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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Nov 26, 2020 at 23:57
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    $\begingroup$ @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) $\endgroup$ Nov 27, 2020 at 0:13
  • $\begingroup$ we show in $\mathsf{ZF}$ that a basis for $\mathbb{R}^\mathbb{N}$ must not have the Baire property when ported along that bijection. $\endgroup$ Nov 27, 2020 at 0:14
  • $\begingroup$ 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.) $\endgroup$
    – Asaf Karagila
    Nov 27, 2020 at 0:31

1 Answer 1


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 can always have the axiom of choice failing, as badly as we want it to fail, while the real numbers, and every set that you would ever cared about, can be well-ordered so that all vector spaces "that matter" have a basis. In other words, the axiom of choice is a global statement, so its negation is not about one set. It's about the existence of a counterexample.

(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.


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