3 ms·
Unless I'm mistaken, the computable numbers are not complete in the Cauchy Sequence sense. Therefore, there can be no hilbert space based on the computable numb
by wyager 10y ago
Unless I'm mistaken, the computable numbers are not complete in the Cauchy Sequence sense. Therefore, there can be no hilbert space based on the computable numbers, because a hilbert space has to be complete in the Cauchy sense. It would only be a pre-hilbert space. I believe this means one can no longer correctly use limits in QM; are there other consequences of this?
- chriswarbo 10y agoVery interesting, and the kind of technical aspect I tried to avoid by not refuting "face-on" since I don't feel qualified to answer this without looking into it further :) Still, if there are reasons to allow uncomputable numbers as elements of a quantum theory, that still doesn't invalidate the larger point that any predictions of those theories are computable; it's just a constraint on which axiom scheme we follow when manipulating our equations. Again, even if a theory involves uncountable sets, it may still have countable models (following either Skolem's "paradox", or a similar argument; since QM operates at much higher level than formal systems, I'm not sure if it's first-order, second-order, higher-order, etc.).