[Math] Closed set in $l^1$ space

Question

Let $$ X := \left \{ (a_n) : \sum_{n=0}^\infty |a_n| < \infty \right\}$$ with the metric $d(a_n,b_n) := \sum_n |a_n-b_n|$. Let $\delta_j^{(n)} := 1$ if $n = j$ and $0$ otherwise. Denote $\delta^{(n)}:=(\delta_j^{(n)})_{j=0}^\infty$ and $E := \{ \delta^{(n)} : n \in \mathbb N\}$.

I want to show that $E$ is closed and bounded but not compact. Boundedness is trivial but I get stuck at closedness. If $(x_j)_{j=0}^\infty$ is the limit of a sequence $(\delta^{(n_k)})_{k=0}^\infty$ in $E$ I get
$$
\forall \epsilon > 0 \exists N \in \mathbb N \forall k \geq N: \sum_{j=0}^\infty |\delta_j^{n_k} – x_j| = |1-x_{n_k}| + \sum_{j \neq n_k} |x_j| < \epsilon.
$$

Answer 1

Your strategy for proving closedness (take an arbitrary limit point then show this point belongs to the set) is the right one under most circumstances. This problem, however, is somehow special.

For one thing, the set $E$ actually has no limit point. To see this, simply note that for $x\neq y\in E$, $\|x-y\|=2$ so there is no Cauchy sequence in $E$, let alone convergent ones. So in particular, there is no limit point to this set $E$.

After you show there is no Cauchy sequence in $E$, you can just take a countable subset in $E$. Since this countable subset has no limit point, we see that $E$ is not sequentially compact, which implies non-compactness in a metric space.