In infinite-dimensional spaces “closed and bounded” does not imply “compact”

In general "closed and bounded" does not imply "compact" in infinite-dimensional spaces. I want to show this for the following set:

Let $A:=\{\phi_n:[0,1]\rightarrow \mathbb{R},x\rightarrow x^n|n\in \mathbb{N}\}$ be in $C^0([0,1])$ with the supremum norm.

I already showed that it is bounded and not compact. But I don't know how to show that $A$ is closed. Can you give me a hint on how to do it?

• If a sequence in that set converges with that norm, so that it converges uniformly, then it also converges pointwise. Show that there is exactly one function which is the pointwise limit of a sequence of $A$. – Mariano Suárez-Álvarez Dec 7 '17 at 22:09

Hint: Let $(\phi_{n_k})_{k\in\mathbb N}$ be a convergent sequence of elements of your set. There are two possibilities:
1. $\{n_k\,|\,k\in\mathbb{N}\}$ is bounded. Since the bounded subsets of $\mathbb N$ are the finite ones, this meanse that, for some $m\in\mathbb N$, $n_k=m$ infinitely often. But then a subsequence of $(\phi_{n_k})_{k\in\mathbb N}$ converges to $\phi_m$ and, since we are assuming that the whole sequence converges, it must converge to $\phi_m$.
2. $\{n_k\,|\,k\in\mathbb{N}\}$ is unbounded. Then some subsequence of the sequence $(n_k)_{k\in\mathbb N}$ tends to $+\infty$. We can, without loss of generality, assume that it is the sequence $(n_k)_{k\in\mathbb N}$ itself. But this is not possible, because$$\lim_{k\to\infty}x^{n_k}=\begin{cases}0&\text{ if }x<1\\1&\text{ if }x=1.\end{cases}$$This is impossible, since this last function is outside $\mathcal{C}^0\bigl([0,1]\bigr)$.
• If there are no sequences that converge, does this imply that A is closed? I'm also not quite sure how to show that $\phi$ is the only possible limit of sequences in A. It makes sense to me, but I don't know how to show it mathematically. – Yasuduck Dec 7 '17 at 22:26