# alternate statement of Bolzano-Weierstrass theorem

###### Theorem.

Every bounded^{}, infinite set^{} of real numbers has a limit point^{}.

###### Proof.

Let $S\subset \mathbb{R}$ be bounded and infinite. Since $S$ is bounded there exist $a,b\in \mathbb{R}$, with $$, such that $S\subset [a,b]$. Let $b-a=l$ and denote the midpoint^{} of the interval^{} $[a,b]$ by $m$. Note that at least one of $[a,m],[m,b]$ must contain infinitely many points of $S$; select an interval satisfying this condition, denoting its left endpoint by ${a}_{1}$ and its right endpoint by ${b}_{1}$. Continuing this process inductively, for each $n\in \mathbb{N}$, we have an interval $[{a}_{n},{b}_{n}]$ satisfying

$$[{a}_{n},{b}_{n}]\subset [{a}_{n-1},{b}_{n-1}]\subset \mathrm{\cdots}\subset [{a}_{1},{b}_{1}]\subset [a,b]\text{,}$$ | (1) |

where, for each $i\in \mathbb{N}$ such that $1\le i\le n$, the interval $[{a}_{i},{b}_{i}]$ contains infinitely many points of $S$ and is of length $l/{2}^{i}$. Next we note that the set $A=\{{a}_{1},{a}_{2}\mathrm{\dots},{a}_{n}\}$ is contained in $[a,b]$, hence is bounded, and as such, has a supremum which we denote by $x$. Now, given $\u03f5>0$, there exists $N\in \mathbb{N}$ such that $$. Furthermore, for every $m\ge N$, we have $$. In particular, if we select $m\ge N$ such that $$, then we have

$$ | (2) |

Since $[{a}_{m},{b}_{m}]\subset (x-\u03f5,x+\u03f5)$ contains infinitely many points of $S$, we may conclude that $x$ is a limit point of $S$. ∎

Title | alternate statement of Bolzano-Weierstrass theorem |
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Canonical name | AlternateStatementOfBolzanoWeierstrassTheorem |

Date of creation | 2013-03-22 16:40:13 |

Last modified on | 2013-03-22 16:40:13 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 7 |

Author | mathcam (2727) |

Entry type | Theorem |

Classification | msc 26A06 |

Related topic | BolzanoWeierstrassTheorem |

Related topic | LimitPoint |

Related topic | Bounded |

Related topic | Infinite |