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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?

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  • $\begingroup$ 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 $\mathbb N$. $\endgroup$
    – Jochen
    Nov 12, 2023 at 12:35
  • $\begingroup$ The correct accent is Arzelà. $\endgroup$
    – Jochen
    Nov 12, 2023 at 12:52

1 Answer 1

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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 $g\in K$.

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