# non-uniformly continuous function

We assert that the real function $x\mapsto \mathrm{sin}\frac{1}{x}$ is not uniformly continuous^{} on the open interval^{}
$(0,\mathrm{\hspace{0.17em}1})$.

For proving this, we make the antithesis that there exists a positive number $\delta $ such that

$$ |

Choose

$${x}_{1}=\frac{1}{\frac{\pi}{2}+2n\pi},{x}_{2}=\frac{1}{\frac{3\pi}{2}+2n\pi}$$ |

where the integer $n$ is so great that $$, $$. Then we have

$$ |

However,

$$f({x}_{1})-f({x}_{2})=\mathrm{\hspace{0.33em}1}-(-1)=\mathrm{\hspace{0.33em}2}.$$ |

This contradictory result shows that the antithesis is wrong.

Title | non-uniformly continuous function |
---|---|

Canonical name | NonuniformlyContinuousFunction |

Date of creation | 2013-03-22 19:00:07 |

Last modified on | 2013-03-22 19:00:07 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 9 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 26A15 |

Related topic | PointPreventingUniformConvergence |

Related topic | ReductioAdAbsurdum |