Bolzano-Weierstrass Theorem/Proof 2
Theorem
Every bounded sequence of real numbers has a convergent subsequence.
Proof
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 $x_{k_1}$.
Sets themselves are not ordered, so if you want to pick the "first element", you would do best to establish that "the set" you are picking $x_{k_1}$ from is in fact a subsequence of $\sequence {x_n}$. As it is, the above statement is too loose to be useful.
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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.
$\blacksquare$
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Also see
Source of Name
This entry was named for Bernhard Bolzano and Karl Weierstrass.