closed linear span of $\{ e_n : n \in \mathbb{N} \}$ in $l^{\infty}$

Question

If $A=\{ e_n : n \in \mathbb{N} \}$ where $e_n$ is a sequence whose $n$^th term is $1$ and $0$ otherwise.

Then I need to find the smallest closed linear subspace in $l^{\infty}$ that contains $A$.

I think $c_0$ is the answer.

My attempt:

I proved that $c_0$ is a closed linear subspace that containing $A$ and this proves one containment, but I couldn't prove that $c_0 \subset \bigcap_{A\subset E_i}{E_i}$.

Answer 1

If $x=(a_1,a_2,a_3,\ldots)\in c_0$, then let

• $x_1=(a_1,0,0,0,0,\ldots)$;
• $x_2=(a_1,a_2,0,0,0,\ldots)$;
• $x_3=(a_1,a_2,a_3,0,0,\ldots)$

and so on. Then $x=\lim_{n\to\infty}x_n$ (and, in order to prove this, the fact that $x\in c_0$ is essential) and each $x_n$ belongs to$$\operatorname{span}\bigl(\{e_n\mid n\in\Bbb N\}\bigr).\tag1$$Therefore every closed subspace of $\ell^\infty$ that contains $(1)$ also contains $c_0$. So, $c_0$ is indeed the smallest closed subspace of $\ell^\infty$ which contains $(1)$.