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If $\mathcal{A}$ is a unital $C^*$-algebra then I want to show that $\operatorname{conv}(\operatorname{Proj}(\mathcal{A}))=\mathcal{P}_1(\mathcal{A}):= \lbrace x \in \mathcal{A^+} : \Vert x \Vert \leq 1 \rbrace$ when $\mathcal{A}= M_n(\mathbb{C})$ for some $n \geq 2$.

Previously I have shown that the set of extreme points of $\mathcal{P}_1(\mathcal{A})$ is equal to $\operatorname{Proj}(\mathcal{A})$. Thus I can write $\operatorname{Ext}(\mathcal{P}_1(\mathcal{A})) = \operatorname{Proj}(\mathcal{A})$ which then implies that

$$\operatorname{conv}(\operatorname{Proj}(\mathcal{A}))= \operatorname{conv} (\operatorname{Ext}(\mathcal{P}_1(\mathcal{A}))$$

Where $\operatorname{conv}$ is the convex hull. By the Krein-Milman Theorem, I then have that

$$\overline{\operatorname{conv}(\operatorname{Ext}(\mathcal{P}_1(\mathcal{A}))}= \mathcal{P}_1(\mathcal{A})$$

So now I want to proceed from here. If $\operatorname{conv}(\operatorname{Ext}(\mathcal{P}_1(\mathcal{A}))$ is a closed set then it coincides with the closure (right?) but I am not sure how to prove this.

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Because you are in finite dimension and $\operatorname{Proj}\mathcal A$ is closed and bounded, it is compact. Caratheodory's Theorem then guarantees that you can use at most $n^2+1$ terms in your convex combinations; a consequence of that is that $\operatorname{conv}(\operatorname{Proj}\mathcal A)$ is closed

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