In my functional analysis class I was recently met with this in the context of C* algebras:

Let A be a C*-Algebra and B is a C*-subalgebra of A and I an ideal of A. We are asked to show that $ B+I $ is itself a C*-subalgebra of A (in particular it is closed).

I am stuck and cannot solve this. Can someone please kindly assist?


This is a corollary of the fact that images of $\ast$-homomorphism of $C^\ast$-algebras are closed.

More precisely, denote by $\pi$ the canonical projection $A\to A/I$. Then $\pi(B)\subset A/I$ is closed and so is $B+I=\pi^{-1}(\pi(B))\subset A$ as the preimage of a closed set under a continuous map.


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