A is said to be elementary if A is isomorphic to some $K(H)$ or $M\_n$.


A C\*-subalgebra $B$ is said to be hereditary if for every $0≤a≤b∈B$ we have $a∈B$.


I wanted to know that is this statement true?


every hereditary C*-subalgebra of a non-elementary simple C*-algebra has infinite dimensions?


If so, could you help me to prove it?



hereditary C*-subalgebra of a non-elementary simple C*-algebra
This is true in the separable case (and more generally) and a consequence of Larry Brown's stable isomorphism theorem (1977 Pacific Journal of Math). A special case of his theorem states: If $A$ is a separable, simple C\*-algebra and $B$ is a hereditary subalgebra of $A$, then $A\otimes K(H)\cong B\otimes K(H).$ One could probably answer your question in the general case from the separable case.


For your question: Suppose $A$ is (separable) non-elementary and simple and $B$ is a hereditary subalgebra. Brown's theorem implies that $A\otimes K(H)\cong B\otimes K(H).$ If $B$ were finite dimensional it would have to be isomorphic to $M\_n$ (otherwise $A$ wouldn't be simple). This would imply that $A$ is stably isomorphic to $K(H)$ which is impossible (for example it would imply that $A$ is Type I, which it can't be because the only separable simple Type I C\*-algebras are the elementary ones).


