How do I prove the interior of subspace $\ell^1$?

Let $E:=\ell^1$ is Banach space with standard norm for $\ell^1$, $P:=\{\bar{x}\in\ell^1: \bar{x}=(x_i)=(x_1,x_2,\ldots),x_i \geq 0, \forall i \in \mathbb{N}\}$ and defined that interior of $P$ is $\{\bar{x}\in\ell^1: \bar{x}=(x_i)=(x_1,x_2,\ldots),x_i > 0, \forall i \in \mathbb{N}\}$, denoted by $\mathrm{int} (P)$. How do I prove that $\mathrm{int} (P)$ is interior of $P$?

Actually, the set $\mathrm{int}(P)$ is not open for the $\ell^1$ topology: if $\bar x =(x_n)\in \mathrm{int}(P)$, then $x_n\to 0$. For any $\delta$, take $n$ such that $|x_n|\lt \delta$: the ball of center $\bar x$ and radius $\delta$ contains an element whose one of coordinates is negative.
If $O$ is a non-empty open subset of $\mathrm{int}(P)$, then for $\bar x\in O$, the ball of center $\bar x$ and radius $\delta$ is contained in $O$ for some $\delta$, hence $x_n -\delta\gt 0$ for each $n$ which is not possible. Therefore, the interior of $P$ for the $\ell^1$ topology is empty.
• I have learned it from my textbook and my problem is which $delta$'s value that I have to choose such that the Ball is contained in P? – Agus Nur Ahmad S Mar 6 '15 at 1:58
• Actually, it turns out that such a $\delta$ does not exist; see edit. – Davide Giraudo Mar 6 '15 at 9:16