# A sequence of points in the spectrum of a subhomogeneous C$^{*}$-algebra can converge to at most finitely many points

Let $$A$$ be a subhomogeneous C$$^{*}$$-algebra (i.e., there is a finite upper bound on the size of the irreducible representations of $$A$$). Let $$\hat{A}$$ denote its spectrum. I heard of a result that states that

If $$(\pi_{n})$$ is a sequence in $$\hat{A}$$, then $$(\pi_{n})$$ can converge to at most finitely many points.

I'd like to know why this statement is true.

For instance, if we look at the algebra

$$B:=\left\{f\in C([0,1],M_{2}(\mathbb{C})):\exists \lambda,\mu,\nu\in\mathbb{C}\text{ with }f(0)=\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}\text{ and }f(1)=\begin{pmatrix}\mu & 0\\ 0 & \nu\end{pmatrix}\right\}$$ it is clear that there is a sequence of $$2\times 2$$ irreducible representations converging to two $$1\times 1$$ irreducible representations (the point evaluations converging to either the upper-left or bottom-right evaluation at $$1$$). But there is not sequence of $$2\times 2$$ irreducible representations converging to $$\lambda,\mu$$, and $$\nu$$, which makes sense to me because we cannot stick all three of them down the diagonal, as we are in $$M_{2}(\mathbb{C})$$.

If $$A$$ is homogeneous, then $$\hat{A}$$ is Hausdorff, so the sequence can converge to at most one point in this case. I guess, in the subhomogeneous case the result has something to do with the fact that there are only finitely many choices for the dimensions of the irreducible representations.

I also know that each subspace $$\hat{A}_{n}:=\{\pi\in\hat{A}:\dim\pi=n\}$$ is Hausdorff, but I'm not sure how to put this together. Any help is appreciated.

See JMG Fell, The Dual Spaces of C$$^*$$-Algebras, Trans. Amer. Math. Soc. 94 (1960), 365-403, Corollary 1 on p. 388:
Let $$A$$ be a C$$^*$$-algebra with dual space $$\hat{A}$$. Let $$T^i$$ be a net of elements of $$\hat{A}$$, all of dimension equal to or less than the integer $$n$$; and let $$S^1, \ldots S^r$$ be distinct elements of $$\hat{A}$$ such that $$T^i\to_i S^k$$ for each $$k$$. Then $$\sum_{k=1}^r \dim S^k\le n.$$