I was wondering if there are commutative Banach rings $(A, \|\!\cdot\!\|)$ with unit which contain an ideal $\mathfrak a \subseteq A$ such that $\mathfrak a$ is not closed w.r.t. to the topology induced by the norm $\|\!\cdot\!\|$ on $A$. I haven't been able to find an example, but I guess there should be one.
Can somebody help me out or has an example in mind? Thanks a lot!
Certainly a Banach algebra can contain a non-closed ideal. Of course a maximal ideal in a Banach algebra with identity must be closed, but that's only for maximal ideals.
Hmm. Say $A$ is the space of bounded sequences of reals, with pointwise operations and the sup norm. Let $I$ be the set of all sequences that are eventually zero (so $x\in I$ if there exists $N$ so $x_j=0$ for all $j>N$). Then $I$ is a non-closed ideal. (Exercise: In fact the closure of $I$ is $c_0$, the set of all $x$ such that $x_j\to0$ as $j\to\infty$.)
Or similarly, say $A=L^1(\mathbb R)$ with convolution for multiplication, and let $I$ be the set of $f$ such that there exists $\delta>0$ with $\hat f=0$ in $(-\delta,\delta)$.
