3
$\begingroup$

This question is from abbott's Understanding Analysis:

Start with the Bolzano--Weierstrass Theorem and use it to construct a proof of the Nested Interval Property.

My try: Let $I_n=[a_n,b_n]$ be closed intervals such that $I_1 \supseteq I_2 \supseteq \dots$. So here $a_n$ is a (nondecreasing) bounded sequence, so it contains a convergent subsequence by hypothesis, so does $b_n$.

But here I am stuck. I have temptation to use Monotone Convergence Theorem, but that would imply that I am assuming Axiom of Completeness, or worse, Nested Interval Property itself!

  • 0
    The Axiom of spremum must be taken as a starting point. – 2017-02-04
  • 0
    The Bolzano-Weierstrass recursion build the intervals that you want. Construct with every choice of the Bolzano-Weierstrass the intervals. – 2017-02-04

1 Answers 1

3

As you noted, there is a subsequence $n_k$ such that $a_{n_k}\to a$. As $I_n$ is closed and contains almost all $a_{n_k}$, we have $a\in I_n$. As this holds for all $n$, $a\in\bigcap I_n$