Bolzano-Weierstrass Theorem/Proof 2

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Every bounded sequence of real numbers has a convergent subsequence.


Let $\sequence {x_n}_{n \mathop \in \N}$ 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:

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

contains infinitely many elements.

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

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

Repeat this process for $\sequence {x_n}_{n \mathop > k_1}$.

As a result we obtain subsequence $\sequence {x_{k_n} }_{n \mathop \in \N}$.

By construction $\sequence {x_{k_n} }_{n \mathop \in \N}$ is a Cauchy sequence and therefore converges.


Also see

Source of Name

This entry was named for Bernhard Bolzano and Karl Weierstrass.