What is the dimension of a vector space without a basis?

Question

It's my understanding that without the axiom of choice, the vector space of $\mathbb{R}$ over $\mathbb{Q}$ lacks a set of basis vectors. Now it seems straight forward that with the axiom of choice that the following holds. $$\text{dim }(\mathbb{R}) = \aleph_1$$ However, what happens if we assume the axiom of choice is false? Does the dimension simply become undefined? As we can't define any basis for the space, it seems like the question of dimension becomes meaningless; you can't have the cardinality of something that doesn't exist. Alternatively, do we have an alternate way of defining dimension for these cases?

Answer 1

Careful: with the axiom of choice we get

$$\dim_{\mathbb{Q}}(\mathbb{R}) = 2^{\aleph_0} = \mathfrak{c}$$

(the cardinality of the continuum). To assert that $2^{\aleph_0} = \aleph_1$ requires the continuum hypothesis.

Without the axiom of choice, if we define the dimension as the size of a basis, then yes, without a basis this particular notion of dimension is undefined. So we might try to use others.

For example, we can talk about the rank, which is the minimal size of a set of generators of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Since $\mathbb{R}$ is uncountable no countable subset generates. And on the other hand $\mathbb{R}$ itself clearly generates. So assuming the continuum hypothesis the rank is $2^{\aleph_0} = \mathfrak{c}$ again. I don't know if it's consistent with ZF that the rank is somewhere strictly between $\aleph_0$ and $2^{\aleph_0}$.

There is also a somewhat different notion of rank given by the maximal size of a linearly independent subset. We can explicitly exhibit a linearly independent subset of $\mathbb{R}$ of cardinality $2^{\aleph_0}$ in ZF (see this MO answer). So it's provable in ZF that the rank in this sense is $2^{\aleph_0}$.

(In the absence of choice there's a subtle distinction to be made between "minimal size of a set of generators" vs. "size of a minimal set of generators" since without Zorn's lemma a minimal set of generators need not exist, and similarly for "maximal size of a linearly independent subset" vs. "size of a maximal linearly independent subset." Either a minimal set of generators or a maximal linearly independent subset must be a basis. I suppose in the absence of choice cardinals don't need to be totally ordered either though. So either way the rank might end up undefined in general. Welp.)

Answer 2

There is no good notion of dimension, in general, when the axiom of choice is omitted. Even if there is a basis, there may be different bases of different cardinality, so there need not be a "smallest" size.

Sometimes it's convenient to say that a space is "infinite dimensional" if it is not finite dimensional, but not necessarily be concrete about whether the dimension is this or that. And since a vector space without a basis is not finite dimensional, that means that any vector space without a basis is infinite dimensional.

But since the use of the notion of dimension in this sort of situation is so sparse, there's no commonly accepted convention, and any author would always caution on the side of simply setting up their own context.