$A=\{e_k : k=1,2,3\ldots\} $ is closed in $\ell1$

Question

$A=\{e_k : k=1,2,3\ldots\} $ is closed in $\ell1$

Let $(x_n)\in A$ such that $(x_n)\to x_0$, where $x_0\in \ell1$

WTS : $x_0 \in A$

We have $||x_n-x_0||_1 \to 0$ in $R$

$\rightarrow \sum_{k\geq 1}|x_n^{k}-x_0^k| \to 0$ in $R$

$|x_n^k-x_0^k| \to 0$ $\;\forall k\in N $

for fixed $n\in N$, because $(x_n)\in A\rightarrow x_n^t = 1$ for some $ t\in N$ and $x_n^k =0 $ $\forall k\neq t$

$\Rightarrow x_0^t=1$ and $x_0^k=0$ forall $k\neq t$

so $x_0 \in A$

Is this proof okay?

EDIT : On a second thought, it is easy to see $||e_k - e_j||_1 = 2 >\frac12 \forall k\neq j$

so it can be shown that the only sequences converging in $A$ are eventually constant ones and hence it is closed (just like a discrete metric space)

Answer 1

A discrete subset of a metric space has no limit points; it is, therefore, a closed subset of that space. The $e_k$ form a discrete subset of $\ell^1$.