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what's wrong about following proof? It's obviously wrong as we never used finite dimension of $\mathcal{H}$...

Lemma: Let $A$ be C*-algebra. Let $\pi: A \to \mathcal{B(H)}$ be irreducible *-representation and $\text{dim}\mathcal{H}=n\in \mathbb{N}$. Then $\pi$ is surjective.

(wrong) proof: The commutant $\pi(A)'$ of $\pi(A)$ is equal to $\mathbb{C}\cdot I$ (property of irreducible representations. $I$ is identity operator). Hence $\pi(A)''=\mathcal{B(H)}$. From Von neumann double commutant theorem we get $\overline{\pi(A)}=\pi(A)''=\mathcal{B(H)}$. But $\pi(A)$ is closed (as homomorphic image of *-algebra isomorphism) in $\mathcal{B(H)}$, hence $\pi(A)=\mathcal{B(H)}$.

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    $\begingroup$ I'm not familiar with the theory anymore, but why is $\pi(A)$ closed? I don't get that part. On the other hand, I believe that in general any finite-dimensional subspace of a normed space is closed, that solves it. $\endgroup$ – Mathematician 42 Apr 27 '18 at 9:22
  • $\begingroup$ @Mathematician42 This is because $\pi(A)=\tilde\pi(A/\ker\pi)$, where $\tilde\pi$ is the induced map on $A/\ker\pi$. But $\tilde\pi$ is injective, hence isometric, hence it's image is complete, and therefore closed. $\endgroup$ – Aweygan Apr 27 '18 at 10:43
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In your second-to-last and last sentences, you're dealing with two different topologies of $\mathcal{B(H)}$. From the von Neumann double commutant theorem we obtain that $\overline{\pi(A)}^\text{SOT}=\pi(A)''=\mathcal{B(H)}$. Now since $\pi$ is a $*$-homomorphism, $\pi(A)$ is closed in the norm-topology. The way the argument is phrased, it seems to imply that closure in the norm topology implies closure in the strong operator topology, which is simply not true.

Nevertheless, in finite-dimensional spaces all vector topologies are the same, which implies the result. Hence I would not say the proof is wrong, but I would say it is incomplete.

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