proof of Ascoli-Arzelà theorem

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

\\bigcup_{f\\in F_{\\epsilon}}B_{4\\epsilon}(f)\\supset F

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

G_{\\epsilon}=\\{g\\in Y_{\\epsilon}^{X_{\\delta}}\\colon\\exists f\\in F\\ \\forall x%\\in X_{\\delta}\\quad d(f(x),g(x))<\\epsilon\\}.

Since $Y_{\\epsilon}^{X_{\\delta}}$ is a finite set, $G_{\\epsilon}$ is finite too: say $G_{\\epsilon}=\\{g_{1},\\ldots,g_{N}\\}$ .Then define $F_{\\epsilon}\\subset F$ , $F_{\\epsilon}=\\{f_{1},\\ldots,f_{N}\\}$ where $f_{k}\\colon X\\to Y$ is a function in $F$ such that $d(f_{k}(x),g_{k}(x))<\\epsilon$ for all $x\\in X_{\\delta}$ (the existence ofsuch a function is guaranteed by the definition of $G_{\\epsilon}$ ).

We now will prove that $F_{\\epsilon}$ is a $4\\epsilon$ -lattice in $F$ .Given $f\\in F$ choose $g\\in{Y_{\\epsilon}}^{X_{\\delta}}$ such that for all $x\\in X_{\\delta}$ it holds $d(f(x),g(x))<\\epsilon$ (this is possible as for all $x\\in X_{\\delta}$ there exists $y\\in Y_{\\epsilon}$ with $d(f(x),y)<\\epsilon$ ).We conclude that $g\\in G_{\\epsilon}$ and hence $g=g_{k}$ for some $k\\in\\{1,\\ldots,N\\}$ . Notice also that for all $x\\in X_{\\delta}$ wehave $d(f(x),f_{k}(x))\\leq d(f(x),g_{k}(x))+d(g_{k}(x),f_{k}(x))<2\\epsilon$ .

Given any $x\\in X$ we know that there exists $x_{\\delta}\\in X_{\\delta}$ such that $d(x,x_{\\delta})<\\delta$ .So, by equicontinuity of $F$ ,

d(f(x),f_{k}(x))\\leq d(f(x),f(x_{\\delta}))+d(f_{k}(x),f_{k}(x_{\\delta}))+d(f(x%_{\\delta}),f_{k}(x_{\\delta}))<4\\epsilon.