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Why is the ker-hull-topology on $Irr(A)$ is the discrete topology?

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 http://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

user62639