If $(f_n)$ is a uniformly convergent sequence in $C(X)$, and $f_n\to f$ uniformly, then $\{f\} \cup \{f_n : n\ge 1\}$ is compact in $C(X)$.

Question

Let $X$ denote a compact metric space. If $(f_n)_{n=1}^\infty$ is a uniformly convergent sequence in $C(X)$, and $f_n\to f$ uniformly, then show that $\{f\} \cup \{f_n : n\ge 1\}$ is compact in $C(X)$.

I've multiple approaches in mind to prove the compactness of $S = \{f\} \cup \{f_n: n\ge 1\}$, but I haven't been able to complete any of them:

1. $S$ is compact iff every sequence in $S$ has a subsequence that converges to a point in $S$.
2. $S$ is compact iff every open cover of $S$ has a finite subcover.
3. $S$ is compact iff $S$ is complete and totally bounded.

How do I approach this problem? I am not sure where to use the compactness of $X$, and the fact that $f_n\stackrel{\text{unif}}{\longrightarrow} f$. For the second approach - I am not sure what open sets in $C(X)$ look like; so I'm not able to get anything useful by writing $S \subset \bigcup_{A\in O} A$ where $O$ is a family of open sets of $C(X)$.

Please let me know what direction to proceed in, and drop some hints - so I can find my way out of this wilderness. Thank you.

Answer 1

In any metric space $(Y,d)$ (or any topological space $Y$ for that matter) any convergent sequence $(y_n)$ with limit $y$ gives a compact set $\{y,y_1,y_2,...\}$ and this is easy to prove from definition of compactness in terms of open covers. Take $Y=C(X)$ with the sup metric.