TITLE: Separating vectors for $C^*$-algebras
QUESTION [5 upvotes]: Let $A$ be a C$^*$-algebra, concretely acting on a Hilbert space $H$.  Suppose that $\xi_0\in H$ is cyclic and separating for $A$ (that is, the map $A\rightarrow H, a\mapsto a(\xi_0)$ is injective with dense range).  Let $M=A''$ the von Neumann algebra generated by $A$.

Need $\xi_0$ still be separating for $M$?  That is, $x\in M, x(\xi_0)=0 \implies x=0$?

Actually, I think I can prove this.  We turn $\mathfrak A = \{ a(\xi_0) : a\in A \}$ into a left Hilbert algebra algebra in the obvious way.  Then run the Tomita-Takesaki machinery (actually not needed in full generality as we start with a state, not a weight).  Then the von Neumann algebra generated by $\mathfrak A$ is nothing but $M$, and so the general theory tells us that $\varphi(x) = \|x\xi_0\|$ will be a faithful weight on $M$, which is what we need.

Is this right?  Surely this argument is far, far more complicated then necessary?

No, I don't think this is right: there seems no reason why the Tomita operator $S:a(\xi_0)\mapsto a^*(\xi_0)$ is preclosed.

REPLY [2 votes]: A counter-example was given on mathoverflow by  Narutaka Ozawa: 

Take a closed nowhere dense subset $C\subset [0,1]$ with positive measure and consider the state $\phi$ on $C([0,1],M_2)$ defined by 
$$\phi(f)=\int_C f(x)_{11}dx+\int_{[0,1]\setminus C} \operatorname{tr}f(x)\,dx$$
Here $f(x)_{11}$ is the $(1,1)$-entry of $f(x)\in M_2$. Then $\phi$ is a faithful state, and the GNS vector is a cyclic separating vector. However it is not separating for the double commutant, which is $L^\infty ([0,1],M_2)$.