I need help understanding the following:
If $A$ is a (complex) banach algebra and $I$ is a proper modular ideal then $\overline{I}$ is also proper.
Proof. Let $u\in A$ be such that $a-ua, a-au \in I$ for all $a \in A$. If $b\in I$ and $\|u-b\|<1$ then $v=1-u+b$ is invertible in $\widetilde{A}$. If $a \in A$ then $av = a-au +ab \in I$ so $A=Av \subset I$. This contradicts the assumption that $I$ is proper and shows that $\|u-b\|\ge 1$ for all $b \in I$. It follows that $u\notin \overline{I}$.
I have tried to make sense of this. Here are my thoughts:
Let $u \in A$ be such that $a-ua, a-au \in I$ for all $a$. Then $u \notin I$ because by assumption $I$ is proper. The goal now is to show that also $u \notin \overline{I}$. This is true if there exists an open ball around $u$ that does not contain any elements of $I$.
Note that $A$ is a subalgebra of its unitisation $\widetilde{A}$ (which is just the product $A \times \mathbb C$) and if $I$ is an ideal in $A$ then $I \times \{0\}$ is an ideal in $\widetilde{A}$. Furthermore, if the open ball $B((u,0),1)$ does not intersect $I \times \{0\}$ in $ \widetilde{A}$ then $B(u,1)$, the projection of $B((u,0),1)$ onto $A$, does not intersect $I$ in $A$.
Now note that if $\|(u,0)-(a,0)\|<1$ then $(0,1) - (u-a,0)$ is invertible in $\widetilde{A}$ hence the ideal generated by $(0,1) - (u-a,0)$ equals $\widetilde{A}$.
Now for the part that's causing me a headache:
If $a$ was an element of $I$ then for $(b,\mu) \in \widetilde{A}$: $$(b,\mu)(-u+a,1) = (-bu +ba + b -\mu u + \mu a, \mu) \notin I \times \{0\}$$
Here I'd want this to be in $I \times \{0\}$ to get the desired contradiction. Can somebody enlighten me please?