Open set in $\ell^2$ Let $a=(a_n)_n\subset(0,\infty)$ be a sequence and $S^{(a)}:=\{(x_n)_n\in\ell^2:\lvert x_n\rvert\  <a_n \forall n \}$.
I want to prove that $S^{(a)}$ is open in $\ell^2$ iff $\inf_{n\in\mathbb{N}} a_n>0$ but don't know where to  start.
 A: HINT: For the left to right direction it’s probably easier to show the contrapositive; suppose that $\inf_{n\in\Bbb N}a_n=0$, and show that $S^{(a)}$ is not open. For this you want to find an $x\in S^{(a)}$ such that no open ball centred at $x$ lies entirely inside $S^{(a)}$. In fact it turns out that you don’t have to search hard for such an $x$: any $x\in S^{(a)}$ will work. Specifically, let $x=\langle x_n:n\in\Bbb N\rangle\in S^{(a)}$, let $\epsilon>0$ be arbitrary, and show that there is a $y=\langle y_n:n\in\Bbb N\rangle\in \ell^2\setminus S^{(a)}$ such that $\|y-x\|_2<\epsilon$. I’ve left a further hint in the spoiler-protected block below.

 You can choose $y$ so that $y_n=x_n$ for all but one $n\in\Bbb N$, if you choose that one $n$ so that $a_n$ is small enough.

For the right to left direction, assume that $\inf_{n\in\Bbb N}a_n>0$, suppose that $x=\langle x_n:n\in\Bbb N\rangle\in S^{(a)}$; then $\sum_{n\in\Bbb N}x_n^2<\infty$, and $|x_n|<a_n$ for each $n\in\Bbb N$, and you want to show that some open nbhd of $x$ in $\ell^2$ is contained in $S^{(a)}$. You might as well try for a basic open nbhd: you want an $\epsilon>0$ such that $y\in S^{(a)}$ whenever $\|y-x\|_2<\epsilon$, i.e., whenever
$$\left(\sum_{n\ge 0}(y_n-x_n)^2\right)^{1/2}<\epsilon\;.\tag{1}$$
Note that if $(1)$ holds, then $|y_n-x_n|<\epsilon$ for each $n\in\Bbb N$. It will also help to remember that $\lim_{n\to\infty}|x_n|=0$.
