# boundedness theorem

Boundedness Theorem.
Let $a$ and $b$ be real numbers with $$, and let $f$ be a continuous^{}, real valued function on $\mathrm{[}a\mathrm{,}b\mathrm{]}$. Then $f$ is bounded above and below on $\mathrm{[}a\mathrm{,}b\mathrm{]}$.

Proof.
Suppose not. Then for all natural numbers^{} $n$ we can find some ${x}_{n}\in [a,b]$ such that $|f({x}_{n})|>n$. The sequence^{} $({x}_{n})$ is bounded^{}, so by the Bolzano-Weierstrass theorem^{} it has a convergent^{} sub sequence, say $({x}_{{n}_{i}})$. As $[a,b]$ is closed $({x}_{{n}_{i}})$ converges^{} to a value in $[a,b]$. By the continuity of $f$ we should have that $f({x}_{{n}_{i}})$ converges, but by construction it diverges. This contradiction^{} finishes the proof.

Title | boundedness theorem |
---|---|

Canonical name | BoundednessTheorem |

Date of creation | 2013-03-22 14:29:18 |

Last modified on | 2013-03-22 14:29:18 |

Owner | classicleft (5752) |

Last modified by | classicleft (5752) |

Numerical id | 6 |

Author | classicleft (5752) |

Entry type | Theorem |

Classification | msc 26A06 |