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Show that in the normed vector space $\mathcal l_1$ (vector space of real convergent sequences under the norm $||r_k||_1= \sum |r_k|$), the set $P=\{\{x_k\} \mid x_k \ge 0 \; \forall k \in \Bbb N\}$ has an empty interior.

My proof so far goes like this: Assume that $\exists \{x_k\} \in \mathcal l_1 \text { such that } \{x_k\} \in int(P)$, we have then that $\{x_k\} \in P$ and that $\exists B(\{x_k\},\epsilon) \subset P$ (by the definition of interior i'm working with), so I know I have to find a sequence $\{y_k\} \in B(\{x_k\},\epsilon)$ with negative terms (a sequence "close enough" to $\{x_n\}$ that is not in $P$, leading to a contradiction). But I can't seem to intuitively find a sequence with those propierties, maybe I'm just taking the long road and it's simplier than that, am I missing something here?

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    Every sequence $\{r_k\}_{k=1}^{\infty}$ in your vector space is assumed to satisfy $\sum_{k=1}^{\infty} |r_k| < \infty$, so its terms must get arbitrarily close to 0. – 2017-01-22

2 Answers 2

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The thing that you are missing is the fact that

$$\sum_{i=1}^\infty |x_k| <\infty$$

implies that $x_k \to 0$ as $k\to \infty$. So for all $\epsilon$, there is $k=k(\epsilon)$ so that $x_k<\epsilon$. Then let

$$y^\epsilon_k = \begin{cases} -\epsilon &\text{if }k = k(\epsilon), \\ x_k &\text{otherwise.}\end{cases}$$

This $y^\epsilon$ is not in $P$ and $\| \{x_k\} - y^\epsilon\| <2\epsilon$

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You need to use the fact that you're working with convergent sequences:

Suppose there is some $(x_k)$ in the interior of $P$. That means that for some $\varepsilon>0$ we have $(y_k)\in P$ for every $(y_k)\in \mathcal{l}_1$ such that $||(x_k)-(y_k)||_1<\varepsilon$. However, since we must have $x_k\to0$, there is some natural number $K$ such that $x_K<\varepsilon/2$. This leads to a contradiction considering the sequence $(z_k)$ defined by $z_k:=x_k$ if $k\neq K$, and $z_K:=-\varepsilon/2$.