# proof of Ascoli-Arzelà theorem

Given $\u03f5>0$ we aim at finding a $4\u03f5$-net in $F$ i.e. a finite set^{} of points ${F}_{\u03f5}$ such that

$$\bigcup _{f\in {F}_{\u03f5}}{B}_{4\u03f5}(f)\supset F$$ |

(see the definition of totally bounded^{}).
Let $\delta >0$ be given with respect to $\u03f5$ in the definition
of equi-continuity (see uniformly equicontinuous) of $F$.
Let ${X}_{\delta}$ be a $\delta $-lattice^{} in
$X$ and ${Y}_{\u03f5}$ be a $\u03f5$-lattice in $Y$.
Let now ${Y}_{\u03f5}^{{X}_{\delta}}$ be the set of functions from
${X}_{\delta}$
to ${Y}_{\u03f5}$ and define
${G}_{\u03f5}\subset {Y}_{\u03f5}^{{X}_{\delta}}$ by

$$ |

Since ${Y}_{\u03f5}^{{X}_{\delta}}$ is a finite set, ${G}_{\u03f5}$ is finite too: say ${G}_{\u03f5}=\{{g}_{1},\mathrm{\dots},{g}_{N}\}$. Then define ${F}_{\u03f5}\subset F$, ${F}_{\u03f5}=\{{f}_{1},\mathrm{\dots},{f}_{N}\}$ where ${f}_{k}:X\to Y$ is a function in $F$ such that $$ for all $x\in {X}_{\delta}$ (the existence of such a function is guaranteed by the definition of ${G}_{\u03f5}$).

We now will prove that ${F}_{\u03f5}$ is a $4\u03f5$-lattice in $F$. Given $f\in F$ choose $g\in Y_{\u03f5}{}^{{X}_{\delta}}$ such that for all $x\in {X}_{\delta}$ it holds $$ (this is possible as for all $x\in {X}_{\delta}$ there exists $y\in {Y}_{\u03f5}$ with $$). We conclude that $g\in {G}_{\u03f5}$ and hence $g={g}_{k}$ for some $k\in \{1,\mathrm{\dots},N\}$. Notice also that for all $x\in {X}_{\delta}$ we have $$.

Given any $x\in X$ we know that there exists ${x}_{\delta}\in {X}_{\delta}$ such that $$. So, by equicontinuity of $F$,

$$ |

Title | proof of Ascoli-Arzelà theorem |
---|---|

Canonical name | ProofOfAscoliArzelaTheorem |

Date of creation | 2013-03-22 13:16:19 |

Last modified on | 2013-03-22 13:16:19 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 12 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 46E15 |