Can anyone give me a relatively simple proof or Some reference for the following fact.(I know that there is a proof of this theorem in Gerard J. Murphy'book: "$C^\*$-Algebras and Operator Theory", but I'm sure that there should be a simple proof of this.


**Every hereditary C\*-subalgebra of a simple $C^\*$-algebra is also simple!**


Maybe this is easy for someone, but it makes me confused for a long time. I am a novice!



Question about hereditary $C^*$-algebra
I don't know much about C\* algebras, but I would guess that there shouldn't be a simple proof of the theorem you state. Here's an algebraic example which seems to be a counterexample to the theorem once the modifier C\* is removed from the statement. Let A be the Weyl algebra, i.e. the algebra generated by x,y with the relation xy-yx=1. It isn't hard to show this algebra is simple and that it has global dimension 1. The subalgebra generated by x also has global dimension 1, but of course it is far from simple.


