Bolzano-Weierstrass Theorem/Proof 2
Contents
Theorem
Every bounded sequence of real numbers has a convergent subsequence.
Proof
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.
$\blacksquare$
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Source of Name
This entry was named for Bernhard Bolzano and Karl Weierstrass.
