[/itex], then [itex]\bigcap_{n\in \mathbb{N}}A_n \neq \emptyset[/itex].
 
  • #3
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It should be:
If [itex]A_n=[a_n,b_n][/itex] is a sequence of closed intervals such that [itex]A_{n+1}\subseteq A_n[/itex] for all [itex]n\in\mathbb{N}[/itex], then [itex]\bigcap_{n\in \mathbb{N}}A_n \neq \emptyset[/itex].

...and not only that: it must be also that [itex]\,b_n-a_n\xrightarrow[n\to\infty]{} 0\,[/itex] , as [itex]\,A_n:=[n,\infty)\,[/itex] would contradict.

DonAntonio
 
  • #4
Bacle2
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Or, more generally, a collection of nested sequence of sets in a complete metric space

with diameter approaching 0 as n-->00 .
 

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