$E=\{e_k|k\in\mathbb{N}\}$ : characterization of $E$? Let $(X, \|.\|)$ be normed linear space consisting of sequences $a=\{a(n)\}_{n=1}^{\infty}$ such that the series $\sum_{n=1}^\infty a(n)$ is absolutely convergent,with $ \|a\|= \sum_{n=1}^\infty |a(n)|$. Let $e_k$ denote the sequence in $X$ whose $k$-th term is 1 and other terms are $0's$ and let $E=\{e_k | k \in \mathbb{N}\}.$ Then 


$X$ is complete in the norm $\|.\|$
$E$ is bounded subset of $X$.
$E$ is closed subset of $X$.
$E$ is compact subset of $X$.
Let $\{ a_n \}_{n=1}^\infty$ be a Cauchy sequence in $X$, i.e. for each $\epsilon &gt; 0$ there exists a $N \in \Bbb N$, such that 
$$ \sum_{j=1}^\infty | a_n(j) - a_m(j) | = \| a_n - a_m \| &lt; \epsilon \quad \text{for all } n,m \geq N \; . \tag{$\ast$}$$
Let $\epsilon &gt; 0$ and $N \in \Bbb N$, such that $(\ast)$ holds. Note that for each $j \in \Bbb N^\times$ the sequence $\{ a_n(j) \}_{n=1}^\infty$ is a Cauchy sequence in $\Bbb R$, and since $\Bbb R$ is complete, the limit 
$$ a(j) := \lim_{n \to \infty} a_n(j) $$
exists. Define $a := \{ a(j) \}_{j=1}^\infty$. We wnat to show that $a_n \to a$ in $X$, i.e. $\| a_n - a \| \to 0$, and $a \in X$. Note that for each $M \in \Bbb N^\times$ 
$$ \sum_{j=1}^M |a_m(j) - a_n(j) | \leq \sum_{j=1}^\infty | a_m(j) - a_n(j) | &lt; \epsilon \quad \text{for all } m,n \geq N \; .$$
By letting $n \to \infty$, we conclude that 
$$ \sum_{j=1}^M | a_m(j) - a(j) | \leq \epsilon \quad \text{for all $m \geq N$ and $M \in \Bbb N^\times$} \; ,$$
and now, by letting $M \to \infty$, we see that 
$$ \| a_m - a \| = \sum_{j=1}^\infty | a_m(j) - a(j) | \leq \epsilon \quad \text{for all } m \geq N \; .$$
This shows that $a_n \to a$ in $X$. It remains to show that $a \in X$, i.e. $\Vert a \Vert &lt; \infty$. Choose a $m \in \Bbb N^\times$, such that $\| a - a_m \| &lt; 1$ (this exists, because $\| a_m - a \| \to 0$). Now we deduce by using the triangle inequality, that 
$$ \|a \| = \| (a-a_m) + a_m \| \leq \underbrace{ \| a - a_m \| }_{&lt; 1} + \underbrace{ \| a_m \| }_{&lt; \infty} &lt; \infty \; ,$$ 
so $a \in X$, and we have shown, that $X$ is complete.
This is clear, because $\| e_k \| = 1$ for each $k \in \Bbb N^\times$.
Observe, that for $e_k, e_\ell \in E$ 
$$ \| e_k - e_\ell \| = \sum_{j=1}^\infty | e_k(j) - e_\ell(j) | = \begin{cases} 2 \, , &amp; k \neq \ell \\ 0 \, , &amp; k = \ell \end{cases} \; .$$
With 1. we conclude, that a sequence $\{ a_n \}_{n=1}^\infty$ in $E$ converges, if and only if there exists a $N \in \Bbb N^\times$ and a $k \in \Bbb N^\times$, such that 
$$ a_n = e_k \quad \text{for all } n \geq N \; ,$$
in which case 
$$ \lim_{n \to \infty} a_n = e_k \in E \; .$$
This shows that $E$ is closed.
Note, that we can not use Heine-Borel here, because $X \neq \Bbb R^n$. But you can easily construct a sequence $\{ a_n \}_{n=1}^\infty$ in $E$, which has no convergent subsequence. This means, that $E$ is not compact.
Complete: for all $a=\{ a(n)\}_{n=1}^∞$ there is sequence $b(k) = \sum _ {i=1}^ k  a(i)*e_i.$ This sequence converges to $a$.
Bounded: for all $e_k: ∥e_k∥ = 1$
Closed: if $a !=e_k$ for any $k$ and we have at least two non-zero components $a(i)$ and $a(j)$. Let $a(i)$ be smaller than $a(j)$ by absolute value. Then $\mid (a(i))/2 \mid$ - surrounding of a doesn't belong to $E$ since it has 2 non -zero components. In case a has only one non-zero component $a(i)$ than $\mid(a(i)-1)/2\mid$ - surrounding of a doesn't belong to $E \Rightarrow$ addition to $E$ is open $\Rightarrow E$ is closed
$E$ is closed and bounded $\Rightarrow$ $E$ is compact.