Continuity of the limit of ${f_n}(x) = \cos(2^n\pi x)$. I am trying to prove that the metric space of continuous functions from [0,1] to $\mathbb{R}$ is not compact by examining $\lim_{n \to \infty} {f_n}(x) = \cos(2^n\pi x)$. I cannot think of a direct proof, so I think maybe it's best to show that the sequence converges to a function that is not Riemann integrable and therefore not continuous (outside the metric space). I'm sort of stuck after that...
(The metric for the space is ${d}(f_1,f_2) = {sup}_{x \in [0,1]}|f_1(x) - f_2(x)|$.)
 A: I. Any metric on a compact space is bounded.Let $(X,d)$ be a compact metric space .If $p$ is any member of $X$ then the open cover $\{B_d(p,r) :r>0\}$ has a sub-cover $\{B_d(p,r) :r\in S\}$ where $S$ is finite. So $X=B_d(p,t)$ where $t=\max S$. By the triangle inequality $d(p',p'')\leq d(p',p)+d(p,p'')<2t$ for all $p',p''\in X$. So $d$ is bounded......II. The $sup$ norm on $C[0,1]$ is not a bounded metric.Consider the constant functions: If $g_r(x)=r$ for all $x\in [0,1]$ then $d(g_r,g_0)=|r|.$..... III. Your sequence $f_n$ does not converge in $C[0,1]$. For example when $x=\sum_{j=0}^{\infty}a_j4^{-j}$ where $a_j=1/2$ when $j$ is even and $a_j=0$ when $j$ is odd, then $f_{2n}(x)$ does not converge.....IV.  The converse theorem, that some metrics for a non-compact metrizable space are  unbounded metrics, is also true, but that's another story.
A: Hint: Every compact metric space is bounded. Is $C[0,1]$ bounded?
A: The sequence
$$
f_n(x) = \cos(2^n \pi x)
$$
does not converge to any function in the infinity norm. In fact, it doesn't even converge for any fixed irrational value of $x$.
What you want to show is that $(f_n)$ has no convergent subsequence. How can you show that? Note that for any dyadic rational $a$, and for any $n$, $f_n(a)$ is $1$. So any subsequence of $f_n(a)$ converges to $1$. Then argue that if some subsequence of $(f_n)$ converges (in the metric space, i.e. in the infinity norm) to $f$, then $f$ must be continuous. Conclude that $f(x) = 1$. Finally, obtain a contradiction by considering $f(2\pi/3)$.

An alternative, silly proof is to argue that the metric space is not bounded in the first place, so it must not be compact. But this proof using $f_n$ is more general; it shows not just that $C([0,1], \mathbb{R})$ is not compact, but even that $C([0,1], [-1,-1])$ is not compact.
A: Consider $f_n(x):[0,1]\to [0,1]$ defined by $f_n(x)=n$ , $\forall n\in \mathbb N$.
