Talk:Naimark's problem

I think that the statement is incorrect.

N. Weaver and C. Akemann showed that it is consistent with ${\mathsf {ZFC}}$ that there exists a counterexample to Naimark's Hypothesis. To show independence, they would have to prove that it is also consistent with ${\mathsf {ZFC}}$ that there exists no counterexample. As far as I know, this was not done.

Also, while it is true that separable C*-algebras are a special case, one should note that they are an extremely important special case. Some people think that non-separable C*-algebras are simply unimportant. (Of course, $B({\mathcal {H}})$ is non-separable, but one should think of it as a von Neumann algebra rather than as a C*-algebra.)

Leonard Huang's clarification response

The statement "There exists a counterexample to Naimark's Hypothesis that is generated by $\aleph _{1}$ elements" is indeed independent of ${\mathsf {ZFC}}$. However, the weaker statement "There exists a counterexample to Naimark's Hypothesis" is only known to be consistent with ${\mathsf {ZFC}}$, and this follows precisely from Weaver's and Akemann's result.

Weaver and Akemann proved that inside any model of ${\mathsf {ZFC}}+(V=L)$ (in which $\diamondsuit$ automatically holds), there exists a counterexample generated by $\aleph _{1}$ elements. They also established that, within ${\mathsf {ZFC}}$ alone, any counterexample must be generated by at least $2^{\aleph _{0}}$ elements. Hence, if the Continuum Hypothesis fails (i.e. $\aleph _{1}<2^{\aleph _{0}}$), then a counterexample generated by $\aleph _{1}$ elements simply cannot exist. However, this does not rule out the existence of a counterexample that is generated by at least $2^{\aleph _{0}}$ elements.

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