proof of Ascoli-Arzelà theorem

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

$$\bigcup _{f\in F_{ϵ}}{B}_{4ϵ}(f)\supset F$$

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

Since $Y_{ϵ}^{X_{\delta }}$ is a finite set, $G_{ϵ}$ is finite too: say $G_{ϵ}=\{g_1,\mathrm{\dots },g_N\}$. Then define $F_{ϵ}\subset F$, $F_{ϵ}=\{f_1,\mathrm{\dots },f_N\}$ 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 $G_{ϵ}$).

We now will prove that $F_{ϵ}$ is a $4ϵ$-lattice in $F$. Given $f\in F$ choose $g\in Y_{ϵ}{}^{X_{\delta }}$ such that for all $x\in X_{\delta }$ it holds (this is possible as for all $x\in X_{\delta }$ there exists $y\in Y_{ϵ}$ with ). We conclude that $g\in G_{ϵ}$ 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