Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 22H / d

For $F_1$, the functions satisfy $\Vert \sin (\pi nx)/n{\Vert }_{\infty }\le 1/n$, so the sequence converges uniformly to zero. Its closure is the sequence together with its limit and is compact. The same argument applies to $F_2$, since $\Vert \sin (\pi nx)/\sqrt{n}{\Vert }_{\infty }\le n^{-1/2}$. Thus $F_1$ and $F_2$ are relatively compact.

The family $F_3$ is not equicontinuous. At $x=0$ its members vanish, but for $x_n=1/(2n)\to 0$,By the Arzela-Ascoli theorem, a relatively compact subset of $C[0,1]$ must be equicontinuous. Hence $F_3$ is not relatively compact.