(a) The attempt works. To see that $S:=\{e_i,i\in\Bbb N\}$ is closed for the $\ell^1$ norm, let $x\notin S$. There are two index $i$ and $j$ such that $x_ix_j\neq 0$. Let $r:=\min\{|x_i|,|x_j|\}$. Then the open ball of center $x$ and radius $r$ is contained in the complement of $S$.
(b) The problem is that we have to check that we have convergence in $\ell^1$ for the subsequence.
As $\ell^1$ is complete, we can check that $B$ is precompact, i.e. given $\delta>0$, we can cover $B$ by finitely many balls of radius $<\delta$. It's equivalent to show both properties hold:
- $B$ is bounded in the $\ell^1$ norm;
- $\lim_{N\to +\infty}\sup_{x\in B}\sum_{k=N}^{+\infty}|x_k|=0$.
Indeed, if a set $S$ is precompact, with $\delta=1$ we get that it's bounded, and $2.$ is a $2\delta$ argument (I almost behaves as a finite set).
Conversely, assume that $1.$ and $2.$ hold and fix $\delta$. Use this $\delta$ in the definition of the limit to get an integer $N$ such that $\sup_{x\in B}\sum_{n=N+1}^{+\infty}|x_n|<\delta$. Then use precompactness of $[-M,M]^N$, where $M=\sup_{x\in B}\lVert x\rVert_1$.
Note that this criterion works for $\ell^p$, $1\leqslant p<\infty$.
In our case, each element of $B$ has a norm $\leqslant 1$, and for all $x\in V$,
$$\sum_{k=N}^{+\infty}|x_k|\leqslant \frac 1N\sum_{k= N}^{+\infty}k|x_k|\leqslant \frac 1N.$$
It is clearly not open: take for instance the all-zero sequence, $\mathbf{0}$, which clearly belongs to $F$.
For any $\varepsilon > 0$, the $\ell_\infty$ ball around $\mathbf{0}$ includes the constant sequence $(\varepsilon)_{n\in\mathbb{N}}$, which does not belong to $F$.
It is also not closed. To show it, it's sufficient to show that its complement is not open either. Consider the sequence $u=(u_n)_n\notin F$ defined by
$$
u_n = \frac{1}{n+1}.
$$
For any $\varepsilon > 0$, the $\ell_\infty$ ball $B(u,\varepsilon)$ around $u$ will intersect $F$: indeed, letting $N\geq 0$ be the smallest integer such that $\frac{1}{N+1} < \varepsilon$, then the sequence $u^\prime$ which agrees with $u$ on the first $N$ terms and is identically $0$ afterwards belongs to $F$; and furthermore $\lVert u - u^\prime \rVert_\infty < \varepsilon$. Thus, $u^\prime\in B(u,\varepsilon)\cap F$, and so $B(u,\varepsilon)\not\subseteq F^c$.
Best Answer
My answer: only 2 is true. It is not closed as $\lim e_n$ does not exist. It is bounded as $||e_n||=1$ It is not compact as one can not find a finite sub-cover containing S. Neither does it contain a convergent subsequence as $\lim e_{n_k} $ does not exist.