Bolzano-Weierstrass Theorem/Proof 2

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Every bounded sequence of real numbers has a convergent subsequence.


Let $\left \langle {x_n} \right \rangle_{n \in \N}$ be a bounded sequence in $\R$.

By definition there are real numbers $c, C \in \R$ such that $c < x_n < C$.

Then at least one of the sets:

$\left\{{x_n : c < x_n < \dfrac{c + C} 2 }\right\}, \left\{{x_n : \dfrac{c + C} 2 < x_n < C }\right\}, \left\{{x_n : x_n = \dfrac{c + C} 2 }\right\}$

contains infinitely many elements.

If the set $\left\{{x_n : x_n = \dfrac{c + C} 2 }\right\}$ is infinite there's nothing to prove.

If this is not the case, choose the first element from the infinite set, say $x_{k_1}$.

Repeat this process for $\left \langle {x_n} \right \rangle_{n > k_1}$.

As a result we obtain subsequence $\left \langle {x_{k_n}} \right \rangle_{n \in \N}$.

By construction $\left \langle {x_{k_n}} \right \rangle_{n \in \N}$ is a Cauchy sequence and therefore converges.


Also see

Source of Name

This entry was named for Bernhard Bolzano and Karl Weierstrass.