I have a simple question I've managed to get myself quite confused about. 


Given a Hilbert space H, what do we know about the cardinality of


(a) the set $P(H)$ of projection operators onto $H$ (equivalently, the set of all closed subspaces of $H$),


and


(b) the set of all Boolean subalgebras of $P(H)$? 


I imagine that if $H$ is separable, we can say more than we could otherwise. 



Cardinality of the set of Boolean subalgebras of the lattice of projections on a Hilbert space
Up to a unitary isomorphism, a Hilbert space is uniquely determined by its dimension, and closed subspaces are Hilbert spaces in their own right. So $P(H)$ is the disjoint union, over all cardinal numbers $\alpha \leq \dim(H)$, of closed subspaces of dimension $\alpha$. And there as many of the latter as unitary operators on a Hilbert space of dimension $\alpha$.


So in finite dimension, $\mathrm{card}(P(H)) = \sum\_{\alpha=1}^{\dim(H)} \mathrm{card}(U(\alpha)) = \mathrm{card}(U(\dim(H))) = \mathrm{card}(\mathbb{R})$.


In infinite dimension, the unitary group $U(H)$ of $B(H)$ injects into the C\*-algebra $B(H)$ of bounded operators on $H$. Also, any element of $B(H)$ is a linear combination of four unitaries, so there is a surjection $U(H)^4 \times \mathbb{C}^4 \to B(H)$. Hence $\mathrm{card}(P(H))=\mathrm{card}(U(H))=\mathrm{card}(B(H))=\max(\dim(H),\mathrm{card}(\mathbb{R}))$.


So regardless of dimension, $P(H)$ has cardinality $\max(\dim(H), 2^{\aleph\_0})$.


Any Boolean subalgebra of $P(H)$ is contained in a maximal one, and those are all induced as the powerset of a choice of orthonormal basis. Hence the cardinality of the set of all Boolean subalgebras of $P(H)$ is $2^{2^{\mathrm{card}(P(H))}}$.


