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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?

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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).

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  • $\begingroup$ Thank you for answering this correctly. My answer was incorrect because of a confusion about the definition of "minimal projection" in a C*-algebra. (I thought the standard definition was $pAp = \mathbb{C}$, but that evidently is not what is meant in the reference I cited.) $\endgroup$
    – Nik Weaver
    Commented Oct 15, 2020 at 1:03
  • $\begingroup$ If you define elementary as $K(H)$ for some (non-zero, potentially non-separable) Hilbert space $H$, then the result also holds for non-separable $C^\ast$-algebras. In fact, if $A$ is simple and contains a finite dimensional hereditary $C^\ast$-subalgebra (which is necessarily a matrix algebra and a full hereditary $C^\ast$-subalgebra) then $A$ is strongly Morita equivalent to a matrix algebra (and thus $\mathbb C$) and hence $A \cong K(H)$ for some (not necessarily separable) Hilbert space. $\endgroup$
    – Jamie Gabe
    Commented Oct 15, 2020 at 9:27

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