# Hints that $\{x \in \ell^{c}: \sum x_{n} = 0\}$ is dense in $\ell^{c}$

Note that $$\ell^{c}=\{x \in \ell^{\infty}: x_{n}=0 \operatorname{ for all but finitely many} n \in \mathbb N\}$$. Now show that:

$$M:=\{x \in \ell^{c}: \sum x_{n} = 0\}$$ is dense in $$\ell^{c}$$

My idea:

let $$x\in \ell^{c}$$ arbitrary, by definition $$\exists N \in \mathbb N:$$ $$\sum\limits_{n=N}^{\infty}x_{n}=0$$ and in the same sense there exists a $$k\leq N$$ so that $$\sum\limits_{n=k}^{N}x_{n}<\epsilon$$. Furthermore, we want to find a sequence $$x^{N}$$ (I assume it will have something to do with the point $$N$$ after which our sequence is $$0$$), so that $$\vert \vert x^{N}-x\vert\vert_{2}<\epsilon$$

But I am struggling to think of a sequence that satisfies $$\sum_{n\geq 1}x_{n}^{N}=0$$, in order for $$x^{N}\in M$$. Any hints, ideas?

• Are you using the $\ell^2$ norm or the $\ell^\infty$ norm? – JonathanZ Jun 3 at 22:26
• I assumed the $\ell^{2}$ norm – SABOY Jun 3 at 22:30

Let $$x \in \ell^{c}$$. Let $$N$$ be a positive integer and consider $$y=x+r_N(1,\frac 1 2,\frac 1 3,...,\frac 1 N,0,0...0)$$ where $$r_N=-\frac {\sum_n x_m} {1+\frac 1 2+\frac 1 3+...+\frac 1 N}$$. Then $$y \in M$$. Also $$r_N \to 0$$ as $$N \to \infty$$. Clearly, $$\|x-y\|^{2}=r_N^{2}(1+\frac 1 {2^{2}}+\frac 1 {3^{2}}+...+\frac 1 {N^{2}}) \to 0$$.
Alternatively yo can use the fact that if the orthogonal complement of a linear subspace of a Hilbert space is $$\{0\}$$ then the subspace is dense. You can now use the answer to your previous question Hint: Computing the orthogonal Complement of $M$ in $\ell^{2}$