I am currently thinking of function-valued random variables. In order to prove a result, I need to approximate by (function-valued) step functions. This naturally leads to the idea of chopping up the function space into finitely many small pieces.


Let $X \subset \mathbb R^d$ be compact and $V$ the set of $1$-Lipschitz functions $X \to \mathbb [0,1]$. Clearly $V$ is equicontinuous and bounded wrt the uniform norm. Since it is also closed it is compact by Arzela-Ascoli. Hence it is totally bounded, meaning:



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> For any $\epsilon >0$ we can express $V$ as the union of finitely many open sets, each with diameter less than $\epsilon$.
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I wonder is there an elementary and not-too-tedious way to prove the above without using the heavy machinery of Arzela-Ascoli? Perhaps we can construct the finitely-many sets directly as balls around some piecewise-defined functions?


The reason I ask is I would like to use the above in the context of optimisation without introducing new terminology (compactness, sequential compactness, equicontinuity et cetera).



Can I get away without using Arzela-Ascoli?
Of course you can, and this is how Arzela-Ascoli is often proved. You may fix a finite $\varepsilon/3$-net $D\subset X$ and partition $[0,1]$ onto disjoint subsets $A\_1,\ldots,A\_N$ of diameter less than $\varepsilon/3$. For any 1-Lipschitz function $f:X\rightarrow [0,1]$ we consider the function $[f]:D\rightarrow \{1,2,\ldots,N\}$ defined as $[f](t)=i$ iff $f(t)\in A\_i$. There are finitely many possible functions of the form $[f]$. Note that if $[f]=[g]$, then $\|f-g\|< \varepsilon$. Indeed, for any $x\in X$ find $t\in D$ such that $\|x-t\|\leqslant \varepsilon/3$, then 
$$
|f(x)-g(x)|\leqslant |f(x)-f(t)|+|g(x)-g(t)|+|f(t)-g(t)|<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon.
$$
So, $V$ is covered by finitely many sets of diameter at most $\varepsilon$, each set is defined as the set of functions $f$ with the same $[f]$. 


You may enlarge them to open subsets of diameter less than $2\varepsilon$, if you wish.


