Sequence in $C[0,1]$ with no convergent subsequence I am trying to show that $C[0,1],$ the space of all real - valued continuous functions  with the sup metric is not sequentially compact with the sup metric by showing that the sequence $f_n = x^n$ has no convergent subsequence. The sup metric $\|\cdot\|$ is defined as 
$$\|f - g \| = \sup_{x \in [0,1]} |f(x) - g(x)|$$
where $|\cdot|$ is the ordinary Euclidean metric. Now I know that $f_n \rightarrow f$ pointwise, where 
$$f = \begin{cases} 0, & 0 \leq x < 1 \\ 1, & x = 1.\end{cases}$$
However $f \notin C[0,1]$ so this means by theorem 7.12 of Baby Rudin that $f_n$ cannot converge to $f$ uniformly. However how does this tell me that no subsequence of $f_n$ can converge to something in $C[0,1]$?
Thanks.
 A: This is (more-or-less) a modification of the hint from t.b's comment.
It suffices to show that any subsequence of $(f_n)$ is not Cauchy. (Since every convergent sequence is a Cauchy sequence.)
To do this, notice that
$$\lVert f_{2n}-f_n \rVert = \sup_{x\in[0,1]} (x^n - x^{2n}) = \sup_{x\in[0,1]} (x^n - (x^{n})^2) = \sup_{t\in[0,1]} (t-t^2)=\frac14.$$
(It is easy to find maximum of the quadratic function $f(t)=t-t^2=t(1-t)$.)
In addition, if we use the monotonicity of the sequence $(f_n)$, we see that
$$k\ge 2n \qquad \Rightarrow \qquad \lVert f_{k}-f_n \rVert \ge \frac14.$$
Now, if we have any subsequence of $(f_n)$ then the above estimate shows that this subsequence is not Cauchy. For any given $k_0$ we can find $k'>k_0$ such that $n_{k'}>2n_{k_0}$ and $\lVert f_{n_{k'}}-f_{n_{k_0}} \rVert \ge \frac 14$.
A: Since $f_n$ converges to $f$ pointwise, all subsequences $f_{n_k}$ also converge pointwise to $f$. So $f_{n_k}$ can not converge uniformly to continuous $g$, because uniform convergence implies pointwise convergence as well, and $f$ is not continuous. 
A: Regarding the motivation for your question (sequential compactness):
First, commenting on your specific question: a topological vector space over the real or complex numbers is never sequentially compact. Only bounded closed subset are possibly (sequentially) compact. Your attempt of proof is, therefore, only relevant if you restrict to bounded sequences (which applies to your example but should be made an explicit assumption).
As a general remark: it is known that compactness and sequential compactness are equivalent in metric spaces, hence in subsets of normed vector spaces (like $C^0$).
It is also known (though a bit more involved when it comes to proving it) that a normed (real or complex) vector space has the so called 'Heine Borel' property (which says that a subset is compact if and only if it is closed and bounded) if and only if it is of finite dimension.
Edit: the following sentence is obviouly not correct (see the comment of t.b. -- thanks for pointing this out), but is a nice illustration of a premature und uncautios conclusion, so I don't delete it ;-):  Since $C^0$ is not finite dimenional, it follows that bounded closed sets of $C^0$ are (sequentially) compact only if they are contained in a finite dimensional subspace. 
The following remains true, though: sets like the closed unit ball (and therefore any set containing an open ball) are not (sequentially) compact in $C^0$ and, in general, bounded sequences need not have a converquent subsequence, and this applies to every infinite dimensional normed vector space.
A: After all of this discussion, here is my understanding of how the problem concludes: We already know that $f_n \rightarrow f$ pointwise, where $f$ is the discontinuous function that I defined on $C[0,1]$. Now go up to a bigger metric space like $B[0,1]$ that contains $C[0,1]$ and equip it with the sup metric. In there we know that $f_n \rightarrow f$ pointwise. This means that for each $x \in [0,1]$, the ordinary sequence of real numbers $f_n(x) \rightarrow f(x)$ and since $\Bbb{R}$ with the Euclidean topology is Hausdorff it follows that the pointwise limit of $f(x)$ is unique. Viz, there is no other function on $B[0,1]$ that converges to $f$ pointwise.
Now I claim that there is no function $g \in C[0,1]$ for which  $f_n$ converges to uniformly. If there were, then $f_n \rightarrow g$ pointwise on $C[0,1]$. Furthermore, a continuous real valued function on a compact set is bounded and so $g$ lives in $B[0,1]$. By what I said about limits being unique, this means that $g = f$. However this is a contradiction because $f \notin C[0,1]$.
It follows that there is no continuous function on $[0,1]$ that $f_n$ converges to. Q.E.D.
