Proof, [???]. (1) $\Rightarrow$ (2): Since $A$ is of type $\vnI$, $A$ admits a non-zero abelian projection $Q \in \text{Proj}(A)$. By Lemma 37.8.12, $Q$ is minimal.

(2) $\Rightarrow$ (3): Let $P \in \text{Proj}(A)$ and $Q \in \text{Proj}(A)$ be a minimal projection. By the comparability theorem, either $P \preceq Q$ or $Q \preceq P$.

If $P \preceq Q$, then there exists $R \in \text{Proj}(A)$ with $P \sim R \le Q$. In which case, $R$ is minimal, and $P$ is also minimal by (5) of Lemma 37.8.4.

If $Q \preceq P$, then there exists $R \in \text{Proj}(A)$ with $Q \sim R \le P$. By (5) of Lemma 37.8.4, $R$ is minimal with $R \le P$.

(3) $\Rightarrow$ (4): By Zorn’s lemma, there exists a maximal orthogonal family $\seqi{P}\subset \text{Proj}(A)$ of minimal projections. Since $\seqi{P}$ is maximal and every non-zero projection admits a non-zero minimal subprojection, $I = \sum_{i \in I}P_{i}$, and $H = \bigoplus_{i \in I}P_{i}H$.

For each $i, j \in I$, by the comparability theorem, either $P_{i} \preceq P_{j}$ or $P_{j} \preceq P_{i}$. In both cases, since both projections are minimal, $P_{i} \sim P_{j}$. Thus there exists a partial isometry $V_{i, j}\in A$ with initial space $P_{i}(H)$ and final space $P_{j}(H)$ such that $P_{i} = V_{i, j}^{*}V_{i, j}$ and $P_{j} = V_{i, j}V_{i, j}^{*}$. By fixing a particular family^[1], assume without loss of generality that $V_{i, j}^{*} = V_{j, i}$ for all $i, j \in I$.

Fix $i_{0} \in I$, let $H_{0} = P_{i_0}H$, and define

then since $H = \bigoplus_{i \in I}P_{i}H$ and each $V_{i, i_0}$ is a partial isometry, $U$ is an isometry with inverse

For each $i \in I$, denote $e_{i} = \one_{\bracs{i}}\in l^{2}(I; \complex)$, then for every $x \in l^{2}(I; H_{0})$,

so $UP_{i}U^{-1}$ is the projection onto the $i$-th component of $l^{2}(I; H_{0})$.

Let $T \in A$, then by Lemma 37.8.13, there exists $\bracsn{\mu_{i, j}}_{i, j \in I}\subset \complex$ such that $T = \sum_{i, j \in I}\mu_{i,j}V_{i, j}$. In which case, for any $x \in l^{2}(I; \complex)$ and $v \in H_{0}$,

For each $i, j \in I$, there exists $\lambda_{i, j}\in \partial B_{\complex}(0, 1)$ such that $V_{i, j}V_{i_0, i}= \lambda_{i, j}V_{i_0, j}$ by (6) of Lemma 37.8.4. For every $i \in I$, let $e_{i} = \one_{\bracs{i}}\in l^{2}(I; \complex)$, then

Moreover, if $v \ne 0$, then $UTU^{-1}(xv) = 0$ for all $x \in l^{2}(I; \complex)$ implies that $\mu_{i, j}=0$ for all $i, j \in I$, and $T = 0$. Thus for any $v \in H_{0} \setminus \bracs{0}$, the mapping

is an injective $*$-homomorphism.

Finally, since $\ol{B_A(0, 1)}$ is weak-operator compact and $U$ is an isometry, $\pi(\ol{B_A(0, 1)})$ is also weak-operator compact, and $\pi(A) \subset B(l^{2}(I; \complex v))$ is a von Neumann algebra. Let $i, j \in I$, then for each $x \in l^{2}(I; \complex)$,

so $\pi(V_{i, j}) = \lambda_{i, j}e_{j}v \otimes e_{i}v$. As $\bracsn{e_iv}_{i \in I}$ is an orthonormal basis for $l^{2}(I; \complex v)$, $\pi(A) = B(l^{2}(I; \complex v))$.

(4) $\Rightarrow$ (1): Identify $A = \pi(A) = B(K)$, and let $T \in Z(A)$. For each $v \in K$ with $\norm{v}_{K} = 1$, $T(v \otimes v) = (v \otimes v)T$, so $Tv = T(v \otimes v)v = (v \otimes v)Tv$, and there exists $\lambda \in \complex$ such that $Tv = \lambda v$.

For any $w \in K$ linearly independent from $v$, there exists $\mu \in \complex$ with $Tw = \mu w$, and $\rho \in \complex$ with $T(v + w) = \rho(v + w)$. In which case, $\rho v + \rho w = \lambda v + \mu w$, so $\lambda = \rho = \mu$, and $T = \lambda I$. Therefore $A$ is a factor.

For any $v \in K$ with $\norm{v}_{K} = 1$, $v \otimes v$ is a minimal, and hence abelian projection by (1) of Lemma 37.8.4. As $v \otimes v \le I$, $A = B(K)$ is of type $\vnI$.$\square$