# proof of Lebesgue number lemma

By way of contradiction^{}, suppose that no Lebesgue number existed. Then there exists an open cover $\mathcal{U}$ of $X$ such that for all $\delta >0$ there exists an $x\in X$ such that no $U\in \mathcal{U}$ contains ${B}_{\delta}(x)$ (the open ball^{} of radius $\delta $ around $x$). Specifically, for each $n\in \mathbb{N}$, since $1/n>0$ we can choose an ${x}_{n}\in X$ such that no $U\in \mathcal{U}$ contains ${B}_{1/n}({x}_{n})$. Now, $X$ is compact^{} so there exists a subsequence $({x}_{{n}_{k}})$ of the sequence of points $({x}_{n})$ that converges^{} to
some $y\in X$. Also, $\mathcal{U}$ being an open cover of $X$ implies that there exists $\lambda >0$ and $U\in \mathcal{U}$ such that ${B}_{\lambda}(y)\subseteq U$. Since the sequence $({x}_{{n}_{k}})$ converges to $y$, for $k$ large enough it is true that $$ ($d$ is the metric on $X$) and $$. Thus after an application of the triangle inequality^{}, it follows that

$${B}_{1/{n}_{k}}({x}_{{n}_{k}})\subseteq {B}_{\lambda}(y)\subseteq U,$$ |

contradicting the assumption^{} that no $U\in \mathcal{U}$ contains ${B}_{1/n}({x}_{n})$. Hence a Lebesgue number for $\mathcal{U}$ does exist.

Title | proof of Lebesgue number lemma |
---|---|

Canonical name | ProofOfLebesgueNumberLemma |

Date of creation | 2013-03-22 13:09:20 |

Last modified on | 2013-03-22 13:09:20 |

Owner | scanez (1021) |

Last modified by | scanez (1021) |

Numerical id | 7 |

Author | scanez (1021) |

Entry type | Proof |

Classification | msc 54E45 |