Are GNS representations the way to build physical Hilbert spaces? Consider a separable $C^*$ algebra $\mathcal A$. The space of states is also separable in the weak* topology, let $S$ be a countable dense subset. 
Denoting with $H_\omega$ the GNS representation of a state $\omega$ we retrieve a representation of $\mathcal A$ on the Hilbert space $H(S)=\bigoplus_{\omega\in S}H_\omega$. This representation is isometric and $H$ is separable.
In the context of quantum mechanics we have built a candidate for the physical Hilbert space just by knowing an algebra of observables. This construction however depends on how we chose our set $S$. My question is:

Are the representations of $\mathcal A$ on $H(S)$ and $H(S')$ unitarily equivalent for any two dense countable subsets $S,S'$ of the state space of $\mathcal A$?

 A: Independently from separability of the algebra, if you restrict the states to pure states the answer is generally negative provided the algebra admits a couple of algebraic  pure states unitarily inequivalent with separable GNS spaces  (and this is the case in nontrivial qft). Consider a such pure state $\omega$. The pure states in its folium are weakly dense in the folium of every other pure state by Fell's theorem. Suppose the Hilbert space of the GNS representation of $\omega$ is separable. You conclude that there is a countable dense set of pure states $S$ in the folium of that state which is weakly dense in all folia of all algebraic  state. The GNS representations of this separable class of pure states are unitarily equivalent, since the states are vector states in the folium of a given  pure state. Therefore:
In the direct sum  of Hilbert spaces labeled over $S$, actually  there is a unique  unitary representation, that of $\omega$,  acting on  each summand.
If you start from a unitarily inequivalent initial pure state $\omega'$ and built up the analogous direct sum, the final direct sum of representations cannot be unitarily equivalent to the initially constructed out of $\omega$.
