Let $H$ be a Hilbert space, and $A$ a $C^*$-subalgebra of $B(H)$ (the bounded operators on $H$). Let $B$ be the strong-operator closure of $A$, so that in particular, $B$ is a von-Neumann-algebra.
According to the Kaplansky-Theorem:
The … in the unit ball of $A$ is s-o dense in the … in unit ball of $B$, where … is:
- unitaries
- self adjoints
Does it hold, that $\operatorname{Proj}(A)$ is strongly dense in $\operatorname{Proj}(B)$ (or even weakly dense).
I.e. does the Kaplansky-Theorem hold for projections? (Or a weaker version of it.)
If not, why not?
No. E.g., let $A=C[0,1]$ acting as multiplication operators on $L^2[0,1]$. The only projections in $A$ are $0$ and $1$. The strong closure contains multiplications by the characteristic functions of all measurable subsets of $[0,1]$.
More generally, infinite-dimensional C*-algebras need not have projections, while von Neumann algebras are spanned by projections.