Exercise 2.5.9

Question

Let $\left(a_{n}ight)$ be a bounded sequence, and define the set
  $$
  S=\left\{x \in \mathbf{R}: x<a_{n} 	ext { for infinitely many terms } a_{n}ight\}
  $$
  Show that there exists a subsequence $\left(a_{n_{k}}ight)$ converging to $s=\sup S$. (This is a direct proof of the Bolzano-Weierstrass Theorem using the Axiom of Completeness.)

Ulisse Mini · Accepted Answer

For every $\epsilon > 0$ there exists an $x \in S$ with $x > s - \epsilon$ implying $|s-x| < \epsilon$. therefore we can get arbitrarily close to $s = \sup S$ so there is a subsequence converging to this value. To make this more rigorous, pick $x_n \in S$ such that $|x_n - s| < 1/n$ then pick $N > 1/\epsilon$ to get $|x_n - s| < \epsilon$ for all $n > N$.

Exercise 2.5.9

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Exercise 2.5.9

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