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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?

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

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