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?
Best Answer
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 $H$.