Let $A$ be a C$^\*$-algebra. Let $Irr(A)=\{[\pi]: \pi$ is an irreducible representation of A}, here is $\rho\in [\pi]$ if there is an unitary operator $V:H\_{\pi}\to H\_{\rho}$ such that $V\pi(a)=\rho(a)V$ for all $a\in A$.


If $A\subseteq K(H)$ is a C$^\*$-subalgebra, $Irr(A)$ is endowed with the discrete topology ($K(H)$ are the compact operators on a complex Hilbert space), then 
$$A\cong \bigoplus\_{[\pi]\in Irr(A)}K(H\_{\pi}).$$


A corollary: $A\subseteq K(H)$ is a $C^\*$-subalgebra, then the ker-hull-topology on $Irr(A)$ is the discrete topology. 


How exactly to conclude that the ker-hull-topology on $Irr(A)$ is always the discrete topology? Maybe this comes from that $Irr(A)$ as index set is discrete.


I don't think this should be difficult, but I don't see it. I asked on math-stackexchange <https://math.stackexchange.com/questions/1486623/ker-hull-topology-on-irra-is-the-discrete-topology-a-is-a-c-subalgebr> but the topic is maybe too specific. If the question is misplaced here, let me know this and I will delate this question. Regards



Why is the ker-hull-topology on $Irr(A)$ is the discrete topology?
I have an answer of my question: It is $$\hat{A}\cong\widehat{ \bigoplus\_{[\pi]\in Irr(A)}K(H\_{\pi})}\cong \coprod\_{[\pi]}\widehat{K(H\_{\pi})}
,$$ everything is endowed with the ker-hull-topology. Now, the $K(H\_{\pi})$ are simple, it follows $\widehat{K(H\_{\pi})}=\{[id\_\pi]\}$ and $\hat{A}$ is a disjoint union of onepoint-sets. It follows that the ker-hull-topology on $\hat{A}$ is the discrete topology.


