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I'm working on the space $\ell_1$ under the $\left \| \cdot \right \|_{\infty}$ and I want to prove the set $$ A=\left \{ (x_n) \in \ell_1: \; \sum_{n=1} ^{\infty} |x_n|\leq 1 \right \} $$ is $(a)$ closed, $(b)$ convex, $(c)$ absorbent, $(d) $ has empty interior. I think i got the $(a),(b),(c)$ part right so I just post the $(c)$ for the sake of correctness:

Let $\{y^{(j)}\}_{j\in\mathbb{N}}$ be a sequence of elements in $A$. i,e, $\displaystyle{\sum_{n=1}^{\infty}\left | y^{(j)}_n \right |}\leq 1$ for every $j \in \mathbb{N }$ and is such that $\displaystyle{\; y^{(j)} \xrightarrow{\Vert \cdot \Vert_{\infty}} y\;}$. We claim that $y=(y_n)_{n \in \mathbb{N}} %=\left ( y_n \right )_{n \in \mathbb{N}} \in A$. For every $\varepsilon >0$ and $n \in \mathbb{N}$, there exist $J_{n,\varepsilon} \in \mathbb{N}$ such that for every $j\geq J_{n,\varepsilon}$, we have $$ \left \| y-y^{(j)} \right \|_{\infty}=\sup_{n \in \mathbb{N}}| y_n-y^{(j)}_n |<\frac{\varepsilon}{2^n} . $$ Then, for a fixed $n \in \mathbb{N}$ if $j\geq J_{n,\varepsilon}$ we have \begin{align*} |y_n|&\leq|y_n-y_n^{(j)}| + |y_n^{(j)}|\\ &\leq||y-y^{(j)}||_\infty + |y_n^{(j)}| \\ &< \frac{\varepsilon}{2^n} + |y_n^{(j)}| \end{align*} Where it follows
\begin{align*} \sum_{n=1}^{\infty}\left | y_n \right | &\leq \sum_{n=1}^{\infty}\left | y_n-y_n^{(j)} \right | + \sum_{n=1}^{\infty} \left | y_n^{(j)} \right | \\ &< \sum_{n=1}^{\infty} \frac{\varepsilon}{2^n} + 1 \\ &=\varepsilon +1 \end{align*} We conclude that $\displaystyle{\sum_{n=1}^{\infty}\left | y_n \right |}<\varepsilon +1$, for every $\varepsilon>0$. So that $y\in A$.

Is this correct? I think it is, however I'm stuck proving the $(d)$ part. I've tried by contradiction.

Suppose there exist an element $x_n \in int(A)$ so there exist $r>0$ such that $B_{\left \| . \right \|_{\infty}}\left ( x,r \right ) \subseteq A$. Is there any way to construct a sequence $(y_n)$ such that $ \sup_{n \in \mathbb{N}}| y_n-x_n |<r$ but $ \sum_{n=1}^{\infty}\left | y_n \right |$ is greater than 1? Any hint would ve appreciate, thanks.

ipreferpi
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  • For part (d), just add $r/2$ to $x$, and show that the absolute sum no longer converges. – user744868 Mar 05 '20 at 02:31
  • I was thinking of that but this only proves that $\sum_{n=1}^{\infty}\left | x_n+\frac{r}{2} \right |\leq \sum_{n=1}^{\infty} \frac{r}{2} $ not that it not converge. WHat if most of the terms of $x_n+\frac{r}{2} $ become small? we can't guarante what will happen then. – ipreferpi Mar 05 '20 at 02:35
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    In order for convergence of the sum to occur, we must have the terms approach $0$. If $\sum |x_n + r/2|$ converges, then $x_n \to -r/2$ and hence $|x_n| \to r/2 \neq 0$, which would apply $\sum |x_n|$ does not converge. – user744868 Mar 05 '20 at 02:37
  • You're right, thank you – ipreferpi Mar 05 '20 at 02:41

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