# the limit of a uniformly convergent sequence of continuous functions is continuous

Theorem. The limit of a uniformly convergent sequence of continuous functions^{} is continuous.

*Proof.* Let ${f}_{n},f:X\to Y$, where $(X,\rho )$ and $(Y,d)$ are metric spaces. Suppose ${f}_{n}\to f$ uniformly and each ${f}_{n}$ is continuous. Then given any $\u03f5>0$, there exists $N$ such that $n>N$ implies $$ for all $x$. Pick an arbitrary $n$ larger than $N$. Since ${f}_{n}$ is continuous, given any point ${x}_{0}$, there exists $\delta >0$ such that $$ implies $$. Therefore, given any ${x}_{0}$ and $\u03f5>0$, there exists $\delta >0$ such that

$$ |

Therefore, $f$ is continuous.

The theorem also generalizes to when $X$ is an arbitrary topological space^{}. To generalize it to $X$ an arbitrary topological space, note that if $$ for all $x$, then
${x}_{0}\in {f}_{n}^{-1}({B}_{\u03f5/3}({f}_{n}({x}_{0})))\subseteq {f}^{-1}({B}_{\u03f5}(f({x}_{0})))$,
so ${f}^{-1}({B}_{\u03f5}(f({x}_{0})))$ is a neighbourhood of ${x}_{0}$. Here ${B}_{\u03f5}(y)$ denote the open ball of radius $\u03f5$, centered at $y$.

Title | the limit of a uniformly convergent sequence of continuous functions is continuous |
---|---|

Canonical name | TheLimitOfAUniformlyConvergentSequenceOfContinuousFunctionsIsContinuous |

Date of creation | 2013-03-22 15:21:58 |

Last modified on | 2013-03-22 15:21:58 |

Owner | neapol1s (9480) |

Last modified by | neapol1s (9480) |

Numerical id | 13 |

Author | neapol1s (9480) |

Entry type | Theorem |

Classification | msc 40A30 |

Related topic | LimitFunctionOfSequence |