# Bolzano-Weierstrass Theorem

This article has been identified as a candidate for Featured Proof status.If you do not believe that this proof is worthy of being a Featured Proof, please state your reasons on the talk page.To discuss this page in more detail, feel free to use the talk page. |

## Theorem

Every bounded sequence of real numbers has a convergent subsequence.

### General Form

Every infinite bounded space in a real Euclidean space has at least one limit point.

## Proof 1

Let $\sequence {x_n}$ be a bounded sequence in $\R$.

By the Peak Point Lemma, $\sequence {x_n}$ has a monotone subsequence $\sequence {x_{n_r} }$.

Since $\sequence {x_n}$ is bounded, so is $\sequence {x_{n_r} }$.

Hence, by the Monotone Convergence Theorem (Real Analysis), the result follows.

$\blacksquare$

## Proof 2

Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $\R$.

By definition there are real numbers $a, b \in \R$ such that $ x_n \in \openint{a}{b}$ for all $n \in \N$.

We will construct a subsequence $\sequence {x_{n_i}}_{i \mathop \in \N}$ and two sequences of real numbers $\sequence {b_i}_{i \mathop \in \N}$, $\sequence {c_i}_{i \mathop \in \N}$ as follows:

Set $a_0 = a$, and $b_0 = b$.

For $i \in \N$, at least one of the sets:

- $\set {x_n: n > n_{i-1} , a_i < x_n < \dfrac {a_i + b_i} 2 }, \set {x_n : n > n_{i-1} , \dfrac {a_i + b_i} 2 < x_n < b_i }, \set {x_n : n > n_{i-1} , x_n = \dfrac {a_i + b_i } 2 }$

contains infinitely many elements.

If the set $\set {x_n : n > n_{i-1} , x_n = \dfrac {a_i + b_i} 2 }$ is infinite, the elements of this set form a subsequence that converges to $\dfrac {a_i + b_i} 2$, and the proof is done.

If instead the set $\set {x_n: n > n_{i-1} , a_i < x_n < \dfrac {a_i + b_i} 2 }$ is infinite, define $x_{n_i}$ as the element of that set with the smallest index.

In this case, set $a_{i+1} = a_i$, and $b_{i+1} = \dfrac { a_i + b_i } 2 $.

If instead the set $\set {x_n : n > n_{i-1} , \dfrac {a_i + b_i} 2 < x_n < b_i }$ is infinite, define $x_{n_i}$ as the element of that set with the smallest index.

In this case, set $a_{i+1} = \dfrac { a_i + b_i } 2 $, and $b_{i+1} = b_i$.

In both cases, we have $x_{n_i} \in \openint{ a_j }{ b_j }$ for all $j \in \N$ with $j < i$, so

- $\size { x_{n_j} - x_{n_i} } < b_j - a_j = \dfrac{ a + b } {2^j}$

Repeat this process so we obtain a subsequence $\sequence {x_{n_i} }_{i \mathop \in \N}$.

Given $\epsilon > 0$, we can choose $j \in \N$ so $\size { x_{n_{i}} - x_{n_{i'}} } < \dfrac{ a + b } {2^j} < \epsilon$ for all $i , i' \geq j$.

By Definition of Real Cauchy Sequence, the subsequence is a Cauchy sequence.

By Cauchy's Convergence Criterion on Real Numbers, the subsequence converges.

$\blacksquare$

## Also known as

Some sources refer to the **Bolzano-Weierstrass Theorem** as the **Weierstrass-Bolzano Theorem**.

It is also known as **Weierstrass's Theorem**, but that name is also applied to a completely different result.

## Also see

## Source of Name

This entry was named for Bernhard Bolzano and Karl Weierstrass.

## Historical Note

The Bolzano-Weierstrass Theorem is a crucial property of the real numbers discovered independently by both Bernhard Bolzano and Karl Weierstrass during their work on putting real analysis on a rigorous logical footing.

It was originally referred to as **Weierstrass's Theorem** until Bolzano's thesis on the subject was rediscovered.

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $5$