Is the following set closed in $\ell_{p}$ for $1\le p$?

Solution 1:

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.