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](https://mathoverflow.net/questions/5351/whats-an-example-of-a-space-that-needs-the-hahn-banach-theorem/5477#5477). I'm not aware of any result for the algebraic dual.


This question was inspired by, and is related to [this question on MO](https://mathoverflow.net/questions/49351/does-the-fact-that-this-vector-space-is-not-isomorphic-to-its-double-dual-require).




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**Edit: Summary of the five answers so far:**


* Todd's answer + comments by François and Asaf:
*in [Läuchli's models of ZF](http://www.ams.org/mathscinet-getitem?mr=763890) 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](http://www.ams.org/mathscinet-getitem?mr=768264) of ZF + [DC](http://en.wikipedia.org/wiki/Dependent_choice) + 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](http://en.wikipedia.org/wiki/Compactness_theorem#Proofs): all vector spaces over finite fields have duals large enough to separate points.*


So DC is too weak, and 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 $\kappa$.



Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?
To add the proof for my claim in Todd's answer, which essentially repeats Läuchli's original [1] arguments with minor modifications (and the addition that the resulted model satisfies $DC\_\kappa$).


We will show that it is consistent to have a model in which $DC\_\kappa$ holds, and there is a vector space over $\mathbb F\_2$ which has no linear functionals.




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Assume that $M$ is a model of $ZFA+AC$ and that $A$, the set of atoms has $\lambda>\kappa$ many atoms, where $\lambda$ is a regular cardinal. Endow $A$ with a structure of a vector space over $\mathbb F=\mathbb F\_2$. Now consider the permutation model $\frak M$ defined by the group of linear permutations of $A$, and by ideal of supports generated by subsets of dimension $\le\kappa$.


Denote by $\operatorname{fix}(X)$ the permutations which fix every element of $X$, by $\operatorname{sym}(X)$ the permutations that fix $X$ as a set, and by $[E]$ the span of $E$ as a subset of $A$. We say that $E\subseteq A$ is *a support* of $X$ if $\pi\in\operatorname{fix}(E)\Rightarrow\pi\in\operatorname{sym}(X)$.


Final word of terminology, since $A$ will play both the role of set of atoms as well the vector space, given $U\subseteq A$ *the complement* will always denote a set complement, whereas *the direct complement* will be used to refer to a linear subspace which acts as a direct summand with $U$ in a decomposition of $A$.


**Claim 1:** If $E$ is a subset of $A$ then $\operatorname{fix}(E)$ is the same as $\operatorname{fix}([E])$.


*Proof:* This is obvious since all the permutations considered are linear. $\square$


From this we can identify $E$ with its span, and since (in $M$) the $[E]$ has the same cardinality of $E$ we can conclude that without loss of generality supports are subspaces.


**Claim 2:** $\frak M$$\models DC\_\kappa$.


*Proof:* Let $X$ be some nonempty set, and $\lt$ a binary relation on $X$, both in $\frak M$. In $M$ we can find a function $f\colon\kappa\to X$ which witness $DC\_\kappa$ in $V$.


Since $\frak M$ is transitive, we have that $\alpha,f(\alpha)\in\frak M$ and thus $\langle\alpha,f(\alpha)\rangle\in\frak M$. Let $E\_\alpha$ be a support for $\lbrace\langle\alpha,f(\alpha)\rangle\rbrace$ then $\bigcup\_{\alpha<\kappa} E\_\alpha$ is a set of cardinality $<\kappa^+$ and thus in our ideal of suports. It is simple to verify that this is a support of $f$, therefore $f\in\frak M$ as wanted. $\square$


**Claim 3:** If $x,y\in A$ are nonzero (with respect to the vector space) then in $M$ there is a linear permutation $\pi$ such that $\pi x=y$ and $\pi y=x$.


*Proof:* Since $x\neq y$ we have that they are linearly independent over $\mathbb F$. Since we have choice in $M$ we can extend this to a basis of $A$, and take a permutation of this basis which only switches $x$ and $y$. This permutation extends uniquely to our $\pi$.


**Claim 4:** If $U\subseteq A$ and $U\in\frak M$ then either $U$ is a subset of a linear subspace of dimension at most $\kappa$, or a subset of the complement of such space.


*Proof:* Let $E$ be a support of $U$, then every linear automorphism of $A$ which fixes $E$ preserves $U$. If $U\subseteq [E]$ then we are done, otherwise let $u\in U\setminus [E]$ and $v\in A\setminus [E]$, we can define (in $M$ where choice exists) a linear permutation $\pi$ which fixes $E$ and switches $u$ with $v$. By that we have that $\pi(U)=U$ therefore $v\in U$, and so $U=A\setminus[E]$ as wanted. $\square$


**Claim 5:** If $U\subseteq A$ is a linear proper subspace and $U\in\frak M$ then its dimension is at most $\kappa$.


*Proof:* Suppose that $U$ is a subspace of $A$ and every linearly independent subset of $U$ of cardinality $\le\kappa$ does not span $U$, we will show $A=U$. By the previous claim we have that $U$ is the complement of some "small" $[E]$. 


Now let $v\in A$ and $u\in U$ both nonzero vectors. If $u+v\in U$ then $v\in U$. If $u+v\in [E]$ then $v\in U$ since otherwise $u=u+v+v\in[E]$. Therefore $v\in U$ and so $A\subseteq U$, and thus $A=U$ as wanted.$\square$


**Claim 6:** If $\varphi\colon A\to\mathbb F$ a linear functional then $\varphi = 0$.


*Proof:* Suppose not, for some $u\in A$ we have $\varphi(u)=1$, then $\varphi$ has a kernel which is of co-dimension $1$, that is a proper linear subspace and $A=\ker\varphi\oplus\lbrace 0,u\rbrace$. However by the previous claim we have that $\ker\varphi$ has dimension $\kappa$ at most, and without the axiom of choice $\kappa+1=\kappa$, thus deriving contradiction to the fact that $A$ is not spanned by $\kappa$ many vectors.




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Aftermath: There was indeed some trouble in my original proof, after some extensive work in the past two days I came to a very similar idea. However with the very generous help of [Theo Buehler](https://mathoverflow.net/users/11081/theo-buehler) which helped me find the original paper and translate parts, I studied Läuchli's original proof and concluded his arguments are sleek and nicer than mine.


While this cannot be transferred to $ZF$ using the Jech-Sochor embedding theorem (since $DC\_\kappa$ is not a bounded statement), I am not sure that Pincus' transfer theorem won't work, or how difficult a straightforward forcing argument would be.


Lastly, the original Läuchli model is where $\lambda=\aleph\_0$ and he goes on to prove that there are no non-scalar endomorphisms. In the case where we use $\mathbb F=\mathbb F\_2$ and $\lambda=\aleph\_0$ we have that this vector space is indeed amorphous which in turn implies that very little choice is in such universe.


**Bibliography:**


1. Läuchli, H. **[Auswahlaxiom in der Algebra.](http://dx.doi.org/10.5169/seals-28602)** *Commentarii Mathematici Helvetici*, vol 37, pp. 1-19.


