# Basis for $\mathbb{R}^\mathbb{N}$ implies axiom of choice?

Let $$\mathbb{R}^\mathbb{N}$$ denote the vector space over $$\mathbb{R}$$ of sequences of real numbers, with multiplication and addition defined by component. It's well-known that though the subspace $$\mathbb{R}^\infty$$ of sequences with only a finite number of nonzero terms has a basis $$\mathbf{e}_1 = (1, 0, 0, 0, \ldots), \mathbf{e}_2 = (0, 1, 0, 0, \ldots)$$, this is not a basis of $$\mathbb{R}^\mathbb{N}$$ (expressing the constant sequence $$(1, 1, 1, \ldots)$$ would require an infinite sum $$\mathbf{e}_1 + \mathbf{e}_2 + \mathbf{e}_3 + \cdots$$, and infinite sums in generic vector spaces are undefined). It's also been proved that the statement that all vector spaces have a basis is equivalent to the axiom of choice.

I'm interested, though, in the specific space $$\mathbb{R}^\mathbb{N}$$. Has it been proved that a basis for this set requires the axiom of choice and cannot be described explicitly? This isn't a homework question or anything; I'm just curious.

• The axiom of choice is an axiom about all sets – even those that are unimaginably big and complicated. So I would not expect that a fact about one particular set implies the axiom of choice in full generality. – Zhen Lin Nov 26 '20 at 23:57
• @ZhenLin Note that there's some disagreement between the title and body. To the OP, the existence of a Hamel basis for $\mathbb{R}^\mathbb{N}$ as an $\mathbb{R}$-vector space is not $\mathsf{ZF}$-provable but also does not imply $\mathsf{AC}$ over $\mathsf{ZF}$. The latter fact is a quick forcing argument: given a countable model $M$ of $\mathsf{ZFC}$ we can get a symmetric extension $N$ which agrees with $M$ up to rank $\omega+2$ but in which choice fails. The former takes a trick; if I recall correctly, fixing an appropriately simple bijection $\mathbb{R}^\mathbb{N}\equiv\mathbb{R}$ (cont'd) – Noah Schweber Nov 27 '20 at 0:13
• we show in $\mathsf{ZF}$ that a basis for $\mathbb{R}^\mathbb{N}$ must not have the Baire property when ported along that bijection. – Noah Schweber Nov 27 '20 at 0:14
• I daresay that math.stackexchange.com/questions/1972321/… and math.stackexchange.com/questions/122571/… combine to a complete answer here. (See also math.stackexchange.com/questions/linked/122571 and math.stackexchange.com/questions/122857/… as very relevant threads.) – Asaf Karagila Nov 27 '20 at 0:31

No single concrete set admitting a certain property would imply the axiom of choice. Period. The axiom of choice is a global statement, and statements about a set with a certain property are local (I'm not talking about a global statement, e.g. "For every set $$A$$, $$A\times X$$ can be well-ordered" implies the axiom of choice for any fixed set $$X$$, that's cheating).
(We actually don't even know if there is a field $$F$$ such that "All vector spaces over $$F$$ have a basis" implies the axiom of choice; speaking of global statements disguised as local statements.)
On the other hand, it is consistent that every set of reals has the Baire property, which implies that every linear $$T\colon\Bbb{R^N\to R^N}$$ is continuous. Alas, being a separable space, there can only be $$2^{\aleph_0}$$ continuous functions; but we can easily show that a basis of $$\Bbb{R^N}$$ must have size $$2^{\aleph_0}$$ as well, and therefore there would be $$2^{2^{\aleph_0}}$$ linear functions just induced by permutations of such a basis. And so, indeed, if all sets of reals have the Baire property, no basis for $$\Bbb{R^N}$$ can exist.