Is the set $A$ open in $\ell^2$

Question

Let $A=\{(a_n)\in\ell^2:|a_n|<1\;\forall n\}$. Is the set $A$ open in $\ell^2=\{(a_n):\sum|a_n|^2<\infty\}$? I think it might be, because it has the form of an open ball, but how to check openness with respect to $\ell^2$ norm $||a||=\left(\sum|a_n|^2\right)^{1/2}$?

Answer 1

It is open. Let $(a_n) \in A$. Since $\sum |a_n|^{2} <\infty$ there exists $m$ such that $|a_n| <\frac 1 2$ for all $n \geq m$. Let $\delta$ be the minimum of the numbers $\frac 1 2, 1-|a_1|,1-|a_2|,..., 1-|a_{m-1}|$. Now $\|(b_n)-(a_n) \|<\delta$ implies that $|b_n-a_n| <\delta$ for each $n$ from which it follows that $|b_n| <1$ for all $n$.

Answer 2

A set $A$ is open iff for every point $a \in A$ exists $\epsilon > 0$ such that $B(a, \epsilon) \subseteq A$, where $B(a, \epsilon)$ is the $\epsilon$-ball around $a$. I claim in this case that your $A$ is open, and I’ll show this by showing that for any $a \in A$, there exists $\epsilon > 0$ such that if $\| x - a \| < \epsilon$, then $x \in A$, i.e. $B(a, \epsilon) \subseteq A$.

Fix some $a \in A$. Since $(a_n)$ is square-summable, we know in particular that $a_n \to 0$, so the sequence $(|a_n|)$ attains a maximum $M < 1$. Choose $$ \epsilon = (1 - M)/2 .$$

If $\| x - a \| < \epsilon$, where $x \in \ell^2$, then \begin{align*} |x_n| & \leq |x_n - a_n| + |a_n| \\ & = \left( |x_n - a_n|^2 \right)^{1/2}+ |a_n| \\ & \leq \left( \sum_n |x_n - a_n|^2 \right)^{1/2} + M \\ & = \| x - a \| + M \\ & < \epsilon + M \\ & < 1. \end{align*} Thus $x \in A$. This proves that $A$ is open.

Answer 3

Denote $\|(a_n)\|_\infty = \sup |a_n|.$ Clearly $\|(a_n)\|_\infty \leq \|(a_n)\|$ and it is well-known that $\| \cdot \|_\infty$ is a norm. This implies that the identity is continuous $(\ell^2, \|\cdot\|) \to (\ell^2, \|\cdot\|_\infty),$ a fortiori the inverse image of the $\|\cdot\|_\infty$-ball of radius one, namely $\{(a_n); |a_n| < 1, \ \forall n\},$ is open relative to $\|\cdot\|.$