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.