Prove that this subset of $l^{2}$ is open

Question

The cause of my problem is this exercise:

Let $(a_n)_n$ be a sequence of strictly positive numbers and consider $$A = \{x = (x_n)_n\in l^2 : |x_n|<a_n \,\,\forall n\in\mathbb{N} \}.$$

Prove that $A$ is open if and only if $\inf_n \,a_n>0$.

I don't know where to start... Can anyone give me some suggestions (other than a simple "use the definition") to solve the exercise? It was given to us in functional analysis class, after having seen the countless consequences of the Hahn-Banach and Banach-Steinhaus theorems.

Answer 1

Let $a_0 = \inf_n a_n$.

If $a_0 > 0$, we claim that given $x = (x_n)_n \in A$, $\inf_n (a_n - |x_n|) = \delta >0$: Otherwise assume $\delta = 0$. Considering $|x_n| < a_n$ for every finite $n$, the only possible case is that there is a subsequence $(x_{n_i})_i$ of $(x_n)_n$ such that $\lim_{i \rightarrow +\infty}(a_{n_i}-|x_{n_i}|)=0$. Since $x \in l^2$, we have $\lim_{n \rightarrow +\infty}x_n=0$, so $\lim_{i \rightarrow +\infty}a_{n_i}=0$, contradicting $a_0 > 0$. Then you can easily prove that the open ball $B(x,\delta)=\{y \in l^2|\lVert y-x \rVert < \delta\}$ is contained in $A$. Thus $A$ is open.

For the other direction, if $A$ is open but $a_0=0$, then for any given $r>0$ there exists $a_i<r$, so $re_i \notin A$ ($e_i$ is the $i$-th unit vector), which means $B(0,r) \nsubseteq A \; \forall r>0$, contradicting $0 \in A$ and $A$ is open.

Answer 2

For a bounded linear operator $T$ and $r>0$ the set $$\{x\in \ell^2\,:\, \|Tx\|_2<r\}$$ is open in $\ell^2.$ Let $P_n$ denote the projection $$P_n(x_1,x_2,\ldots, x_k,\ldots )=(x_1,x_2,\ldots, x_n, 0,0,\ldots )$$ Assume $a:=\inf_ka_k>0.$ Then $$A_n=\{x\in \ell^2\,:\, |x_1|< a_1,|x_2|<a_2,\ldots, |x_n|<a_n, \, \|(I-P_n)x\|_2<a\}$$ is a subset of $A.$ Moreover $A_n$ is open as an intersection of $n+1$ open sets in $\ell^2.$

We have $$A=\bigcup_{n=1}^\infty A_n$$ as for every $x\in \ell^2$ there holds $\|(I-P_n)x\|_2\to 0.$ Hence $A$ is open in $\ell^2.$

If $a=0$ then $A$ has empty interior. Indeed, for $x\in A$ the set $x+\{y\in \ell^2\,:\,\|y\|_2\le r\}$ is not contained in $A$ for any $r>0.$ Indeed for fixed $r>0$ there exists $n$ such that $2a_n<r.$ Then $|x_n+r|>r-a_n>a_n.$