diff --git a/src/op/vn/commutative.tex b/src/op/vn/commutative.tex
index d6ebb8d..743e9ed 100644
--- a/src/op/vn/commutative.tex
+++ b/src/op/vn/commutative.tex
@@ -41,7 +41,7 @@
     Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then $A$ admits a separating vector. 
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Theorem 22.3]{Zhu}}}. ]
-    Let $\seqj{x} \subset H$ be a maximal collection of non-zero vectors such that the spaces $\bracsn{Ax_j|j \in I}$ are mutually orthogonal. Such a collection exists by Zorn's lemma, and must be at most countable given that $H$ is separable. 
+    Let $\seqj{x} \subset H$ be a maximal collection of unit vectors such that the spaces $\bracsn{Ax_j|j \in J}$ are mutually orthogonal. Such a collection exists by Zorn's lemma, and must be at most countable given that $H$ is separable. 
 
     Let $\seq{x_n}$ be an enumeration of such a set, padding by zeroes if necessary, and $x = \sum_{n \in \natp}x_n/2^n$. For any $T \in A$, if $Tx = 0$, then as $\bracsn{Ax_n|n \in \natp}$ are mutually orthogonal, $Tx_n = 0$ for all $n \in \natp$. Since $A$ is commutative, $Ax_n \subset \ker(T)$ for all $n \in \natp$. By maximality of $\seq{x_n}$, $H = [l^2(\natp); \ol{Ax_n}]$, $H \subset \ker(T)$, and $T = 0$. Therefore $x$ is a separating vector. 
 \end{proof}
diff --git a/src/op/vn/topologies.tex b/src/op/vn/topologies.tex
index 6d97ecf..b27c879 100644
--- a/src/op/vn/topologies.tex
+++ b/src/op/vn/topologies.tex
@@ -3,7 +3,7 @@
 
 Let $H$ be a complex Hilbert space. Thanks to its self-duality, there is a natural dual pairing
 \[
-B(H) \times (H \otimes H) \to \complex \quad \dpn{T, \phi \otimes x}{B(H)} =  \dpn{Tx, \phi}{H}
+B(H) \times (H \otimes \ol H) \to \complex \quad \dpn{T, \phi \otimes x}{B(H)} =  \dpn{Tx, \phi}{H}
 \]
 
 Depending on the topology placed on $H \otimes H$, and the corresponding completion, a handful of different topologies arise on $B(H)$. In fact, the above duality produces a predual for $B(H)$, being the trace class operators: 
@@ -111,7 +111,7 @@ Now, a few facts about the more familiar operator topologies:
 
 \begin{proposition}
 \label{proposition:spatial-isomorphism-sot-continuous}
-    Let $A$ be a $C^*$-algebra, $H_1, H_2$ be a complex Hilbert spaces, $\pi_1: A \to B(H_1)$ and $\pi_2: A \to B(H_2)$ be injective representations of $A$, and $U: H_1 \to H_2$ be an unitary equivalence, then the mapping
+    Let $A$ be a $C^*$-algebra, $H_1, H_2$ be complex Hilbert spaces, $\pi_1: A \to B(H_1)$ and $\pi_2: A \to B(H_2)$ be injective representations of $A$, and $U: H_1 \to H_2$ be a unitary equivalence, then the mapping
     \[
     \pi_1(A) \to \pi_2(A) \quad T \mapsto UTU^{-1}
     \]
diff --git a/src/op/vn/type-decomp.tex b/src/op/vn/type-decomp.tex
index f8e10c4..49240a1 100644
--- a/src/op/vn/type-decomp.tex
+++ b/src/op/vn/type-decomp.tex
@@ -25,7 +25,7 @@
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $Q \in \text{Proj}(A)$ with $Q \le P$, then $Q = PQP$. As $PAP = \complex P$, either $PQP = 0$ or $PQP = P$. 
 
-    (2) $\Rightarrow$ (1): Given that there exists no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By \autoref{theorem:vn-projection-norm-dense}, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$. 
+    (2) $\Rightarrow$ (1): Given that there exist no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By \autoref{theorem:vn-projection-norm-dense}, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$. 
 \end{proof}
 
 \begin{lemma}
@@ -277,7 +277,7 @@
         &= \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot v \in l^2(I; \complex v)
     \end{align*}
 
-    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
+    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 $v \in H_0$ with $\norm{v}_{H_0} = 1$, the mapping
     \[
     \pi: A \to B(l^2(I; \complex v)) \quad T \mapsto UTU^{-1}|_{l^2(I; \complex v)}
     \]