# Compactness of a set of functions on an infinite-dimensional function space

Let X be the space of functions $$f:\mathbb{N}\to\mathbb{Z}$$, endowed with the metric $$d(f,g)=\sum_{i=0}^{\infty}{\frac{1}{2^{i}}\mathbb{I}(f(i)\neq g(i))}$$, where $$\mathbb{I}$$ is the indicatorfunction. Now fix $$L\geq 0$$ and let $$K=\{f\in X:f(0)=0, \lvert f(i+1)-f(i)\rvert\leq L\}$$. Does anyone have any ideas how to show that $$K$$ is compact in $$X$$? Probably the sequential definition of compactness is the way to go but how?

• Isn't this a particular case of the Arzelá-Ascoli theorem where $\mathbb N$ and $\mathbb Z$ are endoed with the discrete metric? The metric $d$ desribes the topology of uniform convergence on compact (=finite) subsets of $\mathbb N$. Nov 12, 2023 at 12:35
• The correct accent is Arzelà. Nov 12, 2023 at 12:52

Let $$f_n$$ be a sequence in $$K$$. We will we find a subsequence that converges by a diagonalization argument. Evidently $$f_n(0)=0$$ for all $$n$$. $$f_n(1)$$ is contained in a finite interval $$[-L,L]$$. Since the $$f_n$$ are integer valued in a finite interval there must be some constant subsequence $$f_{n^1_k}$$ of $$f_n$$, e.g. $$f_{n^1_k}(1)=C_1$$. Now $$f_n(2)$$ is contained in the finite interval $$[-2L,2L]$$. Again take a subsequence $$n^2_k$$ of $$n^1_k$$ such that $$f_{n^2_k}(2)=C_2$$ for all $$k$$. We continue in this fashion.
Take $$g_j=f_{n^j_j}$$. If we fix $$i$$, $$g_j(i)$$ is a constant sequence for $$i\leq j$$. Take $$g(i)=g_i(i)$$. Now $$d(g_j,g)=\sum_{i=0}^{\infty} \frac1{2^i}I(g_j(i)\neq g(i))=\sum_{i=j+1}^{\infty} \frac1{2^i}I(g_j(i)\neq g(i))\leq\sum_{i=j+1}^{\infty} \frac1{2^i}$$ which goes to $$0$$ as $$j\to\infty$$. It is not hard to show that $$K$$ is closed which implies that $$g\in K$$.