TITLE: Comparing cardinalities of the spectrum of two masas in $B(H)$
QUESTION [6 upvotes]: If I imagine that (the self-adjoint part of) a C*-algebra $A$ represents the algebra of observables of some quantum system, then certain perspectives on algebraic quantum theory would ask me to imagine that each (maximal) commutative C*-subalgebra $C \subseteq A$ provides a (maximal) "classical snapshot" of this quantum system.  Gelfand duality yields that $C \cong C(X)$ for the compact Hausdorff space $X = \mathrm{Spec}(C)$, so I would picture $X$ as a classical state space that (maximally) "approximates" the would-be quantum state space corresponding to $A$.
I would like to know how different these spaces $X$ can be as the maximal commutative $*$-subalgebra $C \subseteq A$ varies.  Specifically, can it happen that these have different cardinalities?
I'm interested in the particular case $A = B(H)$ for a separable Hilbert space $H$ and two of its well-known masas: the continuous one $C \cong L^\infty[0,1] \subseteq A$ and discrete one $D \cong \ell^\infty(\mathbb{N}) \subseteq A$.  Thus I ask:

Q: Is there a bijection between the Gelfand spectra $\mathrm{Spec}(C)$ and $\mathrm{Spec}(D)$ for the continuous and discrete masas $C,D \subseteq B(H)$?

It's possible to describe these spectra in more explicit terms using Boolean algebra.  Note that each of these masas is an (A)W*-algebra.  By a combination of Gelfand and Stone dualities (see section 2 of this paper for a bit more detail), the spectrum of a commutative AW*-algebra $K$ is the Stone space of the complete Boolean algebra $\mathrm{Proj}(K)$ of projections in $K$, whose points are the ultrafilters of $\mathrm{Proj}(K)$.
The continuous masa $C \cong L^\infty[0,1]$ has $\mathrm{Proj}(C)$ isomorphic to the Boolean algebra of measurable subsets of $[0,1]$ modulo the null sets.  I have just learned through the magic of Wikipedia that this is called the random algebra; I will denote it by $B$.
The discrete masa $D \cong \ell^\infty(\mathbb{N})$ has $\mathrm{Proj}(D)$ isomorphic to the power set Boolean algebra $2^\mathbb{N}$.  (Note that an ultrafilter on the Boolean algebra $2^\mathbb{N}$ is alternatively referred to as an ultrafilter on the set $\mathbb{N}$.)
Thus my question is equivalent to:

Q': Is there a bijection between the sets of ultrafilters on the random algebra $B$ and the power set algebra $2^\mathbb{N}$?

I am aware that $\mathrm{Spec}(D)$ has spectrum homeomorphic to the Stone-Cech compactification $\beta\mathbb{N}$ of the discrete space $\mathbb{N}$, and that this space has various properties that depend on set-theoretic assumptions.  Now that I know what the random algebra is called, I see that it bears a relationship to forcing.  Thus I can imagine that the answer to my question could be independent of ZFC.  Nevertheless, as I am not asking exactly what the cardinality of this spectrum is, but whether it is in bijection with some other (possibly complicated) spectrum, I have an ounce of hope that this can indeed be decided in ZFC.
(By the way, the classification of the possible masas of $B(H)$ implies that, if the answer to my question is affirmative, then the spectra of all masas of $B(H)$ are in bijection with one another.)

REPLY [5 votes]: Yes. The spectra of $\ell_\infty$ and $L_\infty$ have the same cardinality, namely $2^{\mathfrak{c}}$.
Indeed, every infinite, compact $F$-space space (in particular, an extremely disconnected compact space such as the spectrum of $L_\infty$) contains a copy of $\beta \mathbb{N}$ (which happens to be the spectrum of $\ell_\infty$). This is 14N(5) in

L. Gillman and M. Jerison, Rings of continuous functions, van Nostrand Reinhold, New York, 1960.

and it is quite elementary. The Hausdorff–Pospíšil theorem implies that $|\beta\mathbb{N}|=2^{\mathfrak{c}}$. Thus,  $2^{\mathfrak{c}}\leqslant |{\rm spec}\, L_\infty|$.
I claim that the cardinality of ${\rm spec}\, L_\infty$ cannot be bigger than $2^{\mathfrak{c}}$. Indeed, $L_\infty^*$ is a bidual of a separable Banach space, hence by Goldstine's theorem, it is separable in the weak*-topology. Every separable space has cardinality at most $2^{\mathfrak{c}}$. Thus 
$|{\rm spec}\, L_\infty|\leqslant |L_\infty^*|\leqslant 2^{\mathfrak{c}}.$