How to Prove the Nested Interval Theorem for R^n?

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The discussion focuses on proving the nested interval theorem for R^n, which asserts that the intersection of a decreasing sequence of non-empty, bounded closed sets is non-empty. The user presents their approach by selecting points from each set to form a sequence and applying the Bolzano-Weierstrass theorem to ensure the existence of a convergent subsequence. They argue that the limit point of this subsequence must lie within each closed set due to the nested nature of the sets. The user seeks assistance in formalizing this argument, emphasizing the importance of the closed sets in maintaining the limit point within the intersection. The discussion highlights the critical role of closed sets and convergence in proving the theorem.
eckiller
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Hi,

I am trying to prove the nested interval theorem for R^n. It is stated as follows:

Let (K_i : i in N) be a decreasing sequence of bounded closed sets in R^n and each K_i is non empty. Then the intersection of all K_i for i in N is not empty.

This is what I have so far:

Since K_i is nonempty, there exists a p_i in K_i for i in N. Pick these points to form the sequence (p_i : i in N). Because (p_i : i in N) is bounded, it has at least one convergent subsequence by the Bolzano-Weierstrass theorem.

I want to claim that the limit point of this subsequence is in the intersection of all K_i. But I am having trouble formalizing this part of the argument. Any help?
 
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The key is these sets are nested. So p_i \in K_i \subset K_{i-1} \subset K_{i-2} \subset \ldots \subset K_1

So for j\ge i, p_j \in K_i. So the tail of the convergent subsequence is in K_i. So the limit in is K_i since it is closed. Since this is true for an arbitrary i it is true for all i. So the limit is in all of the K_i's so it is in the intersection.

Hope that helps,
Steven
 
Excellent, many thanks.
 

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