I am currently reading "Local Quantum Physics" by R. Haag. In Chapter III, Theorem 2.1.9 says:
Let $\mathcal{R}$ be a factor, $P_i$ projectors belonging to $\mathcal{R}$. Then precisely one of the following relation holds \begin{equation} P_1>P_2,\quad P_1\sim P_2,\quad P_1<P_2. \end{equation}
The proof of this theorem in the book is omitted, but there is a clue with Lemma 2.1.10:
If $\mathcal{R}$ is a factor, $A\in \mathcal{R}$, $B\in \mathcal{R}'$ then $AB=0$ implies either $A=0$ or $B=0$.
My questions:
How can I prove the Lemma 2.1.10? Since $\mathcal{R}$ is a factor, $\mathcal{R}\cap \mathcal{R}' = \{\lambda \mathbb{1}\}$, which "looks like" they are kind of "well separated" and thus Lemma 2.1.10 is reasonable. However I cannot prove the statement.
How can I prove the Theorem 2.1.9 from Lemma 2.1.10? I already found the proof using "Comparability Theorem": there is a central projection $Z$ such that $ZP_1 \leq ZP_2$ and $(\mathbb{1}-Z)P_1\geq (\mathbb{1}-Z)P_2$. Showing the "Comparability Theorem" uses Zorn's Lemma, hence I think that showing Theorem 2.1.9 with Lemma 2.1.10 also uses Zorn's Lemma(or equivalent AC-type statements), but I don't know how.
Terminologies: a von Neumann algebra $\mathcal{R}$ in the algebra of bounded linear operators acting in a Hilbert space $\mathcal{H}$, $\mathfrak{B}(\mathcal{H})$, is a factor when its center is trivial. Two projectors $P_1, P_2$ are equivalent w.r.t. $\mathcal{R}$, i.e. $P_1\sim P_2$, if there is an operator $V\in \mathcal{R}$ with $P_1=V^*V$ and $P_2=VV^*$. $P_1>P_2$ if there is a subspace of $P_1\mathcal{H}$ whose projector $P_{11}\sim P_2$. $\mathcal{R}'$ is the commutant of $\mathcal{R}$, i.e. the elements in $\mathfrak{B}(\mathcal{H})$ which commutes with $\mathcal{R}$.
A crucial notion here is the central carrier of an operator. Given $\def\R{\mathcal R}T\in\R$, its central carrier is the projection $$ C_T=[\R T\def\H{\mathcal{H}} \H]. $$ For any $S\in\R$, it is clear that $S(\R T\H)\subset \R T\H$. This implies that $SC_T=C_TSC_T$. If we do this for selfadjoint $S$ we get that $SC_T=C_TS$, and as the selfadjoints span the whole algebra, this shows that $C_T\in\R'$. If now $S\in\R'$, then $S\R T\H=\R TS\H\subset \R T \H$. By repeating the previous reasoning, $C_T\in\R''=\R$. So $C_T\in\R\cap\R'$.
Proof of Lemma 2.1.10. If $A\in\R$, $B\in\R'$ and $AB=0$, we get $B\R A \H=\R A B\H=0$ and so $BC_A=0$. Since $\R$ is a factor, either $C_A=1$, in which case $B=0$, or $C_A=0$, in which case $A=0$.
Proof of Theorem 2.1.9. We may assume without loss of generality that $P_1,P_2$ are both nonzero. The proof uses Zorn's Lemma indeed. We consider the family of pairs $\{(p_j),(q_j)\}$ such that each net is pairwise orthogonal, $p_j\leq P_1$, $q_j\leq P_2$, and $p_j\sim q_j$ for all $j$. Then a maximal family will satisfy either $\sum p_j=P_1$ or $\sum q_j=P_2$, or both, giving the three possibilities (the needed partial isometries will be the sum of the corresponding partial isometries at each $j$).
The key to make the above work is to show that if $P,Q\in\R$ are nonzero projections, then there exists a nonzero partial isometry $V$ such that $V^*V\leq Q$ and $VV^*\leq P$. For this, suppose that $PRQ=0$ for all $R\in\R$. Then $PRQ\H=0$, which shows that $PC_Q=0$. AS $\R$ is a factor, if $C_Q=1$ we get $P=0$, a contradiction, and if $C_Q=0$ we get $Q=0$, also a contradiction. It follows that there exists $R\in\R$ such that $PRQ\ne0$. We can write the polar decomposition $PRQ=V|PRQ|$, with $V$ a nonzero partial isometry. Now $V^*V$ is the orthogonal projection onto the range of $(PRQ)^*=QR^*P$, which is itself inside the range of $Q$; thus $V^*V\leq Q$. Similarly, $VV^*$ is the orthogonal projection onto the range of $PQR$, and so $VV^*\leq P$.