# If a vector space has a basis then its dual vector space has a basis

Consider the following statement:

If a vector space has a basis then its dual vector space also has a basis.

It is not an axiom of ZF. It clearly follows from the Axiom of Choice. But it is also strictly weaker than the Axiom of Choice.

I have convinced myself that this statement holds in a variety of models of set theory with atoms (e.g. it holds in the first Fraenkel-Mostowski model, in the second Fraenkel-Mostowski model, in the ordered Fraenkel-Mostowski model, etc.).

I am wondering:

What is the strength of the above statement?

I am also interested in any reasonable variant of the question --- e.g. what if we restrict to a particular field $$\mathbf{K}$$? In case $$\mathbf{K} = 2$$, the statement can be rephrased as:

For any set $$A$$ the vector space $$\mathcal{P}(A)$$ has a basis.

Can the general statement be reduced to this special one over two-element field $$2$$?

Let us work out some examples in ZFA. For simplicity, I shall assume that our vector spaces are over the field of real numbers.

1. Let $$A$$ be the set of all atoms in the first Fraenkel-Mostowski model. Then the vector space $$\mathbb{R}^A$$ consists of all symmetric functions $$f \colon A \rightarrow \mathbb{R}$$. Let us assume that the support of $$f$$ is $$A_0$$. Then $$f$$ has to be constant on $$A \setminus A_0$$ and arbitrary on $$A_0$$. Therefore, the set of vectors: $$\{1, a_1^*, a_2^*, \cdots \} \approx A \sqcup 1$$ where:
• 1 is the constant function, i.e. $$1(a)=1$$
• $$a^*$$ is the characteristic function, i.e. $$a^*(b) = [a = b]$$

is a basis for $$\mathbb{R}^A$$

1. Let $$Z_*$$ be the set of all atoms in the second Fraenkel-Mostowski model (thought of as non-zero integer numbers with symmetry given as multiplication by $$-1$$). A symmetric function $$f \colon Z_* \rightarrow \mathbb{R}$$ must satisfy $$f(-n) = f(n)$$ for all but finitely many $$n$$. Let us consider classical (i.e.~in the real world) vector space $$\mathbb{R}^{\mathcal{N}_*}$$, where $$\mathcal{N}_*$$ is the set of positive natural numbers. Let us consider standard vectors $$e_1, e_2, \cdots$$ and extend them (in any way) to a basis of $$\mathbb{R}^{\mathcal{N}_*}$$. Let us denote the set of these additional vectors by $$\Lambda$$, i.e. $$\{e_1, e_2, \cdots\} \cup \Lambda$$ is a basis for $$\mathbb{R}^{\mathcal{N}_*}$$. For every vector $$\lambda \in \Lambda$$ define a symmetric function $$\overline{\lambda} \colon Z_* \rightarrow \mathbb{R}$$ as follows: $$\overline{\lambda}(z) = \lambda(|z|)$$ I claim that the set of vectors: $$\{\overline{\lambda} \colon \lambda \in \Lambda\} \cup \{z^* \colon z \in Z_*\} \approx \Lambda \sqcup Z_*$$ forms a basis for $$\mathbb{R}^{Z_*}$$ Indeed, it is clear that this set generates $$\mathbb{R}^{Z_*}$$. To see that the vectors are linearly independent consider the following two cases:
• $$\overline{\lambda} = \sum_{i=1}^k a_i\overline{\lambda_i} + \sum_{i=1}^l b_i z^*_i$$ where all vectors are distinct. In particular, this implies that: $$\overline{\lambda}(n) = \sum_{i=1}^k a_i\overline{\lambda_i}(n) + \sum_{i=1}^l b_i z^*_i(n)$$ for every positive natural number $$n$$. But this reduces to: $$\lambda(n) = \sum_{i=1}^k a_i \lambda_i(n) + \sum_{i=1}^l b_i e_{z_i}(n)$$ where $$e_{z_i} \equiv 0$$ if $$z_i < 0$$. But this leads to the contradiction with the assumption that $$\{e_1, e_2, \cdots\} \cup \Lambda$$ are linearly independent.

• $$z^* = \sum_{i=1}^k a_i\overline{\lambda_i} + \sum_{i=1}^l b_i z^*_i$$ where all vectors are distinct and $$z < 0$$ (the case $$z> 0$$ is symmetric). In particular, this implies that: $$z^*(-n) = \sum_{i=1}^k a_i\overline{\lambda_i}(-n) + \sum_{i=1}^l b_i z^*_i(-n)$$ for every positive natural number $$n$$. This reduces to: $$e_{-z}(n) = \sum_{i=1}^k a_i \lambda_i(n) + \sum_{i=1}^l b_i e_{-z_i}(n)$$ This leads to the contradiction with the assumption that $$\{e_1, e_2, \cdots\} \cup \Lambda$$ are linearly independent.

1. Let $$Q$$ be the set of all atoms in the ordered Fraenkel-Mostowski model. Consider a symmetric function $$f \colon Q \rightarrow \mathbb{R}$$. Then there is a finite decomposition $$Q = I_1 \sqcup I_2 \sqcup \cdots I_n$$ on intervals $$I_k$$, such that $$f$$ is constant on each $$I_k$$. Therefore, the set of vectors: $$\{1, p_1^*, p_2^*, \cdots, p_1^{<}, p_2^{<}, \cdots \} \approx Q \sqcup Q \sqcup 1$$ where:
• 1 is the constant function, i.e. $$1(a)=1$$
• $$p^*$$ is the characteristic function, i.e. $$p^*(q) = [p = q]$$
• $$p^{<}$$ is the open down set of $$p$$, i.e. $$p^{<}(q) = [p > q]$$

generates $$\mathbb{R}^Q$$. Moreover, these vectors are linearly independent: if $$S$$ is any non-empty finite set of the above vectors, then there exists vector $$m \in S$$ and an atom $$p$$ with $$m(p) = 1$$ such that for every $$s \in S \setminus \{m\}$$ we have that $$s(p) = 0$$, so $$m$$ cannot be a linear combination of $$S \setminus \{m\}$$.

• Since you mentioned power sets as $\mathbb{F}_2$-vector spaces: Have you investigated how this compares to the ultrafilter principle? Jun 23, 2021 at 11:35
• If you are interested in bases of vector spaces, working with permutation models is the wrong way to go, since "Every vector space has a basis" might be weaker than AC in ZFA. Jun 23, 2021 at 11:43
• @JohannesHahn, in the first Fraenkel-Mostowski model UF does not hold, whereas in the ordered Fraenkel-Mostowski model UF principle holds. On the other hand, in both of these models the statement from the question holds. Jun 23, 2021 at 12:05
• @AsafKaragila, in both the first and the ordered Fraenkel-Mostowski model MC does not hold. So, I do not think it should be a big issue. Jun 23, 2021 at 12:10
• Yes, but how do you plan on moving the statement to ZF from ZFA? In ZFA the power set of an ordinal can be well-ordered (at least in the standard "ZFC holds in the pure part" situation). In ZF the power set of an ordinal will fail to be well-orderable if AC failed. Especially when asking about power sets, this seems to be at least somewhat relevant. Jun 23, 2021 at 12:13

## 1 Answer

Here are two statements you might find interesting:

1. With base field $$k\subseteq\mathbb C$$ your axiom implies that any free $$\mathbb Z$$-action has a choice of representatives.
2. For each prime $$p,$$ there is an $$\omega$$-categorical theory such that the associated permutation model contains a set $$X$$ with no basis for $$\mathbb F_p^X$$ - this addresses a suggestion in the comments.

I’ll use ZFA throughout, blithely ignoring the question of whether your axiom implies AC over ZF.

A free $$\mathbb Z$$-action on a set $$X$$ is (generated by) a bijection $$T:X\to X$$ such that $$T^ix=x$$ implies $$i=0.$$ Let $$k\subseteq \mathbb C.$$ Let $$X$$ be a set such that $$k^X$$ has a basis $$\mathcal B.$$ We will show that every free $$\mathbb Z$$-action on $$X$$ has a choice of orbit representatives.

Why might this be interesting? It seems like a non-trivial consequence because the conclusion does not directly refer to linear algebra. The conclusion for all $$X$$ implies the axiom $$\omega\zeta\mathrm{AC}$$ introduced in [1], Howard-Rubin Form 119, which is equivalent to: every free $$\mathbb Z$$-action with a countable number of orbits has a set of representatives. It might not seem like a huge amount of strength, but then the evidence you’ve gathered suggests your axiom is quite weak (not pejoratively!), at least over ZFA.

Restricting to a single orbit, we are given a free transitive $$\mathbb Z$$-action generated by $$T:A\to A$$ and we want to pick an element of $$A,$$ uniformly in $$(A,T).$$

Let $$\delta_b\in (k^X)^*$$ for $$b\in\mathcal B$$ be the dual vectors, so every $$v\in k^X$$ is a unique $$k$$-linear sum $$\sum_{b\in\mathcal B} \delta_b(v) b$$ with finitely many non-zero coefficients. Let $$[A]$$ be the indicator function of $$A.$$ For each $$a\in A$$ define $$f_a\in k^X$$ by $$f_a(T^na)=n$$ and $$f_a(x)=0$$ for $$x\not\in A.$$ Define $$c(a)=\sum_{b\in \mathcal B} |\delta_b(f_a)|.$$ Then $$c(T^na)\to\infty$$ as $$|n|\to\infty$$ for fixed $$a,$$ because $$|\delta_b(f_{T^na})| = |\delta_b(f_a) - n\delta_b([A])|\to\infty$$ for any $$b$$ such that $$\delta_b([A])\neq 0.$$

The set $$C=\operatorname{argmin} c\subset A$$ is therefore finite, and we can take the least element in the ordering generated by $$a $$\square$$

Take the $$\omega$$-categorical theory to be the Fraïssé limit of the class of bilinear forms $$B:X\times \Phi\to \mathbb F_p,$$ where $$X$$ and $$\Phi$$ are finite-dimensional vectors spaces over $$\mathbb F_p.$$ These are structures in the language with unary symbols $$X,\Phi,\mathbb F_p,$$ field operations, two sets of vector space operations, and a binary operation for $$B.$$ The finite support permutation model $$N$$ of this structure will therefore contain a bilinear form which I’ll also call $$B:X\times\Phi\to\mathbb F_p.$$

I’ll need a “unique minimal support” property. Unique support arguments often require a lot of care. To make life as easy as possible I’ve tried to prove only the exact statement that the argument needs. Define an “acl support” to be a pair $$(X_0,\Phi_0)$$ where $$X_0$$ is a finite subspace of $$X$$ and $$\Phi_0$$ is a finite subspace of $$\Phi.$$ (“acl” stands for algebraically closed, the same as definably closed here). A set $$x\in N$$ is “supported by” $$(X_0,\Phi_0)$$ if every automorphism that fixes $$(X_0,\Phi_0)$$ elementwise also fixes $$x.$$

Lemma. Every $$f\in \mathbb F_p^X$$ in $$N$$ is supported by a unique minimal acl support $$\operatorname{supp}(f).$$

Proof: $$f$$ is supported by $$(X_0,\Phi_0)$$ if and only if: $$f(x)=f(x’)$$ whenever $$x$$ and $$x’$$ have the same $$1$$-type over $$(X_0,\Phi_0).$$ These $$1$$-types (extending the partial type of elements of $$X$$) have an explicit description: there are exactly $$|X_0|+|\Phi_0|.$$ corresponding to the points of $$X_0$$ and the $$1$$-types defined by the formula $$(\bigwedge_{x_0\in X_0}x\neq x_0)\wedge (\bigwedge_{\phi\in\Phi_0}B(x,\phi)=\hat x(\phi))$$ for each $$\hat x\in \Phi_0^*.$$

Suppose $$f$$ is supported by both $$(X_0,\Phi_0)$$ and $$(X_1,\Phi_1).$$ We will show that $$f$$ is supported by the intersection of these supports, which clearly implies the Lemma. Consider $$x,x’\in X$$ with the same $$1$$-type over $$(X_0\cap X_1,\Phi_0\cap \Phi_1)$$; we want to show $$f(x)=f(x’).$$ The case $$x\in X_0\cap X_1$$ is immediate. If $$x\in X_0\setminus X_1$$ then we can replace $$x$$ by $$x’’$$ where $$x’’$$ has the same $$1$$-type over $$(X_1,\Phi_1)$$ but $$x’’\not\in X_0.$$ Since $$f$$ is supported by $$(X_1,\Phi_1),$$ we know $$f(x)=f(x’’).$$ So we can reduce to the case $$x\not\in X_0\cup X_1,$$ and by a similar argument we can reduce to the case $$x’\not\in X_0\cup X_1.$$

Now pick $$x’’\in X\setminus (X_0\cup X_1)$$ such that $$B(x’’,\phi)=B(x,\phi)$$ for all $$\phi\in\Phi_0,$$ and $$B(x’’,\phi)=B(x’,\phi)$$ for all $$\phi\in\Phi_1.$$ Then $$f(x)=f(x’’)$$ because $$f$$ is supported by $$(X_0,\Phi_0).$$ And $$f(x’’)=f(x’)$$ because $$f$$ is supported by $$(X_1,\Phi_1).$$ Therefore $$f(x)=f(x’’)$$ as required. $$\square$$

Corollary. If a finite set $$\mathcal S\subset\mathbb F_p^X$$ is supported by $$(X_0,\Phi_0),$$ then every element of $$\mathcal S$$ is supported by $$(X_0,\Phi_0).$$

Proof. Let $$Z=\bigcup_{f\in\mathcal S} \operatorname{supp}(f).$$ Then $$Z$$ is a finite subset of $$X\cup \Phi$$ supported by $$(X_0,\Phi_0).$$ Suppose for contradiction that there exists $$z\in Z\setminus (X_0\cup \Phi_0).$$ There are infinitely many $$z’$$ with the same $$1$$-type as $$z$$ over $$(X_0,\Phi_0).$$ So we can find an automorphism fixing $$(X_0,\Phi_0)$$ elementwise but sending $$z$$ to some $$z’\not\in Z,$$ a contradiction. $$\square$$

Suppose for contradiction that $$N$$ has a basis $$\mathcal B$$ of $$\mathbb F_p^X.$$ Work in the full universe of ZFCA. $$\mathcal B$$ is supported by some $$(X_0,\Phi_0),$$ which can be taken to be vector subspaces but are not assumed minimal or unique in any sense.

Pick two elements of $$\Phi$$ that map to linearly independent vectors in $$\Phi/\Phi_0.$$ Their span $$\Phi_1$$ has dimension $$2.$$ Let $$L$$ be the set of $$p+1$$ one-dimensional subspaces of $$\Phi_1.$$ Each $$\ell\in L$$ determines a function $$f_{\ell}\in \mathbb F_p^X$$:

• $$f_{\ell}(x)=0$$ if $$B(x,\phi)=0$$ for all $$\phi\in \ell$$
• $$f_{\ell}(x)=1$$ otherwise

For any $$x\in X$$ the kernel $$\{\phi\in\Phi_1: B(x,\phi)=0\}$$ either has dimension $$1,$$ in which case $$f_\ell(x)=0$$ for exactly one $$\ell,$$ or has dimension $$2,$$ in which case $$f_\ell(x)=0$$ for all $$\ell.$$ Thus $$\sum f_\ell = 0.$$

Let $$F_0\subset \mathbb F_p^X$$ denote the space of functions that are supported by $$(X_0,\Phi_0).$$ Let $$F_\ell$$ denote the space of functions that are supported by $$(X_0,\operatorname{span}(\Phi_0\cup \ell)).$$ Then $$F_\ell\cap F_{\ell’}=F_0$$ for distinct $$\ell,\ell’$$ - consider the unique minimal acl support of functions in $$F_\ell\cap F_{\ell’}.$$

Let $$\delta_b(f_\ell)$$ denote the coefficient of the basis function $$b\in\mathcal B$$ when expanding $$f_\ell$$ in the basis $$\mathcal B.$$ $$f_\ell$$ is not in the span of $$F_0,$$ because it is possible to find $$x,x’$$ satisfying $$f_\ell(x)=0$$ and $$f_\ell(x’)=1$$ with the same $$1$$-type over $$(X_0,\Phi_0),$$ specifically $$x,x’\not\in X_0$$ and $$B(x,\phi)=B(x’,\phi)=0$$ for all $$\phi\in\Phi_0.$$ By the Corollary applied to the set $$\{b:\delta_b(f_\ell)\neq 0\},$$ every $$b$$ with $$\delta_b(f_\ell)\neq 0$$ lies in $$F_\ell.$$ Pick any $$\ell$$ and $$b\in F_\ell\setminus F_0$$ with $$\delta_b(F_\ell)\neq 0.$$ For $$\ell’\neq\ell,$$ we have $$b\not\in F_{\ell’}$$ and hence $$\delta_b(f_{\ell’})=0.$$ We can finally state the contradiction to the assumption of a basis of $$\mathbb F_p^X$$: $$0\neq \delta_b(f_\ell)=\delta_b(\sum_{\ell’} f_{\ell’}) = 0.$$

[1] van Douwen, Eric K., Horrors of topology without AC: A nonnormal orderable space, Proc. Am. Math. Soc. 95, 101-105 (1985). ZBL0574.03039.

• thank you (+1) --- this looks very interesting to me, but I need a few moments to dig through the details. Jul 23, 2021 at 14:35