Please help me understand this argument from the proof of the above Proposition.
Proposition 3: Let $\mathcal{A} \subset \mathcal{L}(H)$ be a Von Neumann algebra (VNA) of operators on a Hilbert space $H$. Then there exists a central projection $G\in \mathcal{Z}(\mathcal{A})$ for which $\mathcal{A}_G$ has a cyclic element and $\mathcal{A}_{I-G}$ has a separating element. The subscript here refers to the reduction of the algebra with respect to the projection.
Proof: By Zorn's lemma we choose a maximal sequence of elements $(x_i)_I$ in $\mathcal{A}$ subject to the two conditions
- The projections $E_i \in \mathcal{A}$ to the subspaces $\overline{\mathcal{A}'x_i}$ are pairwise orthogonal.
- The projections $E_i' \in \mathcal{A'}$ to the subspaces $\overline{\mathcal{A}x_i}$ are pairwise orthogonal.
Let $E=\sum E_i \in \mathcal{A}$ and $E'= \sum E_i' \in \mathcal{A}'$. Further let $F=I-E$ and $F'=I-E'$ and let $F_1$, $F_1'$ be their respective central supports so that $$ F_1F = FF_1 = F\qquad \text{ and }\qquad F_1'F' = F'F_1' = F'. $$ Claim: If $F_1F_1' \neq 0$, then $(F')_{F_1}\neq 0$. That is, $F_1F' \neq 0$.
Question: How to see the above claim?
Attempt: We know that $\text{Im}(F) \subset \text{Im}(F_1)$ and $\text{Im}(F') \subset \text{Im}(F'_1)$ from the support hypothesis. The hypothesis in the claim gives us that $\text{Im}(F_1) \cap \text{Im}(F_1') \neq 0$.
But why can't $\text{Im}(F')$ be perpendicular to this intersection? I wonder.
Hope someone can help. I'm new to all this Boolean algebra stuff and I find the logic of it very complicated.
$$F_1F'_1\ne0\iff F_1\mathcal A'F'\ne0\iff\mathcal A'F_1F'\ne0\iff F_1F'\ne0.$$