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If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The existence of a non-zero linear functional can be shown by taking a basis of $V$ and specifying the values of the functional on the basis.

To find a basis of $V$, the axiom of choice (AC) is needed, and indeed, it was shown by Blass in 1984 that in Zermelo-Fraenkel set theory (ZF) it is equivalent to the axiom of choice that any vector space has a basis. However, it's not clear to me that the existence of a non-zero element of $V^*$ really needs the full strength of AC. I couldn't find a reference anywhere, so here is my question:

Consider the following statement:

(D) For any vector space $V$ that is not finite-dimensional, $V^*\neq \{0\}$.

Is (D) equivalent to AC in ZF? If not, is there some known axiom that is equivalent to (D) in ZF?

Note that this question is about the algebraic dual $V^*$. There are examples of Banach spaces, for example $\ell^\infty/c_0$, where it is possible (in the absence of the Hahn-Banach theorem, itself weaker than AC) for their topological dual to be $\{0\}$; see this answer on MO. I'm not aware of any result for the algebraic dual.

This question was inspired by, and is related to this question on MO.

Edit: Summary of the three five answers so far:

• Todd's answer + comment comments by François and Asaf: in Blass' model Läuchli's models of ZF there is an infinite dimensional vector space $V$ such that all proper subspaces are finite dimensional. In particular, $V$ does not have a basis and $V^*=\{0\}$. Also, according to Asaf, in these models Dependent Choice can still hold up to an arbitrarily large cardinal.

• Amit's answer + comment by François: in Shelah's model of ZF + DC + PB (every set of real numbers is Baire), $\Bbb R$ considered as a vector space over $\Bbb Q$ has a trivial dual.

• François's answer : (see also godelian's answer) + Andreas' answer in ZF + the following is equivalent to BPIT, : all vector spaces over finite fields have duals large enough to separate points.

So DC is too weak, and BPIT BPT is strong enough for finite fields (and in fact equivalent to a slightly stronger statement). How far does Choice fail in Blass' model? Update: according to Asaf Karagila, $DC_{\kappa}$ can hold for arbitrarily large $\kappa$.

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If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The existence of a non-zero linear functional can be shown by taking a basis of $V$ and specifying the values of the functional on the basis.

To find a basis of $V$, the axiom of choice (AC) is needed, and indeed, it was shown by Blass in 1984 that in Zermelo-Fraenkel set theory (ZF) it is equivalent to the axiom of choice that any vector space has a basis. However, it's not clear to me that the existence of a non-zero element of $V^*$ really needs the full strength of AC. I couldn't find a reference anywhere, so here is my question:

Consider the following statement:

(D) For any vector space $V$ that is not finite-dimensional, $V^*\neq \{0\}$.

Is (D) equivalent to AC in ZF? If not, is there some known axiom that is equivalent to (D) in ZF?

Note that this question is about the algebraic dual $V^*$. There are examples of Banach spaces, for example $\ell^\infty/c_0$, where it is possible (in the absence of the Hahn-Banach theorem, itself weaker than AC) for their topological dual to be $\{0\}$; see this answer on MO. I'm not aware of any result for the algebraic dual.

This question was inspired by, and is related to this question on MO.

Edit: Summary of the three answers so far:

• Todd's answer + comment by François: in Blass' model of ZF there is an infinite dimensional vector space $V$ such that all proper subspaces are finite dimensional. In particular, $V$ does not have a basis and $V^*=\{0\}$.

• Amit's answer + comment by François: in Shelah's model of ZF + DC + PB (every set of real numbers is Baire), $\Bbb R$ considered as a vector space over $\Bbb Q$ has a trivial dual.

• François's answer: in ZF + BPIT, all vector spaces over finite fields have duals large enough to separate points.

So DC is too weak, and BPIT is strong enough for finite fields. How far does Choice fail in Blass' model?

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