In a recent lecture my professor proved the existence of hilbertbases in arbitrary hilbert spaces using Zorn's Lemma.

If I only want to use countable choice (and not the full axiom of choice which is equivalent to Zorn's Lemma), for which hilbert spaces can I prove the existence of hilbert bases?


If I were a gambling man? Almost nothing that you couldn't prove without the axiom of choice itself.

See, the point is that either:

  1. Your Hilbert space is separable (or any well-orderable dense set), in which case you can do this by well-ordering a countable well and using induction to construct the basis. (Also, note that there aren't that many separable Hilbert spaces...)

  2. Or your Hilbert space is not separable, in which case there's no reason to believe that using countable choice (which is quite darn weak as far as choice goes, it won't even prove the Baire Category Theorem) would be helpful in doing something that would require transfinite recursion over some uncountable length and uncountably many arbitrary choices.

The key point here is that as far as analysis goes, countable choice (or rather dependent choice, which is what most laymathematicians would mean when they actually think about countable choice) is good for making sure the basics of $\Bbb R$ and other separable stuff works fine (e.g. continuity by sequences and by $\varepsilon$-$\delta$, or having a reasonable definition for the Borel measure). And pretty much anything else requires much more, in terms of choice.

Eric points out in the comments that countable choice would be needed in order to prove that $\bigoplus_{n=1}^\infty\cal H_n$ has a basis where each of them is separable, as that would require choosing a basis for each $\cal H_n$.

But that would more or less be the only situation where I expect countable choice to actually be needed. As to whether or not such situation occurs "naturally" where you truly can't choose the basis uniformly is beyond my pay grade.

  • $\begingroup$ To give a somewhat contrived counterexample to your gamble, I expect that countable choice is what it takes to prove that a direct sum of countably many separable Hilbert spaces has a basis. $\endgroup$ – Eric Wofsey Nov 20 '17 at 22:22
  • $\begingroup$ @Eric: That sounds almost reasonable, let me rephrase. $\endgroup$ – Asaf Karagila Nov 20 '17 at 22:23

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