Using Ascoli-Arzela theorem to prove a set is compact

Question

Let $D= \{f_k(x) = \sin kx/(1+k);\quad k= 0,1,2,\dots\quad x \in [0,1]\}$. Show that $D$ is compact.

I know that I should approach this kind of question by Ascoli-Arzela theorem (proving $D$ is equicontinuous, bounded and closed). For equicontinuity, do we always use $3\epsilon$ technique? Thus, first find the limit $f$ (I am not sure about $f$ is $0$.) And I am confused about proving that it is closed. In general, I don't know how to handle this question which contains specific functions.

Answer 1

You can actually proceed directly: let $f_k$ be a sequence in $D$, then argue that either it has infinitely many copies of the same element of $D$, or else it converges to $0$.

Arzela-Ascoli would basically require the same argumentation in this case.

Answer 2

Note that each $f_k$ is 1-lipschitz. In particular, this implies that, for each $\varepsilon > 0$, you can write $(|x-y| < \varepsilon) \Rightarrow (|f_k(x)-f_k(y)| < \varepsilon)$ for each $ k \ge 1$, therefore the family is equicontinuous.

Answer 3

Note that with $k=0$ this sequence contains its own limit. This is the second most trivial situation for a compact set, the most trivial being a set of finitely many points.