Is the following set closed in $\ell_{p}$ for $1\le p$? 
Let $S=\left\{ \left\{a_n\right\}_{n=1}^\infty\in\ell_{p}\mid
 \sum_{n=1}^\infty {a_n}=1, a_n\ge0 \ \forall \ n\in\Bbb N \right\}$
Is the set $S$ closed in the normed space
  $\left(\ell_{p},\|\cdot\|_{p}\right)$ for $1\le p$ ?

I was able to conclude that S is actually closed in $\left(\ell_{1},\|\cdot\|_{1}\right)$ by proceeding the following way:  
Let $\left\{y_{n}\right\}_{n=1}^\infty$ be a sequence (of sequences) such that $y_n=\left\{y_{k}^{n}\right\}_{k=1}^\infty\in S\ \forall \ n\in\Bbb N$ and $y_{n}\rightarrow x$ as $n\rightarrow\infty$, where $x=\left\{x_k\right\}_{k=1}^\infty\in \ell_{1}$
If we prove that $x\in S$, then S would be closed. Given any fixed $k\in\Bbb N$, we can deduce from the convergence of $y_{n}$ to $x$ as $n\rightarrow\infty$ that $y_{k}^{n}\rightarrow x_k$ as $n\rightarrow\infty$. Since $y_{k}^{n}\ge0\  \forall \ k\in\Bbb N$ then $x_k\ge0$.  
Therefore $x_k\ge0\ \forall \ k\in\Bbb N$.
Now, since $y_n\rightarrow x$ as $n\rightarrow\infty$ and $\|y_n\|_{1}=1\ \forall \ n\in\Bbb N$, by continuity of the norm we obtain that $1=\|y_n\|_{1}\rightarrow \|x\|_{1}$ as $n\rightarrow\infty$. Then, $\|x\|_{1}=1$, which implies that $\|x\|_{1}=\sum_{k=1}^\infty {|x_k|}=\sum_{k=1}^\infty {x_k}=1$.
Therefore $x\in S$.
However, I haven't been able to prove or disprove the closedness of $S$ for the case in which $p\gt1$. I was trying to follow the same proof, but I can't conclude that $\sum_{k=1}^\infty {x_k}=1$ (at most I can conclude that $\sum_{k=1}^\infty {x_k^{p}}=1$ by continuity of the norm in $\ell_{p}$). Since the proof failed I started to think that the set might not be closed for the case in which $p\gt1$, but I haven't been to prove that it isn't closed either.
Any hints or ideas would be highly appreciated.
 A: For $p>1$, the set is not closed. To see this, note that
$$
x_n = (1/n, \dots ,1/n,0\dots)
$$
is an element of $S$ (the number $1/n$ appears $n$ times). 
Now, we have
$$
\Vert x_n \Vert_{\ell^p} = 1/n \cdot n^{1/p} = n ^{1/p -1}\to 0,
$$
for $p>1$. But $0$ is not an element of the set. 
EDIT: Here is more on the general principle involved: Your set is defined using the linear functional
$$
\Phi : (x_n)\mapsto \sum x_n =\sum 1\cdot x_n. 
$$
Then, it is not hard to see that this functional is bounded (continuous) on $\ell^1$, but not on $\ell^p$ (one way to see this is that the constant sequence $(1,\dots)$ is in $\ell^\infty$, but in no other $\ell^q$ space). In fact, $\Phi$ is not even well defined on $\ell^p$ for $p>1$. 
Now it is generally true that if you have a continuous function $f$, then the set $f^{-1}(A)$ is closed if $A$ is closed itself. For linear functionals on Banach spaces, most of the time the converse is also True, i.e. a set defined through "closed conditions" on a linear functional will most of the time not be closed if the functional is not continuous. Of course, this is only a heuristic, but the proof above shows that this heuristic is correct in this case. 
