I need help understanding the proof of Lemma $3$ in the paper Commutators of Operators by Paul R. Halmos.
Lemma $3$. Every Hermitian operator on an infinite-dimensional Hilbert space leaves invariant at least one large subspace with a large orthogonal complement.
A subspace $H$ of a Hilbert space is large if $H$ contains infinitely many orthogonal copies of its orthogonal complement, or, in other words, if $\dim H \ge \aleph_0 \dim (H^\perp)$.
Proof. The underlying Hilbert space, if it is not already separable, can be expressed as a direct sum of separable, infinite-dimensional subspaces invariant under the given operator. There is, therefore, no loss of generality in restricting attention to separable Hilbert spaces.
Q1. I understand that $\mathcal H$ can be written as an uncountable direct sum of separable subspaces, but why does it suffice to consider the separable case? Suppose $A \in \mathcal B(H)$ is Hermitian and $\mathcal H = \bigoplus_{n=1}^\infty \mathcal H_n$ where $\{\mathcal H_n\}_{n=1}^\infty$ is a family of $A$-invariant separable subspaces of $\mathcal H$. So, $A_i:= A\vert_{\mathcal H_i}: \mathcal H_i \to \mathcal H_i$ is also Hermitian for every $i\ge 1$. Assuming we have worked out the separable case; for every $i\ge 1$ there exists a large $A_i$-invariant subspace $M_i \le H_i$ such that $H_i - M_i$ (orthogonal complement in $H_i$) is also large. Consider $M = \bigoplus_{n=1}^\infty M_i$. It is clear that $M$ is $A$-invariant. How do I show that $M$ and $M^\perp$ are also large?
If $A$ is Hermitian and $E$ is the spectral measure of $A$, and if, for every Borel subset $M$ of the real line, $E(M) = \mathbf{0}$ or $\mathbf{I}$, then $A$ is a scalar multiple of $\mathbf{I}$. It follows easily that if, for every $M$, the dimension of the range of $E(M)$ is finite or co-finite, then $A$ differs from a scalar multiple of $\mathbf{I}$ by a finite-dimensional operator. In the contrary case both $E(M)$ and $\mathbf{I} - E(M)$ have infinite-dimensional ranges for some $M$. In either case, the conclusion of the lemma is obvious.
Q2. Could someone please explain the details of the above proof? It seems very cryptic and not at all straightforward.
Thank you!
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