a space has empty algebraic interior Can someone help me to continue? Let $p \in [1,\infty)$, $l_{p}^{+}=\{((x_{n})_{n \ge 1}) \in l_{p}|x_{n}\ge 0, \forall n\ge 1  \}$. I am trying so show that $l_{p}^{+}$ has empty algebraic interior. I thought to prove it like that, but I dont know if I am on the right way. Suppose that it would be non empty. So there exists $(a_{n})$ in the algebraic interior of $l_{p}^{+}$. Then the set $(l_{p}^{+})-(a _{n})$ is absorbing; so for every sequence $(x_{n})$ in $l_{p}$, there is $\delta \ge 0$, such as $\lambda (x_{n})_{n}+(a_{n})_{n}) \in l_{p}^{+}$, $\forall \lambda \in [0,\delta]$. Here I stucked...
 A: Hello from 7 years in the future; I come bearing an answer to this long-forgotten question. The key is to show that given any sequence $(a_n)$ in $l_p^+$, there is a nonpositive sequence $(b_n)$ with asymptotically slower decay than $(a_n)$, yet which still manages to be in $l_p$. Then $(a_n)+\lambda(b_n)$ will land outside $l_p^+$ for all $\lambda >0$, which certifies $(a_n)$ is not in the algebraic interior of $l_p^+$, and therefore proves the algebraic interior is empty since $(a_n)$ was arbitrary.
So we fix a sequence $(a_n)$ in $l_p^+$ and show it is not in the algebraic interior of $l_p^+$. Note that since $a_n$ is in $l_p$ and $p<\infty$, we know that $a_n \rightarrow 0$ as $n \rightarrow \infty$. Thus, for each natural $k$, there is a number $N_k$ so large that $a_n < 2^{-k}/k$ whenever $n \geq N_k$. Thus, we can choose $M_1 = N_1$, and for each $k \geq 2$, let $M_k = \max(N_k,M_{k-1}+1)$. This trick just gives us a strictly increasing sequence of naturals $(M_k)$ which satisfies $a_{M_k} < 2^{-k}/k$.
Now, for each natural $k$, we define $b_{M_k} = -2^{-k}$, and if $m$ is not equal to $M_k$ for any $k$, we let $b_m = 0$. We note that $(b_n)$ is in $l_p$ (as $(-2^{-n})$ is in $l_p$, and $b_m$ is this same sequence with a bunch of zeros interspersed in it), so we are done if we show that $(a_n)+\lambda(b_n)$ is not in $l_p^+$ for any $\lambda>0$.
So we fix $\lambda>0$; it suffices to show there is an $n$ such that $a_n+\lambda b_n < 0$. Since $\lambda>0$, we can choose $k$ so large that $1/k < \lambda$; then we let $n = M_k$. We now have
$$a_n + \lambda b_n = a_{M_k} + \lambda b_{M_k} = a_{M_k} - \lambda 2^{-k} < 2^{-k}/k - \lambda 2^{-k} < \lambda 2^{-k}-\lambda 2^{-k} = 0,$$
which is what we wanted.
