Tracial topological rank of hereditary

Question

Let $A$ be a unital simple $C^*$-algebra with $TR(A) \leq k$ (tracial topological rank). I can prove that for any unital hereditary $C^*$-subalgebra $B$ of $A$, $TR(B) \leq k$.

If $TR(A) \nleq k - 1$, when we say $TR(A) = k$. So, let $A$ be as above with $TR(A) = 0$, is it true to say that $TR(B) = 0$ for every hereditary $C^*$-subalgebra $B$ of $A$?

Answer 1

Since $TR(A) = 0 $, we have $RR(A) = 0 $ (real rank of $A$). Also, for any hereditary $C^*$-subalgebra $B$ of $A$, we have $RR(B) = 0$.(Automatically, $B$ is unital). We know that for a unital simple $C^*$-algebra with $TR(A) \leq k$, we have $TR(B) \leq k$ for any unital hereditary $C^*$-subalgebra $B$ of $A$.