Sequence of ideals in C*-algebra such that its intersection are not zero but intersection of dense sets are zero

Question

Let $$A_0 \supset A_1 \supset A_2 \supset A_3 \supset\ldots$$ be a sequence of $C^*$-algebras, such that $A_n$ is an ideal in $A_m$ (while $n>m$), and $K_i \subset A_i$ be a dense ${}^*$-subalgebra in $A_i$, such that $$K_0 \supset K_1 \supset K_2 \supset K_3 \supset\ldots$$ Is it possible, that $\bigcap_i K_i = 0$ but $\bigcap A_i \neq 0$? I know that it's possible if the condition "$A_i$ is a $C^*$-algebra" replaced by "$A_i$ is a Banach space", so I think it's possible, but I can't construct an example. Thanks!

Answer 1

Consider $A = C([0,1], \mathbb{C})$ and let every $A_i$ equal $A$. Let $f \in A$ be any strictly positive one-to-one function; for instance, $f(x) = 1+x$ would do. Let $K_i$ be the *-subalgebra of $A$ generated by the function $f^{2^i}$, that is, all functions of the form $$g(x) = a_1 f(x)^{2^i} + a_2 f(x)^{2 \cdot 2^i} + \dots + a_n f(x)^{n \cdot 2^i}, \quad a_1, \dots, a_n \in \mathbb{C}, \quad n \ge 1.$$ Clearly $K_i \supset K_{i+1}$. Also, since $f$ is one-to-one and positive, so is $f^{2^i}$. Thus $K_i$ separates points and vanishes nowhere so by Stone-Weierstrass $K_i$ is dense in $A$. Yet the intersection of all $K_i$ is just 0.