Cauchy subsequence problem Consider the set $S_0$ of all continuous functions $f : [0,1] \to [0,1]$.
Define a metric on the set $S_0$ by setting
$$
\rho(f,g)=\sup\limits_{x\in{[0,1]}}|f(x)-g(x)|.
$$
(a) Give an example of a sequence in  $S_0$ that does not contain a Cauchy subsequence with respect to the metric $\rho$.
(b) For a given $\epsilon> 0$ consider $S_\epsilon \subset S_0$ consisting of the set of functions $f \in S_0$
such that
$$
|f(x)-f(y)|\leq|x-y|^{\epsilon}\qquad \rm{for~ all} ~x,y\in[0,1]
$$
Show from first principles that any sequence in  $S_\epsilon$ contains a subsequence that is
Cauchy with respect to the metric $\rho$.
I cannot find an example in (a). And in (b), if $f : [0,1] \to [0,1]$ is finite dimentional with respect to metric $\rho$. Then by proving f is bounded and closed, we can get f is compact. But f is infinite dimensional, so how to tackle this problem?

Hint for (a): Try a sequence of functions $f_n$ such that, e.g., $f_n(1/n) = 1$ but $f_n(1/m) = 0$ for all other $m$.
Hint for (b): if you have a sequence $f_n$ in $S_\epsilon$ such that $f_n(r)$ is Cauchy for each rational $r$, show that $f_n$ is Cauchy in the metric $\rho$.
