The set $\{x=(x_n)\in\ell^2:\sum_n|\sin x_n|\lt\infty\}$ is dense in $\ell^2$

Question

I want to show that the set $\{x=(x_n)\in\ell^2:\sum_n|\sin x_n|\lt\infty\}$ is dense in $\ell^2$.

So we have to show for given $\epsilon\gt0$ and any $a\in\ell^2$, there is $x=(x_n)\in\ell^2$ such that

$$||x-a||_2\lt\epsilon\text{ with }\sum_n|\sin(x_n)|\lt\infty$$

I really don't know where to start, so any hint would be helpful. And I would also like to know whether there are any classifications of dense sets in $\ell^2$ and more generally in $\ell^p$ and how weird can a dense set be?

Answer 1

Your set is simply $\ell^1$: This is because if $(x_n) \in \ell^2$, then it converges to $0$. Near zero, i.e. for the highest indices we have $\sin(x_n) = O(x_n)$ which means it behaves the like the sequence itself.

Now let $y \in \{x=(x_n)\in\ell^2:\sum_n|\sin x_n|\lt\infty\} =: X$ be arbitrary. Then choose the sequence $(y^{(j)}) \subseteq \ell^1$ given by $$ y^{(j)}_n := \begin{cases} y_n , ~\text{ if }n \leq j \\ 0, ~\text{ if }n >j \end{cases}. $$ Now: $$ \lVert y_n^{(j)} - y \rVert_{\ell^2}^2 = \sum_{n = j+1}^\infty \lvert y_n \rvert^2 = \sum_{n = 1}^{\infty} \lvert y_n \rvert^2 - \sum_{n = 1}^j \lvert y_n \rvert^2 \overset{j \rightarrow \infty}{\longrightarrow} 0 $$ This proves density.

You could also simply observe that $X\supseteq c_{00} := \lbrace (x_n): x_n =0 \text{ for all but a finite amount of }n \rbrace$. And the latter is dense in $\ell^p$ ($1 \leq p < \infty$) because of the above argumentation. About weird dense sets I do not know a lot about, but they pretty much all behave similar to $c_{00}$.

Answer 2

Every element of $\ell^1\cap \ell^2$ is inside your set. Since $\ell^1\cap \ell^2$ is dense in $\ell^2$ (actually $\ell^1\subset \ell^2$ so the intersection is $\ell^1$), the result follows.