# Is the set of points in the irreducible decompositions of this C$^{*}$ -algebra's representations closed?

Suppose $$X$$ and $$Y$$ are compact Hausdorff spaces. Let $$\varphi\colon C(X)\to M_{n}(C(Y))$$ be any $$*$$-homomorphism. If $$\pi$$ is an irreducible representation of $$M_{n}(C(Y))$$, then $$\pi$$ is unitarily equivalent to a point evaluation $$\textrm{ev}_{y}$$. The $$*$$-homomorphism $$\textrm{ev}_{y}\circ\varphi\colon C(X)\to M_{n}(\mathbb{C})$$ is a representation of $$C(X)$$. Since it's a finite-dimensional representation, we can find a unitary $$u_{y}\in M_{n}(\mathbb{C})$$ and a set of points $$X_{y}=\{x^{y}_{1},\ldots,x^{y}_{n}\}\subset X$$ such that for all $$f\in C(X)$$, $$(\varphi\circ f)(y)=(\textrm{ev}_{y}\circ\varphi)(f)=u_{y} \begin{pmatrix} f(x^{y}_{1}) & 0 & \cdots & 0\\ 0 & f(x^{y}_{2}) & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & f(x^{y}_{n}) \end{pmatrix} u_{y}^{*}.$$

My question is:

Is the set $$\widetilde{X}:=\bigcup_{y\in Y}X_{y}$$ closed in $$X$$?

This question is in a similar vein to one of my earlier questions: Closeness of points in the irreducible decomposition of a C$$^{*}$$-algebra representation