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.

  • $\begingroup$ 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? $\endgroup$ – Agus Nur Ahmad S Mar 6 '15 at 1:58
  • $\begingroup$ Actually, it turns out that such a $\delta$ does not exist; see edit. $\endgroup$ – Davide Giraudo Mar 6 '15 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.