Is the Nested Interval Theorem Flawed in My Textbook?

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The Nested Interval Theorem states that if a sequence of closed intervals An = [an, bn] is nested such that An+1 is a subset of An, then the intersection of all intervals is non-empty, provided that the lengths of the intervals approach zero as n approaches infinity. The discussion highlights a misunderstanding regarding the theorem, where the intervals are incorrectly assumed to always intersect at the original closed interval when the endpoints are constant. It emphasizes that for the theorem to hold, the lengths of the intervals must converge to zero. Additionally, it clarifies that a counterexample, such as intervals defined as A_n = [n, ∞), would violate the theorem's conditions. The conclusion reinforces the necessity of both the nested property and the condition on interval lengths for the theorem to be valid.
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The Nested interval theorem: If An = [an, bn] is a sequence of closed intervals such that An+1 \subseteq An for all n \in N, then _{n \in n}\bigcapA = ∅.
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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pyfgcr said:
The Nested interval theorem: If An = [an, bn] is a sequence of closed intervals such that An+1 \subseteq An for all n \in N, then _{n \in n}\bigcapA = ∅.
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 A_n=[a_n,b_n] is a sequence of closed intervals such that A_{n+1}\subseteq A_n for all n\in\mathbb{N}, then \bigcap_{n\in \mathbb{N}}A_n \neq \emptyset.
 
micromass said:
It should be:
If A_n=[a_n,b_n] is a sequence of closed intervals such that A_{n+1}\subseteq A_n for all n\in\mathbb{N}, then \bigcap_{n\in \mathbb{N}}A_n \neq \emptyset.


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

DonAntonio
 
Or, more generally, a collection of nested sequence of sets in a complete metric space

with diameter approaching 0 as n-->00 .
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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