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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 five answers so far:

  • Todd's answer + comments by François and Asaf: in 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 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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This is probably a naive question, but is there no chance that (D) is provable in ZF? –  Thierry Zell Dec 14 '10 at 14:14
    
@Thierry: I don't know. If (D) holds in ZF, then it would be a very obscure fact, because I've seen in many texts it explicitly mentioned that AC is needed to prove (D), as one needs a basis. So I doubt that it can be done in ZF, but I would also be very surprised if the full AC is needed. –  Konrad Swanepoel Dec 14 '10 at 14:23
    
Great question! –  Amit Kumar Gupta Dec 14 '10 at 18:49

6 Answers 6

up vote 8 down vote accepted

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.


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.


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 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. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.
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Nice! Thanks for adding as an answer. –  Konrad Swanepoel Nov 3 '11 at 11:13

This is a very partial answer (really in response to Thierry's question) which indicates that it is not provable in ZF that $V^\ast \neq \{0\}$ for every vector space $V$. This answer piggybacks on an answer Andreas Blass gave here, which gives a model of ZF in which the automorphism group of a vector space over $\mathbb{F}_2$ can be the cyclic group of order 3, which is really quite exotic.

So, I will prove that if every vector space $V$ over $\mathbb{F}_2$ (of dimension greater than 1) has a nontrivial dual, then $V$ has a nontrivial involution, which would run counter to Andreas's model. Indeed, suppose there exists a surjective linear map $f: V \to \mathbb{F}_2$. There exists an element $x \in V$ such that $f(x) = 1$. There also exists a surjective map $V/\langle x \rangle \to \mathbb{F}_2$, hence a surjective map $g: V \to \mathbb{F}_2$ such that $g(x) = 0$, and thus there exists an element $y \in \ker(f)$ such that $g(y) = 1$. It follows that we have a surjective linear map

$$\langle f, g \rangle: V \to \mathbb{F}_2 \times \mathbb{F}_2$$

say with kernel $W$. This epimorphism splits, so we have an identification

$$V \cong W \oplus \mathbb{F}_{2}^{2}$$

and clearly now we can exhibit a non-identity involution on the right side which acts as the identity on $W$ and permutes two basis elements of the 2-dimensional summand.

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Awesome! Thanks, Todd. –  Thierry Zell Dec 15 '10 at 2:43
    
Very nice. Do you know how badly AC fails in Blass's model? –  Konrad Swanepoel Dec 15 '10 at 11:42
5  
The permutation model used by Andreas is in fact due to H. Läuchli. [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37 1962/1963] This is the standard model for the existence of a vector space without a basis. (Note that there is nothing special about $\mathbb{F}_2$ or rather $\mathbb{F}_4$ in the construction.) The vector space in question has the curious property that all of its proper subspaces are finite dimensional; this gives a very quick proof of the desired result. –  François G. Dorais Dec 15 '10 at 12:21
2  
@Konrad: Something that I am working on nowadays shows that in Lauchli models it is consistent to have $DC_\kappa$ for arbitrarily high $\kappa$. I'm not sure that Todd's proof will carry though. –  Asaf Karagila Jun 6 '11 at 19:45
1  
@Konrad: Of course it carries over! It is very simple, a simple generalization of Blass construction (generalizing Lauchli) still yields a vector spaces without non-scalar automorphisms, and from here the proof by Todd carries over completely. –  Asaf Karagila Jun 6 '11 at 19:48

Some restricted forms of (D) are weaker than the Axiom of Choice. Fix a field $F$ and consider the stronger statement:

For every $F$-vector space $V$ and every nonzero $v_0 \in V$ there is a $F$-linear functional $f:V\to F$ such that $f(v_0) = 1$.

When $F$ is a finite field, this is a consequence of the Ultrafilter Theorem or, equivalently, the Compactness Theorem for propositional logic.

To see this, consider the following propositional theory with one propositional variable $P(v,x)$ for each pair $v \in V$ and $x \in F$. The idea of the theory is that $P(v,x)$ should be true if and only if $f(v) = x$. The axioms for the theory are:

  1. $\lnot(P(v,x) \land P(v,y))$ for all $v \in V$ and distinct $x, y \in F$
  2. $\bigvee_{x \in F} P(v,x)$ all $v \in V$
  3. $P(v,x) \land P(w,y) \rightarrow P(v + w, x + y)$ for all $v, w \in V$ and $x, y \in F$
  4. $P(v,x) \rightarrow P(yv,yx)$ for all $v \in V$ and $x, y \in F$
  5. $P(v_0,1)$

Axiom schemes 1 & 2 ensure that the $P(v,x)$ describe the graph of a function $f:V \to F$. Axiom schemes 3 & 4 ensure that the function $f$ thus described is $F$-linear. Finally, the last axiom 5 ensures that $f(v_0) = 1$. It is clear that every finite subset of the axioms is satisfiable, therefore, by the Compactness Theorem, the whole theory is satisfiable.

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Very interesting approach. I've been wondering how to adapt this to an infinite field, but it seems the big problem is axiom 2. So if I try to do it for e.g. $F=\Bbb F_2(x)$, then I need an existential quantifier over the natural numbers. –  Konrad Swanepoel Dec 15 '10 at 13:50

It seems that very little is known about the strength of statements like this. A few years ago, I encountered (in a pedagogical context) the slightly stronger looking statement "every non-zero vector in any vector space has a non-zero image under some element of the dual space." Looking in the canonical place, "Consequences of the Axiom of Choice" by Howard and Rubin, I didn't find this statement in the section on vector spaces, but I found an equivalent statement in the section on fields, form 284: A system of linear equations over a field $F$ has a solution in $F$ if and only if every finite sub-system has a solution in $F$.

Howard and Rubin also have the version of 284 restricted to finite fields; it's 14BN, where the 14 indicates it's equivalent to the Boolean Prime Ideal Theorem. Unfortunately, there seems to be essentially no more information about 284.

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I just came across this page on Timothy Gowers' website wherein he states:

there are models of set theory without the axiom of choice that contain infinite-dimensional vector spaces V such that V* really is {0}.


Update: As Konrad points out, there's no reference on his page for the statement above. So I emailed Mr. Gowers and he gave me the following answer which I believe is correct:

Consider $V = \mathbb{R}$ regarded as a vector space over $\mathbb{Q}$. A non-trivial element of $V^{\ast}$ is a non-zero $\mathbb{Q}$-linear map from $\mathbb{R} \to \mathbb{Q}$. Such a map, being a non-trivial map to $\mathbb{Q}$, will not be $\mathbb{R}$-linear, but it would still be additive because it's $\mathbb{Q}$-linear. I believe it holds in ZF that every additive Lebesgue measurable map $\mathbb{R} \to \mathbb{R}$ is $\mathbb{R}$-linear (there's a proof of this statement here; I don't believe the Lebesgue density theorem requires choice$^{\dagger}$, and none of the other steps in the proof require choice). Thus an additive non-$\mathbb{R}$-linear map would give us a non-measurable function, and hence a non-measurable set. This argument so far hasn't used choice, so at this point it suffices to find a model of ZF with no non-measurable sets of reals: just take Solovay's model.

$^{\dagger}$ This is the only point I'm not sure of.

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Unfortunately he doesn't give any reference.... –  Konrad Swanepoel Dec 15 '10 at 11:41
    
The proof that additive measurable maps are linear uses $\sigma$-additivity (or at least $\sigma$-subadditivity); see the 4th last line of the proof you link to. I think Countable Choice is needed to show $\sigma$-subadditivity. But CC holds in Solovay's model (even Dependent Choice), so $\sigma$-additivity should hold there (otherwise what's the use talking about measurable sets?), so this proof should go through there. –  Konrad Swanepoel Dec 15 '10 at 12:49
    
Since the consistency of Solovay's model needs an inaccessible cardinal, this argument doesn't give the consistency of the negation of (D) relative to the consistency of ZF; you need the consistency of the existence of an inaccessible cardinal. –  Konrad Swanepoel Dec 15 '10 at 12:51
1  
@Konrad: If you use Baire category instead of Lebesgue measure, you can avoid the inaccessible. See this old answer of mine - mathoverflow.net/questions/16666/… –  François G. Dorais Dec 15 '10 at 12:54
    
@François: aha, very interesting; I missed that question. So to elaborate: Shelah has a model equiconsistent with ZF where all sets are Baire and where DC holds, and it is known in ZF+DC that an additive function $\Bbb R\rightarrow \Bbb R$ with the Baire property must be $\Bbb R$-linear (I found it as a special case of Theorem 6 of M. R. Mehdi, On convex functions, J. London Math. Soc. 39 (1964) 321–326, but it could be older.) –  Konrad Swanepoel Dec 15 '10 at 13:35

EDIT: According to François's comment below, this answer only works for finite fields instead of fields of positive characteristic, as I originally intended. I would like to leave it for a while in community wiki and see if someone else's can use these ideas to give a further step. END EDIT

Here is another proof that the ultrafilter theorem is enough to deduce statement (D) for vector spaces over finite fields. The idea is to use partial functionals defined on finite-dimensional subspaces and use a consistency principle to deduce the existence of a functional defined in the whole space. This can be done by using the following theorem, which is equivalent to the prime ideal theorem (and hence to the ultrafilter lemma):

THEOREM: Suppose for each finite $W \subset I$ there is a nonempty set $H_W$ of partial functions on I whose domains include $W$ and such that $W_1 \subseteq W_2$ implies $H_{W_2} \subseteq H_{W_1}$. Suppose also that, for each $v \in I$, {$h(v): h \in H_{\emptyset}$} is a finite set. Then there exists a function $g$, with domain $I$, such that for any finite $W$ there exists $h \in H_W$ with $g|_W \subseteq h$.

This is theorem 1 in this paper by Cowen, where he proves the equivalence with the prime ideal theorem (a simple proof using compactness for propositional logic is given by the end of the paper, and is close to what François had in mind). This is essentially also the "Consistency principle" as appearing in Jech's "The axiom of choice", pp. 17, since although Jech's formulation uses only two-valued functions, the proof he gives there, through the ultrafilter lemma, actually works when the functions are $n$-valued.

Now, for an infinite-dimensional vector space $V$ over a finite field, fix a nonzero $v_0 \in V$ and define the sets $H_W$ for $W \subset V$ as follows: if $W \subseteq U$ for a finite $U$, consider the set $S_{U}$ of all functionals defined on the (finite-dimensional) subspace generated by $U$ and such that $v_0 \in U \implies f(v_0)=1$. Then $H_W$ is the union of all $S_U$ for finite $U \supseteq W$. By the previous theorem, we have a function $f: V \to \mathbb{F}$, and the restriction property shows that $f$ is linear and $f(v_0)=1$.

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I don't understand the generalization. Unless the field really is $\mathbb{F}_p$, there won't be any nonzero functionals that only take values in $\mathbb{F}_p$. –  François G. Dorais Oct 16 '11 at 22:28
    
@François: You are completely right; f will only be linear with respect to multiplication by scalars in F_p. I totally missed this obvious fact. So in the end this is just really another proof for the finite case. I will edit accordingly and then delete the answer. –  godelian Oct 16 '11 at 22:43
    
Done. Also, I'm wondering if for finite fields statement (D) might not be actually equivalent to theorem 1 (and hence to BPI). Perhaps the proof in Cowen's paper can give some hints if that's the case. –  godelian Oct 16 '11 at 23:04

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