Is the set $A$ open in $\ell^2$ Let $A=\{(a_n)\in\ell^2:|a_n|&lt;1\;\forall n\}$. Is the set $A$ open in $\ell^2=\{(a_n):\sum|a_n|^2&lt;\infty\}$? 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 $||a||=\left(\sum|a_n|^2\right)^{1/2}$?
A set $A$ is open iff for every point $a \in A$ exists $\epsilon &gt; 0$ 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 $\epsilon &gt; 0$ such that if $\| x - a \| &lt; \epsilon$, 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 $M &lt; 1$. Choose $$ \epsilon = (1 - M)/2 .$$

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