Open/closed, bounded, compact sets of sequences

Question

Given the sets $$E := \{x = (x_n)_n ∈ \ell^1 | n\in N \Rightarrow 0 < x_n < 1/n\}$$ $$K := \{x = (x_n)_n ∈ \ell^1 |n\in N \Rightarrow 0 ≤ x_n ≤ 1/n\}$$ I have to prove or disprove that $E$ is open, bounded, compact in $(\ell^1,\|.\|_1)$ and that $K$ is closed, bounded, compact.

Answer 1

The set $E$ is not bounded: for each $m\in\mathbb N$ and we can define $$ x=(1,\tfrac12,\tfrac13-\tfrac14,\tfrac14-\tfrac18,\ldots,\tfrac1m-\tfrac1{2^{m-1}},0,\ldots). $$ Then $$ \|x\|_1=\sum_{k=1}^m\frac1k-\sum_{k=2}^{m-1}\frac1{2^k}\geq-1+\sum_{k=1}^m\frac1k. $$ So there exist $x\in E$ with norm arbitrarily large. As $E\subset K$, $K$ is also unbounded and thus not compact.

• The set $E$ is not closed: consider the elements $$ (1-\tfrac1m,0,0,\ldots). $$ The limit is $(1,0,0,\ldots)$, which is not in $E$.

• The set $K$ is closed. A Cauchy sequence in $K$ is Cauchy in $\ell^1$, which is complete, so the limit exists. As norm convergence implies entrywise convergence, the entries of the limit satisfy the same inequalities as the entries int the Cauchy sequence.

• The set $E$ is not open. Let $x\in E$ be the element given by $$ x_n=\begin{cases} 2^{-n}-4^{-n},&\ n=2^k\ \text{ for some } k\\[0.3cm] 0,&\ \text{ otherwise} \end{cases} $$ Now, for each $m\in\mathbb N$, let $x^{(m)}\in \ell^1$ be the element given by $$ x^{(m)}_n=\begin{cases} x_n,&\ n=2^k,\ k\ne m\\[0.3cm] 2^{-m},&\ n=2^m\\[0.3cm] 0,&\ \text{ otherwise} \end{cases} $$ Then $x^{(m)}\not\in E$ (because of the $2^m$ coordinate), but $\|x^{(m)}-x\|_1=4^{-m}\to0$