Let $B$ be a closed subalgebra of a $C^{\ast}$-algebra $A$.
Is it true that every closed ideal of $B$ is of the form $I\cap B$ for some closed ideal $I$ of $A$.
Let $i$ be natural inclusion from $B$ to $A$. Let $J$ be a closed ideal of $A$. Set, $I=\overline{i(J)}$, closure of $i(J)$ in $B$. Does this works?
No, $A$ can be simple and the subalgebra $B$ not simple.