Added projections.

src/op/banach/definitions.tex

@@ -18,6 +18,17 @@
     For each $x \in A$, let $L_x \in L(A; A)$ be defined by $y \mapsto xy$, and let $\norm{x}_1 = \norm{L_x}_{L(A; A)}$, then $\norm{x}_1 \le \norm{x}_A$ and $\norm{1}_1 = 1$. On the other hand, $\frac{\norm{x}_A}{\norm{1}_A} \le \norm{x}_1$, so $\norm{\cdot}_1$ is equivalent to $\norm{\cdot}_A$.
 \end{proof}
 
+\begin{definition}[Centre]
+\label{definition:banach-algebra-centre}
+    Let $A$ be a Banach algebra, then
+    \[
+    Z(A) = \bracsn{x \in A|xy = yx \forall y \in A}
+    \]
+
+    is the \textbf{centre} of $A$. 
+\end{definition}
+
+
 \begin{definition}[Homomorphism]
 \label{definition:banach-algebra-homomorphism}
     Let $A, B$ be Banach algebras and $\phi: A \to B$, then $\phi$ is a \textbf{homomorphism} if:

src/op/banach/invertible.tex

@@ -9,7 +9,7 @@
 
 \begin{lemma}[Neumann Series]
 \label{lemma:neumann-series}
-    Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
+    Let $A$ be a unital Banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
     \[
     x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
     \]

src/op/example/bounded.tex

@@ -50,21 +50,34 @@
         \item $\ker P = \ker V$.
     \end{enumerate}
 
-    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$. 
+    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$, and
+    \begin{enumerate}[start=4]
+        \item $V$ is a partial isometry from $\ker(T)^\perp$ to $\ol{T(H)}$.
+        \item $P$ and $V$ are contained in the von Neumann algebra generated by $T$. 
+    \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 12.8]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 12.8, Theorem 18.9]{Zhu}}}. ]
-    Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). For each $x \in H$, 
+    (1): Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). 
+    
+    (2): For each $x \in H$, 
     \[
         \norm{Px}_H^2 = \dpn{Px, Px}{H} = \dpn{P^*Px, x}{H} = \dpn{T^*Tx, x}{H} = \norm{Tx}_H^2
     \]
 
-    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry (2). Moreover, for any $x \in H$, $Tx = V_0Px = VPx$ (3).
+    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry. 
-
-    Finally, since the initial space of $V$ is $\ol{P(H)}$, $\ker(V) = P(H)^\perp = \ker(P)$ (4). 
     
-    It remains to show uniqueness. Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
+    (3): For any $x \in H$, $Tx = V_0Px = VPx$. 
+    
+    
+    (4), (5): Since the initial space of $V$ is $\ol{P(H)} = \ker(T)^\perp$, $\ker(V) = P(H)^\perp = \ker(P)$. 
+    
+    (Uniqueness): Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
 
-    Now, since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+    Since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+
+    (6): Let $A$ be the von Neumann algebra generated by $T$. By \autoref{theorem:existence-of-projections-vna}, $A$ is a unital $C^*$-algebra, so $P = \sqrt{T^*T} \in A$. To see that $V \in A$, it is sufficient to apply the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+
+    To this end, let $S \in A'$, then $TS = ST = SVP$ and $VSP = VPS = TS$, so $SV$ and $VS$ agree on $\ol{P(H)}$. Since $\ker(V) = \ker(P) = \ol{P(H)}^\perp$, $SV|_{\ker(P)} = 0$. On the other hand, as $SP = PS$, $S(\ker(P)) \subset \ker(P) = \ker(V)$, so $VS|_{\ker(P)} = 0$ as well. Therefore $V \in A'' = A$. 
 \end{proof}
 
 

src/op/notation.tex

@@ -5,6 +5,7 @@
     \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
     \hline
     $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
+    $Z(A)$ & Centre of a Banach algebra. & \autoref{definition:banach-algebra-centre} \\ 
     $G(A)$ & Invertible group of a unital algebra. & \autoref{definition:banach-algebra-invertible} \\
     $G_0(A)$ & The identity component of $G(A)$. & \autoref{definition:identity-component} \\
     $I(A)$ & The index group of $A$. & \autoref{definition:index-group} \\
@@ -20,7 +21,11 @@
     $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
     $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
     $U(T)$ & Cayley transform of $T$. & \autoref{definition:cayley-transform-bounded} \\ 
-    $E_{x, y}$ & $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \autoref{definition:spectr}
+    $E_{x, y}$ & $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \autoref{definition:spectral-measure} \\ 
+    $\text{Proj}(A)$ & Projections in $A$. & \autoref{definition:vn-projection-lattice} \\
+    $Z(P)$ & Central support of $P \in \text{Proj}(A)$. & \autoref{definition:central-support-vna} \\ 
+    $P \sim Q$ & $P, Q \in \text{Proj}(A)$ are Murray-von Neumann equivalent & \autoref{definition:murray-von-neumann-equivalent} \\ 
+    $P \preceq Q$ & $P$ is Murray-von Neumann subequivalent to $Q$. & \autoref{definition:murray-von-neumann-subequivalent} \\ 
 
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\

src/op/vn/commutative.tex

@@ -48,7 +48,9 @@
 
 \begin{corollary}
 \label{corollary:maximal-abelian}
-    
+    Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a maximal abelian von Neumann algebra, then $A$ admits a cyclic vector. 
 \end{corollary}
-
+\begin{proof}[Proof, {{\cite[Corollary 22.4]{Zhu}}}. ]
+    By \autoref{proposition:maximal-commutative-vn}, $A = A'$. By \autoref{theorem:commutative-has-separating}, $A$ admits a separating vector. By \autoref{proposition:cyclic-separating-commutant}, this separating vector for $A$ is a cyclic vector for $A' = A$. 
+\end{proof}
 

src/op/vn/index.tex

@@ -6,4 +6,5 @@
 \input{./vn.tex}
 \input{./commutative.tex}
 \input{./spec.tex}
-\input{./fc.tex}
+\input{./fc.tex}
+\input{./projection.tex}

src/op/vn/projection.tex

@@ -3,20 +3,61 @@
 
 \begin{definition}[Projection Lattice]
 \label{definition:vn-projection-lattice}
-    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then the set of all projections $\text{Proj}(A)$ in $A$ is the \textbf{projection lattice} of $A$.
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\text{Proj}(A)$ be the set of all projections in $A$, then:
+    \begin{enumerate}
+        \item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto the closed subspace generated by ${\bigcup_{Q \in S}Q(H)}$, then $P = \sup(S) \in A$. 
+        \item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto $\bigcap_{Q \in S}Q(H)$, then $P = \inf(S) \in A$. 
+        \item $\text{Proj}(A)$ is order complete. 
+    \end{enumerate}
+
+    The set $\text{Proj}(A)$ is the \textbf{projection lattice} of $A$. 
+\end{definition}
+\begin{proof}
+    (1): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, the closed subspace generated by $\bigcup_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+
+    (2): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, $\bigcap_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+\end{proof}
+
+\begin{definition}[Central Support]
+\label{definition:central-support-vna}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $Z(P) = \inf_{Q \in \text{Proj}(Z(A)), Q \ge P}Q$ is the \textbf{central support} of $P$. 
 \end{definition}
 
 \begin{proposition}
-\label{proposition:vn-projection-lattice-complete}
+\label{proposition:central-support-vna}
-    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $\text{Proj}(A)$ is order complete. 
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$. For each $T \in A$, let $R(TP)$ be the orthogonal projection onto $\ol{TP(H)}$, then
+    \[
+    Z(P) = \sup_{T \in A}R(TP)
+    \]
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 24.6]{Zhu}}}. ]
-    By \autoref{theorem:existence-of-projections-vna}. 
+    Since $Z(P) \in Z(A)$, $Z(P)(H)$ is a reducing subspace of every operator in $A$. As $P \le Z(P)$, $TP(H) \subset T(Z(P)(H)) \subset Z(P)(H)$, so $Z(P) \ge R(TP)$ for all $T \in A$, and $Z(P) \ge \sup_{T \in A}R(TP)$. 
+    
+    On the other hand, for each $S, T \in A$, $S(TP(H)) \subset \bigcup_{R \in A}RP(H)$. As $A$ is a von Neumann algebra, the range of $\sup_{T \in A}R(TP)$ is a reducing subspace for every operator in $A$. Therefore $\sup_{T \in A}R(TP) \in Z(A)$, and $Z(P) \le \sup_{T \in A}R(TP)$.
 \end{proof}
 
 
+
+\begin{lemma}
+\label{lemma:projection-mental-gymnastics}
+    Let $H$ be a complex Hilbert space and $P, Q \in B(H)$ be projections, then:
+    \begin{enumerate}
+        \item $\ker(PQ) = \ker(Q) + \ker(P) \cap Q(H)$. 
+        \item If $PQ = QP$, then $PQ$ is a projection with $PQ(H) = P(H) \cap Q(H)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $x \in \ker(PQ)$, then $Q(x) \in \ker(P)$, so $x = Q(x) + (1 - Q)(x) \in  \ker(Q) + \ker(P) \cap Q(H)$.
+    
+    (2): Since $PQ = QP$, $(PQ)^2 = P^2Q^2 = PQ$ and $(PQ)^* = Q^*P^* = QP = PQ$, $PQ$ is a projection. As $PQ(H) = P(Q(H)) \subset P(H)$ and $PQ(H) = Q(P(H)) \subset Q(H)$, $PQ(H) \subset P(H) \cap Q(H)$. On the other hand, $PQ$ is the identity on $P(H) \cap Q(H)$, so $PQ(H) = P(H) \cap Q(H)$. 
+\end{proof}
+
+
+
+
+
 \begin{definition}[Murray-von Neumann Equivalent]
-\label{definition:murrey-von-neuman-equivalent}
+\label{definition:murray-von-neumann-equivalent}
     Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
     \begin{enumerate}
         \item There exists $V \in A$ such that $P = V^*V$ and $Q = VV^*$. 
@@ -31,4 +72,140 @@
     (2) $\Rightarrow$ (1): By \autoref{proposition:partial-isometry-characterisation}, $P = V^*V$ is a projection onto $\ker(V)^\perp$, and $Q = VV^*$ is a projection onto $\ker(V^*)^\perp = V(H)$.
 \end{proof}
 
+\begin{definition}[Murrey-von Neumann Subequivalent]
+\label{definition:murray-von-neumann-subequivalent}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then $P$ is \textbf{Murrey-von Neumann subequivalent} to $Q$, denoted $P \preceq Q$, if there exists $R \in \text{Proj}(A)$ such that $P \sim R$ and $R \le Q$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:mvn-equivalent-adjoint}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $T \in A$, and $P, Q \in \text{Proj}(A)$ be orthogonal projections onto $\ol{T(H)}$ and $\ol{T^*(H)}$, respectively, then $P \sim Q$. 
+\end{lemma}
+\begin{proof}
+    Let $T = VQ$ be the \hyperref[polar decomposition]{theorem:hilbert-polar-decomposition} of $T$, then $V$ is a partial isometry from $\ol{T^*(H)}$ to $\ol{T(H)}$. Since $V \in A$, $P \sim Q$. 
+\end{proof}
+
+\begin{theorem}[Kaplansky's Formula]
+\label{theorem:kaplansky-formula}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then
+    \[
+    [(P \vee Q) - Q] \sim [(P - P \wedge Q)]
+    \]
+\end{theorem}
+\begin{proof}
+    Using \autoref{theorem:existence-of-projections-vna}, assume without loss of generality that $I \in A$. In which case, by (1) of \autoref{lemma:projection-mental-gymnastics},
+    \[
+    
+        [(I - Q)P](H)^\perp = \ker(P(I - Q)) = \ker(Q)^\perp + [\ker(Q) \cap \ker(P)]
+    \]
+
+    and since $P \vee Q$ and $Q$ commute, 
+    \begin{align*}
+        [(I - Q)P](H) &= [\ker(Q)^\perp + [\ker(Q) \cap \ker(P)]]^\perp \\
+        &= \ker(Q) \cap [\ker(Q) \cap \ker(P)]^\perp \\
+        &= \ker(Q) \cap [\ker(Q)^\perp + \ker(P)^\perp]\\
+        &= (I - Q)(H) \cap [Q(H) + P(H)]\\
+        &= (I - Q)(H) \cap [(Q \vee P)(H)] \\
+        &= (P \vee Q)(I - Q)(H) = [(P \vee Q) - Q](H)
+    \end{align*}
+
+    by (2) of \autoref{lemma:projection-mental-gymnastics}. Similarly, 
+    \[
+    [P(I - Q)](H)^\perp = \ker((I - Q)P) = \ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]
+    \]
+
+    so
+    \begin{align*}
+        [P(I - Q)](H) &= [\ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]]^\perp \\
+        &= \ker(P)^\perp \cap  [\ker(Q) + \ker(P)] \\
+        &= P(H) \cap [(I - Q)(H) + (I - P)(H)] \\
+        &= P(H) \cap [(I - Q) \vee (I - P)](H) \\
+        &= P(H) \cap [I - (P \wedge Q)](H) \\
+        &= [P - (P \wedge Q)](H)
+    \end{align*}
+
+    Therefore
+    \[
+    [(P \vee Q) - Q](H) = [(I - Q)P](H)  \sim P(I - Q)(H) = [P - (P \wedge Q)](H)
+    \]
+    
+    by \autoref{lemma:mvn-equivalent-adjoint}. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:central-support-mvn-eq}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$ with $P \sim Q$, then $Z(P) = Z(Q)$. 
+\end{lemma}
+\begin{proof}
+    Let $V \in A$ with $P = V^*V$ and $Q = VV^*$, then $V$ is a partial isometry with initial space $P(H)$ and final space $Q(H)$. In which case, since $Z(P) \ge P$ and $Z(P) \in Z(A)$, 
+    \[
+    Z(P)Q = Z(P)VV^* = VZ(P)V^* = VV^* = Q
+    \]
+
+    and $Z(P) \ge Q$, and $Z(P) \ge Z(Q)$. By symmetry, $Z(Q) \ge Z(P)$, so $Z(P) = Z(Q)$. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:central-support-mvn}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $Z(P)Z(Q) \ne 0$. 
+        \item $PAQ \ne \bracsn{0}$. 
+        \item There exists non-zero projections $P_0 \le P$ and $Q_0 \le Q$ such that $P_0 \sim Q_0$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 24.7]{Zhu}}}. ]
+    (1) $\Rightarrow$ (2): For each $T \in A$, let $R(T)$ be the orthogonal projection onto $\ol{T(H)}$. By \autoref{proposition:central-support-vna}, 
+    \[
+    Z(P) = \sup_{T \in A}R(TP) \quad Z(Q) = \sup_{T \in A}R(TQ)
+    \]
+
+    so $Z(P)Z(Q) = \sup_{S, T \in A}R(SP)R(TQ) \ne 0$. Thus there exists $S, T \in A$ such that $R(SP)R(TQ) \ne 0$. As such, there exists $x, y \in H$ with
+    \[
+    0 \ne \dpn{SPx, TQy}{H} = \dpn{QT^*SPx, y}{H}
+    \]
+
+    so $PAQ \ne 0$. 
+
+    (2) $\Rightarrow$ (3): Let $T \in A$ with $PTQ \ne 0$. Let $P_0 = R(PTQ)$ and $Q_0 = R(QT^*P)$, then $0 \ne P_0 \le P$, $0 \ne Q_0 \le Q$, and $P_0 \sim Q_0$ by \autoref{lemma:mvn-equivalent-adjoint}. 
+
+    (3) $\Rightarrow$ (1): By \autoref{lemma:central-support-mvn-eq}, $Z(P_0) = Z(Q_0)$, so
+    \[
+    Z(P)Z(Q) = Z(P) \wedge Z(Q) \ge Z(P_0) \vee Z(Q_0) \ne 0
+    \] 
+\end{proof}
+
+
+\begin{theorem}["Cantor-Bernstein"]
+\label{theorem:murray-von-neumann-subequivalent-partial-order}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$. If $P \preceq Q$ and $Q \preceq P$, then $P \sim Q$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Lemma 25.1]{Zhu}}}. ]
+    Let $U, V \in A$ be partial isometries such that $P = U^*U$, $UU^* \le Q$, $Q = V^*V$, and $VV^* \le P$. Denote $Q_0 = Q$ and $P_0 = P$. For each $n \in \natz$, inductively define $P_{n+1} = VQ_nV^*$ and $Q_{n+1} = UP_nU^*$, then:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $n \in \natz$, $P_n, Q_n \in \text{Proj}(A)$.
+        \item For each $n \in \natz$, $P_n \le P$ and $Q_n \le Q$.
+        \item For each $n \in \natz$, $P_{n+1} \le P_n$ and $Q_{n+1} \le Q_n$. 
+    \end{enumerate}
+
+    As $\seq{P_n}, \seq{Q_n} \subset \text{Proj}(A)$ are non-increasing sequences, by \autoref{theorem:existence-of-projections-vna}, there exists $P_\infty, Q_\infty \in \text{Proj}(A)$ such that $P_n \to P_\infty$ and $Q_n \to Q_\infty$ in the strong operator topology as $n \to \infty$. 
+
+    For each $n \in \natz$, $U(P_n - P_{n+1})U^* = Q_{n+1} - Q_{n+2}$, so
+    \begin{align*}
+    [U(P_n - P_{n+1})]^*[U(P_n - P_{n+1})] &= (P_n - P_{n+1})P(P_n - P_{n+1}) = P_n - P_{n+1} \\
+    [U(P_n - P_{n+1})][U(P_n - P_{n+1})]^* &= U(P_n - P_{n+1})^2U^* = Q_{n+1} - Q_{n+2}
+    \end{align*}
+
+    and $P_n - P_{n+1} \sim Q_{n+1} - Q_{n+2}$. Similarly, $Q_n - Q_{n+1} \sim P_{n+1} - P_{n+2}$. As $P_{n+1} = VQ_nV^*$ for all $n \in \natz$, $P_\infty \sim Q_\infty$ after passing through a strong-operator limit. 
+    
+    For each $N \in \natz$, $\sum_{n = 0}^N (P_n - P_{n+1}) = P - P_{N+1}$, so $P = P_\infty + \sum_{n = 0}^\infty (P_n - P_{n+1})$. Similarly, $Q = Q_\infty + \sum_{n = 0}^\infty (Q_n - Q_{n+1})$. Therefore
+    \begin{align*}
+        P &= P_\infty + \sum_{n = 0}^\infty (P_{2n} - P_{2n+1}) + \sum_{n = 0}^\infty (P_{2n + 1} - P_{2n+2}) \\
+        &\sim Q_\infty + \sum_{n = 0}^\infty (Q_{2n + 1} - Q_{2n+2})  + \sum_{n = 0}^\infty (Q_{2n} - Q_{2n+1}) = Q
+    \end{align*}
+
+    because $\sim$ is preserved through direct sums.
+\end{proof}
+
 

src/op/vn/vn.tex

@@ -6,7 +6,7 @@
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a strong-operator closed $C^*$-subalgebra, then:
     \begin{enumerate}
         \item For any bounded directed family $\cf \subset A_{sa}$, $\sup(\cf) = \sotlim_{T \in \cf}T \in A_{sa}$. 
-        \item For any family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$. 
+        \item For any directed family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$. 
     \end{enumerate}
 
     and
@@ -46,7 +46,7 @@
 
     for all $S \in \cf$ with $S \ge T$. As such a $T$ exists for all $\eps > 0$, $R = \sotlim_{T \in \cf}T$. 
 
-    (2): Assume without loss of generality that $\mathcal{P}$ is directed. By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
+    (2): By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
     \[
     \dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
     \]
@@ -89,7 +89,7 @@
 
     By (3) applied to $TT^*/\norm{TT^*}_{B(H)}$, the orthogonal projection onto $\ol{T(H)}$ is in $A$. 
 
-    (5): Let $\mathcal{P}$ be the set of all projections in $A$, then $\mathcal{P} \subset A_{sa}$ is bounded and directed. By (2), $P = \sup_{Q \in \mathcal{P}}Q \in A$, which is the maximum projection in $A$. 
+    (5): Let $\mathcal{P}$ be the set of all projections in $A$, and $\cf \subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$. For each $F \in \cf$, let $P_F$ be the projection onto $\braks{\sum_{P \in F}P}(H)$, then $P_F \ge P$ for all $P \in F$ and $P_F \in A$ by (4). Since $\bracsn{P_F}_{F \in \cf}$ is a bounded and directed family of projections, $\sup_{F \in \cf}P_F \in A$ by (1). As $\sup_{F \in \cf}P_F \in \mathcal{P}$, it is the maximum projection in $A$. 
 
     Let $T \in A$, then by (4), $P$ is greater than the projection onto $\ol{T(H)}$, so $PT = T$. On the other hand, since $PT^* = T^*$, $TP = T$ as well. Therefore $P$ is the multiplicative identity in $A$. 
 \end{proof}