The set of sequences $e_j= (0,...,0,1,0,...)$ is closed in $\ell_1$.

Question

If $e_j=(0,...,0,1,0,...) $ is the sequence with $1$ in the $j-th$ entry and $0$ in the others then prove that the set $A=\{ e_j \; : \; j \in \mathbb{N} \}$ is closed in $\ell_1$. What I have tried is taking a convergent sequence in $A$, but I don't know how to argument that the sequence has its limit in $A$. I conjecture that the sequence will be eventually constant and then the limit is in $A$.

Answer 1

Note first that $\|e_i - e_j \|_1 = 2$ for all $i \neq j$, thus if any two elements of $A$ are closer than $2$ to each other they are equal. Suppose $(a_n)_{n \in \mathbb{N}}$ is convergent in $A$, then it is Cauchy in $\ell_1$ and for $\epsilon=2$ there exists $N \in \mathbb{N}$ such that for all $n,m \geq N$ we have $\|a_n - a_m \|_1 < 2$. As each element of the sequence lies in $A$ then this implies $a_n = a_m$ for all $n,m \geq N$, thus there exists $j \in \mathbb{N}$ such that $a_n=e_j$ for all $n \geq N$. This implies $a_n \rightarrow e_j \in A$ as $n \rightarrow \infty$ and thus $A$ is closed.