Why is compactness required in Arzela Ascoli? My question comes from the necessity of proving that one of the hypothesis in the Arzela Ascoli theorem fails in the example:
$f_n(x)=\frac{1}{1+(x-n)^2}; x\in [0,\infty)$
I have tried proving it is not uniformly equicontinuous nor uniformly bounded but keep failing, so my only assumption is that Arzela Ascoli cannot be applied because $[0,\infty)$ is closed but not bounded, so it is not compact in $\mathbb{R}$. Yet, I cannot think of a way to show the theorem fails in this example. 
 A: This sequence is equi-continuous and uniformly bounded. The only reason Arzela-Ascoli Theorem is not applicable here is because $[0,\infty)$ is not compact. 
Note that $0 \leq f_n(x) \leq 1$. So the sequence is uniformly bounded. Let su prove equi-continuity: $|f_n(x)-f_n(y)| \leq \frac {|x-y| (|(x-n)+(y-n)|} {(1+(x-n)^{2}) (1+(y-n)^{2})}$. Use the inequality $|a| \leq \frac  1 2 (1+a^{2})$ with $a=x-n$ and $a=y-n$ to see that the sequence is  equi-continuous. 
Note that $f_n(x) \to 0$ for each $n$ whereas $f_n(n)=1$. This shows that the sequence has no uniformly convergent subsequence. 
A: You have a sequence $(f_n)_{n\in\mathbb N}$ of functions which is uniformly bounded and equicontinuous. Asserting that the theorem fails in that situation means that the conclusion of the theorem doesn't hold (and not that one of the hypothesis doesn't hold); in other words, $(f_n)_{n\in\mathbb N}$ has no subsequence which is uniformly convergent.
In order to see why, not that, for each $x\in[0,\infty)$, $\lim_{n\to\infty}f_n(x)=0$, but $(\forall n\in\mathbb N):f_n(n)=1$.
