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There are some strange consequences if one rejects the axiom of choice [e.g. $\mathbb{R}$ can be a countable union of countable sets, not every commutative ring with unit has a maximal ideal, not every vector space has a (Hamel) basis,...].

Also, there are some strange consequences if you assume the axiom of choice (like the Banach-Tarski paradox).

I was wondering: What consequences might occur if one assumes the axiom of countable choice?

It is obvious, that $\mathbb{R}$ is not a countable union of countable sets since this must be countable then.

But are there any other similar results as in the case of the axiom of choice (or maybe the same results but with stricter assumptions)?

EDIT: Let me ask more particularly: Can you prove the existence of a Schauder basis for every separable Hilbert space just using ACC?

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  • $\begingroup$ Everything in "equivalents" will be false(unprovable, the negation may hold), there may be surjective function whose range greater than the domain, pretty sure Banach–Tarski paradox is not provable and more $\endgroup$ – Holo Feb 25 at 19:12
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    $\begingroup$ It is consistent with countable choice that every set of reals is Lebesgue measurable. Under this assumption there can't be a basis of $\mathbb R/\mathbb Q$, nor can Banach-Tarski paradox hold. So neither of those two can follow from countable choice. $\endgroup$ – Wojowu Feb 25 at 19:12
  • $\begingroup$ Didn't I write a bunch of answers about that? $\endgroup$ – Asaf Karagila Feb 25 at 19:24
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    $\begingroup$ "$\mathbb R$ is a countable union of countable sets" is not a consequence of the negation of the axiom of choice. (It is, however, consistent with that negation and the other axioms of ZF.) $\endgroup$ – Andreas Blass Feb 25 at 20:51
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    $\begingroup$ To add on the remark on @Andreas, I think that you mean "e.g." where you wrote "i.e." $\endgroup$ – Asaf Karagila Feb 25 at 23:25
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Existence of Schauder bases in separable Hilbert spaces can be proven without any appeal to the axiom of choice, countable or not. Here is the outline.

Let $v_1,v_2,\dots$ be a dense countable subset of the Hilbert space $H$ in question. Let is look at every element $v_i$ of the sequence and look if it is in the linear span of $v_j,j<i$. If it is, we discard it, and if not, we keep it. This way, we get a (possibly finite; for simplicity I'll assume there are infinitely many of them, the proof is the same otherwise) sequence $w_1,w_2,\dots$ of elements of $H$ which form a linearly independent set, and which moreover linearly span a subspace of $H$ containing each $v_i$, so the linear span is in $H$. By applying the Gram–Schmidt process (which is completely constructive, no choice needed) we can conclude there is a sequence $u_1,u_2,\dots$ which is orthonormal and topologically spans $H$. We claim this sequence is a Schauder basis.

Let $v\in H$ be arbitrary. Uniqueness of representation in terms of basis is clear: if $v=\sum_{i=1}^\infty a_iu_i$, then it's easy to see $\langle v,u_i\rangle=a_i$. So we are left with existence. Recall the vectors $v_i$ were dense in $H$, so there is a sequence $v_{i_n}$ convergent to $H$ (no choice needed - e.g. let $i_n$ be the least such that $||v-v_{i_n}||<1/n$. By construction, $v_{i_n}$ is in the linear span of $u_k$, so there is a unique sequence of reals $a_{n,k}$, finitely many of which are nonzero, such that $$v_{i_n}=\sum_{k=1}^\infty a_{n,k}u_k.$$ The sequence $v_{i_n}$ is Cauchy, hence so is the sequence $a_{n,k}=\langle v_{i_n},a_k\rangle$ for each fixed $k$, thus it converges to some number $b_k$. One can show that then we have $v=\sum_{k=1}^\infty b_ku_k$ (this is very similar to the proof that $\ell_2$ is a Hilbert space), which shows existence. Hence $u_1,u_2,\dots$ is a Schauder basis of $H$.

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  • $\begingroup$ Yes, of course, that's so obvious! I should be damned! And for the non-separable case you need AC, and ACC will not be enough, right? But with ACC you get, at least, that a countable Hilbert-ONB and Separability is equivalent, right? $\endgroup$ – YoungMath Feb 27 at 21:03
  • $\begingroup$ I do believe ACC is not enough to show every Hilbert space has an orthonormal basis, but bases exist for spaces with well-orderable dense subset (by essentially the same argument). Equivalence of countable orthonormal basis and separability is provable without ACC (right to left I have just shown, left to right is easy, just take finite combinations with rational coefficients). $\endgroup$ – Wojowu Feb 27 at 21:12
  • $\begingroup$ Yeah, that's what I meant. You need the full AC (in particular Zorn's Lemma) to prove that. And you're right, finite combinations should be enough. Thanks! $\endgroup$ – YoungMath Feb 27 at 21:19
  • $\begingroup$ However, is there anything you can ACC use for in functional analysis? $\endgroup$ – YoungMath Feb 27 at 21:20
  • $\begingroup$ In this thread you can see a proof that the uniform boundedness principle, usually proven from Baire category theorem which is equivalent to dependent choice, can be proven using just countable choice. $\endgroup$ – Wojowu Feb 27 at 21:28

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