Show that a set is not closed in $l_p(\mathbb{N})$

Question

I have to figure out if $F = \{ (x_n)_n \in l_p(\mathbb{N}) \; | \; \sum_{n = 0}^{\infty} x_n = 0 \}$ is a closed set in $l_p(\mathbb{N})$ or not. For $p=1$, I've shown that it's a closed set. But for $p>1$, I really think it's not the case, but I don't find a counter-example. Anyone could help me? Thank you !

Answer 1

Let $p>1$. Take the sequence $y=(-1, 0, 0, \dots )$. This sequence is in the complement of your set. If your set was closed, then there would exist a ball of radius $\varepsilon>0$ such that it does not contain any element of your set. Consider now the the following sequence

$$ y^N = (\underbrace{-1 + \frac{1}{N}, \frac{1}{N}, \dots, \frac{1}{N}}_{N \text{ times}}, 0, 0, \dots).$$ The $y^N$ are elements of your set and the $l^p$ distance is $$ \Vert y - y^N \Vert_{l^p} = \left( N \cdot \frac{1}{N^p} \right)^{1/p} = N^{(1-p)/p} \rightarrow 0, \quad \text{for } N\rightarrow \infty $$ This implies that your set is not closed.

Added: A nice way to show that $F$ is closed for $p=1$ is to note that the linear operator $$ T: l^1(\mathbb{N}) \rightarrow \mathbb{R}, \ (x_n)_{n\geq 1} \mapsto \sum_{n\geq 1} x_n$$ is continuous (as $\vert T((x_n)_{n\geq 1}) \vert \leq \Vert (x_n)_{n\geq 1} \Vert_{l^1}$) and thus $$ F = T^{-1}(\{0\}) $$ is closed.