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

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SUMMARY

The Nested Interval Theorem for R^n states that for a decreasing sequence of bounded closed sets (K_i) in R^n, where each K_i is non-empty, the intersection of all K_i is non-empty. The proof involves selecting points (p_i) from each K_i, which form a bounded sequence. By the Bolzano-Weierstrass theorem, this sequence has a convergent subsequence whose limit lies in the closed sets K_i, thus proving the theorem. The key aspect is the nested nature of the sets, ensuring that the limit point remains within all K_i.

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  • Understanding of the Nested Interval Theorem
  • Familiarity with R^n and closed sets
  • Knowledge of the Bolzano-Weierstrass theorem
  • Basic concepts of convergence in metric spaces
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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 [tex]p_i \in K_i \subset K_{i-1} \subset K_{i-2} \subset \ldots \subset K_1[/tex]

So for [tex]j\ge i[/tex], [tex]p_j \in K_i[/tex]. So the tail of the convergent subsequence is in [tex]K_i[/tex]. So the limit in is [tex]K_i[/tex] 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 [tex]K_i[/tex]'s so it is in the intersection.

Hope that helps,
Steven
 
Excellent, many thanks.
 

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