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My questions are motivated by

this question

which asks, in the absence of AC, whether a subspace of a vector space with a basis must have a basis.

  1. Does every real vector space embed isomorphically into a vector space with a basis?

If $V$ is a vector subspace of a vector space with a basis, then clearly the linear functionals on $V$ separate points. If $V$ is a vector space s.t. the the linear functionals on $V$ do not separate points, then by modding out the intersection of the kernels of all linear functionals you get a vector space that has no non zero linear functional.

  1. Is there a non zero real vector space on which there is no non zero linear functional?

There are models of ZF in which every linear functional on every Banach space is continuous. In ZFC there are complete linear metric spaces on which every non zero linear functional is discontinuous. So I assume that (2) has a negative answer, which would imply that (1) also has a negative answer.

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Nice question! I'll take the opportunity to advertise Todd Trimble's answer there:… that gives a vector space over $\mathbb{F}_2$ that does not have non-trivial linear forms. Ever since, I am very curious about the case of $\mathbb{R}$... – Thierry Zell Nov 14 '11 at 17:49
I will take the opportunity started by Thierry to advertise my answer on this math.SE question which has a proof of Lauchli's theorem in a modern context which shows that it is possible for a fixed field to have a vector space which has only scalar endomorphisms, this implies of course that there are no linear functionals, since those would induce non-scalar automorphism of the vector space. – Asaf Karagila Nov 14 '11 at 18:30
up vote 15 down vote accepted

To complement François' answer, here's a fairly explicit example of a real vector space admitting no nonzero linear functional in a model of ZFDC + all sets of reals have the Baire property (which is equiconsistent with ZF).

The space is $\mathbb{R}^\mathbb{N} / E_1$, where $E_1$ is the equivalence relation of eventual agreement of real sequences. This can be given a real vector space structure in the obvious way. Now, if $f: \mathbb{R}^\mathbb{N} / E_1 \to \mathbb{R}$ is a linear function, it descends to a linear function $g: \mathbb{R}^\mathbb{N} \to \mathbb{R}$ by $g(x) = f([x]_{E_1})$. Since $g$ is Baire measurable and $E_1$-invariant, by generic ergodicity of $E_1$ it is constant on a comeager set. This constant has to be $0$. But then $g$ is a Polish group homomorphism and is automatically continuous, and thus equals $0$ everywhere.

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Thanks, Clinton, that's a very neat construction. – François G. Dorais Nov 14 '11 at 18:57
Thanks, François and Clinton, for two very informative answers, and thank you, Asaf, for clarifying what François said. I accept Clinton's answer because I can (almost) understand the proof. – Bill Johnson Nov 14 '11 at 19:12

Hans Läuchli [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, MR0143705, DOI:10.5169/seals-28602] has constructed a model of ZFA wherein there is vector space (over any given field, see comments) which is not finite dimensional but all of its proper subspaces are finite dimensional. In particular, there cannot be any nontrivial linear functionals since the kernel of such a linear functional would have codimension 1.

This can be transferred to ZF via the Jech-Sochor Embedding Theorem.

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Francois: This is not accurate. Lauchli only proved the case where there are infinitely many atoms, and the field is countable. Of course the generalization to other fields is almost trivial. It is not true, however, that his proof does that. – Asaf Karagila Nov 14 '11 at 18:26
Thanks for the correction, Asaf. I did not remember that hypothesis from the paper. Do you remember where he uses that hypothesis? (It doesn't seem that useful to assume countability, but I might be overlooking some details.) – François G. Dorais Nov 14 '11 at 18:44
He does not uses this hypothesis directly. However since the model he is using is not "out of the box" model of ZFA he can only assume that he has infinitely many "atoms" but cannot determine how many. Since the vector space must have at least the cardinality of the field, he cannot assure that he has enough atoms for larger fields than any countable field. However nowadays (as in my proofs on both threads mentioned in the comments to the questions) we can simply assume to have enough atoms for a vector space over a fixed field. – Asaf Karagila Nov 14 '11 at 18:49
I see. Thanks for the info, I'll try to remember that next time I refer to this paper. – François G. Dorais Nov 14 '11 at 18:54
Läuchli's striking example reminds me of extreme rigidity results about topological vector spaces in ZFC. – Bill Johnson Nov 14 '11 at 19:00

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