For every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , can we prove the existence without the "axiom of choice " ? Is the existence equivalent to axiom of choice ?

  • $\begingroup$ do you see how to do it with AC? $\endgroup$ – mm-aops Oct 12 '14 at 14:13
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    $\begingroup$ Of course, taking a basis, we can define such a linear functional. This uses axiom of choice though, for infinite dimensional spaces. I'm not sure about doing so without the axiom of choice. $\endgroup$ – Seth Oct 12 '14 at 14:16
  • $\begingroup$ @Nate: Thanks. :P $\endgroup$ – Asaf Karagila Oct 12 '14 at 15:00
  • $\begingroup$ Dear @Seth, given a $\mathbb K$-space with basis $\mathcal{B}$, I think the map $x=\sum k_ib_i \mapsto \sum k_i$ is a nontrivial linear functional. Here, $\sum k_ib_i$ is the representation of $x$ using this basis. Am I right? $\endgroup$ – rgm May 5 '18 at 21:25

No, you can't prove this without the axiom of choice.

Given any field, $F$, we can extend the universe of set theory, such that over $F$ there is a vector space which is not finitely generated, but every subspace is finitely generated. In particular, this means that there are no linear functionals from this vector space to the field (because a linear functional would have a kernel whose codimension is $1$).

But we can do even more. We can even show that there might not be a linear functional from $\Bbb R$ to $\Bbb Q$ in some models where the axiom of choice fails. In particular in models where we have automatic continuity (e.g. models where all sets of reals are Lebesgue measurable; or all sets of reals have the Baire property). In such models if $\varphi\colon\Bbb R\to\Bbb Q$ is a group homomorphism then it has to be continuous, and therefore $0$.

Finally, whether or not the existence of nontrivial functionals may or may not imply the axiom of choice. To my knowledge this is still open. I'd imagine that the answer would be "almost positive", in the sense that it's negative but some augmentation of this (e.g. every non-zero vector has a linear functional mapping it to $1$) might turn out equivalent to the axiom of choice.


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