A circle algebra is an algebra of form $M_{n_1}(C(X_1))\oplus ...\oplus M_{n_k}(C(X_k))$ where $X_j$ are compact subsets of $S^1\simeq \{z\in \mathbb C:|z|=1\}$.
An $A\mathbb T$-algebra is an inductive limit of circle algebras.
Elliott theorem states that if $K$-groups of two unital simple $A\mathbb T$-algebras with real rank $0$ are isomorphic, then the two algebras are isomorphic. I want to see why is real rank $=0$ necessary here, but I failed to find a non-real-rank-$0$ unital simple $A\mathbb T$-algebra.
The only way I know to construct a simple $A\mathbb T$ algebra is:
step1. choose an arc $X\subset S^1$ with $\{x_1,...\}$ dense in $X$
step2. for $a\in C(X)$ let $\pi_i(a)=a(x_1)1_{C(X)}\oplus ...\oplus a(x_i)1_{C(X)}\in M_i(C(X))$
step3. define (unital homomorphism) $h_i:A_i\to A_{i+1}$ such that $h_i(a)=a\oplus (\pi_i\otimes 1_{M_{l(i)}})(a)$, where $A_i=M_{l(i)}(C(X))$ and $A_{i+1}=M_{(i+1)\cdot l(i)}(C(X))$
Then $A=\lim_n A_n$ is a simple unital $A\mathbb T$ algebra. But it seems that such $A$ has real rank $0$.