Notation 2.1. Let X be a set and also let F be a field.

If $f\in F^{X}$ , where $F^X$ is the set of all functions from X to F, then $\operatorname {\mathrm {{supp}}}(f)$ denotes the support of f, i.e., $\operatorname {\mathrm {{supp}}}(f)=\{x\in X:f(x)\neq 0_{F}\}$ , where $0_{F}$ is the additive identity of F.

If $A\subseteq X$ , then the element $\chi _{A}$ of $F^{X}$ denotes the characteristic function of A, i.e., $\chi _{A}(x)=1_{F}$ if $x\in A$ and $\chi _{A}(x)=0_{F}$ if $x\in X\setminus A$ , where $1_{F}$ is the multiplicative identity of F.

$[X]^{<\omega }$ denotes the set of finite subsets of X and, for $n\in \omega $ , $[X]^{n}$ denotes the set of n-element subsets of X.

Proof. (i) $\Rightarrow $ (ii) Let F be any field, and also let V be an infinite-dimensional vector space over F. Let $\mathcal {A}=\{W_{n}:n\in \omega \}$ be an AD2 family of infinite-dimensional subspaces of V with cardinality $\aleph _{0}$ (the mapping $n\mapsto W_{n}$ is a bijection). Let $U_{0}=W_{0}$ and for $n\in \omega \setminus \{0\}$ , let

$$ \begin{align*}U_{n}=W_{n}\setminus (W_{0}+\cdots+W_{n-1}).\end{align*} $$

Since $\mathcal {A}$ is AD2, it follows that for all $n\in \omega $ , $U_{n}\ne \emptyset $ ; in particular, $\langle U_{n}\rangle $ is an infinite-dimensional subspace of V for all $n\in \omega $ . By $\mathsf {MC}^{\aleph _{0}}$ , let f be a multiple choice function for the denumerable, disjoint family $\mathcal {B}=\{U_{n}:n\in \omega \}$ . Then

$$ \begin{align*}Z={\Big\langle}\bigcup\{f(U_{n}):n\in\omega\}{\Big\rangle}\end{align*} $$

is an infinite-dimensional subspace of V since, for every $n\in \omega $ , any choice set for $\{U_{i}:i<n+1\}$ is a linearly independent subset of V with cardinality $n+1$ (and hence Z has arbitrarily large finite linearly independent subsets).

Furthermore, $Z\not \in \mathcal {A}$ and $\mathcal {A}\cup \{Z\}$ is AD2. Indeed, if $Z\in \mathcal {A}$ , then $Z=W_{n}$ for some $n\in \omega $ . But then, for any $u\in f(U_{n+1})\subseteq Z$ , we have $u\not \in W_{n}=Z$ , which is a contradiction.

For the second assertion, let $\{W_{n_{1}},W_{n_{2}},\ldots ,W_{n_{k}}\}\subset \mathcal {A}$ , where $n_{1}<n_{2}<\cdots <n_{k}$ . Fix $j\in \{1,\ldots ,k\}$ . Since $f(U_{n})\subset W_{n}\setminus (W_{0}+\cdots +W_{n-1})$ for all $n\in \omega \setminus \{0\}$ , it follows that

(1) $$ \begin{align} W_{n_{j}}\cap\left(\left(\sum_{n_{i}\ne n_{j}}W_{n_{i}}\right)+Z\right)=W_{n_{j}}\cap\left(\left(\sum_{n_{i}\ne n_{j}}W_{n_{i}}\right)+ \left\langle \bigcup\{f(U_{i}):i\leq n_{k}\} \right\rangle \right). \end{align} $$

Let

$$ \begin{align*}W_{n_{j}}\cap \left(\sum\{W_{i}:i\in (n_{k}+1)\setminus\{n_{j}\}\right)=\langle u_{1},u_{2},\ldots,u_{m}\rangle,\end{align*} $$

(recall that $\mathcal {A}$ is AD2). Then it is not hard to verify that

$$ \begin{align*}{W_{n_{j}}}{\cap}\left(\left(\sum_{n_{i}\ne n_{j}}W_{n_{i}}\right){+} \left\langle \bigcup\{f(U_{i}):i\leq n_{k}\} \right\rangle \right) &\subseteq \Big\langle \{u_{1},u_{2},\ldots,u_{m}\} \\ & \cup \left(\bigcup\{f(U_{i}):i\leq n_{k}\}\right) \Big\rangle ,\end{align*} $$

and thus, by Equation (1), we have

$$ \begin{align*}W_{n_{j}}\cap\left(\left(\sum_{n_{i}\ne n_{j}}W_{n_{i}}\right)+Z\right)\subseteq \left\langle \{u_{1},u_{2},\ldots,u_{m}\}\cup \left(\bigcup\{f(U_{i}):i\leq n_{k}\}\right) \right\rangle .\end{align*} $$

Therefore, $W_{n_{j}}\cap ((\sum _{n_{i}\ne n_{j}}W_{n_{i}})+Z)$ is finite-dimensional.

On the other hand, it is easy to see that

$$ \begin{align*}Z\cap\left(\sum_{j=1}^{k}W_{n_{j}}\right)\subseteq \left\langle \bigcup\{f(U_{j}):j\leq n_{k}\} \right\rangle ,\end{align*} $$

and hence $Z\cap (\sum _{j=1}^{k}W_{n_{j}})$ is finite-dimensional. We leave the details of the above observations to the interested reader. Thus $\mathcal {A}\cup \{Z\}$ is an AD2 family in V which properly contains $\mathcal {A}$ . Hence, $\mathcal {A}$ is not MAD2.

(ii) $\Rightarrow $ (i) Assume the hypothesis. Let $\mathcal {A}=\{A_{i}:i\in \omega \}$ be a denumerable family of non-empty sets. Without loss of generality, we assume that $\mathcal {A}$ is disjoint and that every member of $\mathcal {A}$ is infinite.

Let $A=\bigcup \mathcal {A}$ and also let

$$ \begin{align*}V=\{f\in\mathbb{Z}_{2}^{A}:|\operatorname{\mathrm{{supp}}}(f)|<\aleph_{0}\},\end{align*} $$

i.e., $V=\{f\in \mathbb {Z}_{2}^{A}:|f^{-1}(\{1\})|<\aleph _{0}\}$ . Then V with pointwise operations is an infinite-dimensional vector space over $\mathbb {Z}_{2}$ (and note that $\{\chi _{\{a\}}:a\in A\}$ is a basis for V).

For each $i\in \omega $ we let

$$ \begin{align*}V_{i}=\{f\in V:\forall x\in A\setminus A_{i}(f(x)=0)\},\end{align*} $$

(so an element f of V is in $V_{i}$ if and only if $\operatorname {\mathrm {{supp}}}(f)\subset A_{i}$ ). It is clear that for every $i\in \omega $ , $V_{i}$ is an infinite-dimensional vector subspace of V (and furthermore, note that $\{\chi _{\{a\}}:a\in A_{i}\}$ is a basis for $V_{i}$ ). We let

$$ \begin{align*}\mathcal{V}=\{V_{i}:i\in\omega\}.\end{align*} $$

$\mathcal {V}$ is an AD2 family in V since, if J is a finite subset of $\omega $ (with at least two elements) and $i\in J$ , then $V_{i}\cap (\sum _{j\in J\setminus \{i\}}V_{j})=\{0\}$ . Furthermore, $|\mathcal {V}|=\aleph _{0}$ , and thus (by our hypothesis) $\mathcal {V}$ is not MAD2 in V. So there exists an infinite-dimensional subspace W of V such that $W\not \in \mathcal {V}$ and $\mathcal {W}=\mathcal {V}\cup \{W\}$ is AD2 in V.

Since the field of scalars (i.e., $\mathbb {Z}_{2}$ ) is finite and $\mathcal {W}$ is AD2, we have that

(2) $$ \begin{align} \forall J\in[\omega]^{<\omega}\left(|W\cap \left(\sum_{j\in J}V_{j}\right)|<\aleph_{0}\right). \end{align} $$

By (2) and the fact that W is infinite-dimensional (and thus W is infinite), it follows that for every $i\in \omega $ , $W\nsubseteq \sum _{j\leq i}V_{j}$ , and thus there exists a strictly increasing sequence $(n_{i})_{i\in \omega }$ of natural numbers and a sequence $(F_{n_{i}})_{i\in \omega }$ of finite sets which have the following two properties:

(a) for every $i\in \omega $ , $F_{n_{i}}\subseteq W\cap (\sum _{j\leq n_{i}}V_{j})$ and

(b) for every $i\in \omega $ and every $f\in F_{n_{i}}$ , $\operatorname {\mathrm {{supp}}}(f\upharpoonright A_{n_{i}})\ne \emptyset $ .

We define

$$ \begin{align*}h=\left\{\left(A_{n_{i}}, \bigcup\{\operatorname{\mathrm{{supp}}}(f\upharpoonright A_{n_{i}}):f\in F_{n_{i}}\}\right):i\in\omega\right\}.\end{align*} $$

Then h is a multiple choice function for the infinite subfamily $\{A_{n_{i}}:i\in \omega \}$ of $\mathcal {A}$ (and hence h is a partial multiple choice function for $\mathcal {A}$ ).

The conclusion now follows from the fact that $\mathsf {MC}^{\aleph _{0}}$ is equivalent to $\mathsf {PMC}^{\aleph _{0}}$ (see Remark 2.8). This completes the proof of “(ii) $\Rightarrow $ (i)” and of the theorem.⊣

Proof. Let V be an infinite-dimensional vector space over a well-ordered field $F=\{r_{\beta }:\beta <\lambda \}$ , having a well-ordered basis $B=\{b_{\alpha }:\alpha <\kappa \}$ , where $\lambda $ is a well-ordered cardinal number (possibly finite) and $\kappa $ is an infinite well-ordered cardinal number. Then $[F]^{<\omega }$ and $[B]^{<\omega }$ are well orderable (recall that, in $\mathsf {ZF}$ , $|[\mu ]^{<\omega }|=\mu $ for any infinite well-ordered cardinal $\mu $ —see [Reference Levy11, Proposition 4.21(ii)]).

Let $\mathcal {A}=\{W_{n}:n\in \omega \}$ be an AD2 family of infinite-dimensional subspaces of V with cardinality $\aleph _{0}$ , and also let $U_{0}=W_{0}$ , and for $n\in \omega \setminus \{0\}$ , $U_{n}=W_{n}\setminus (W_{0}+\cdots +W_{n-1})$ .

For every $n\in \omega $ , let

$$ \begin{align*}J_{n}= \left\{\{b_{\alpha_{1}},\ldots,b_{\alpha_{k}}\}\in [B]^{<\omega}:\exists\{r_{\beta_{1}},\ldots,r_{\beta_{k}}\}\in [F]^{<\omega}\left(\sum_{j=1}^{k}r_{\beta_{j}}b_{\alpha_{j}}\in U_{n}\right)\right\}.\end{align*} $$

Since $J_{n}$ is non-empty (for $U_{n}\ne \emptyset $ for all $n\in \omega $ ), and $[B]^{<\omega }$ and $[F]^{<\omega }$ are well-ordered, we may pick the smallest $\{b_{\alpha _{1}},\ldots ,b_{\alpha _{k}}\}\in J_{n}$ , and for this element of $J_{n}$ , the smallest $\{r_{\beta _{1}},\ldots ,r_{\beta _{k}}\}\in [F]^{<\omega }$ (with respect to some prescribed well orderings of $[B]^{<\omega }$ and $[F]^{<\omega }$ ) such that the vector

$$ \begin{align*}u_{n}=\sum_{j=1}^{k}r_{\beta_{j}}b_{\alpha_{j}}\end{align*} $$

is an element of $U_{n}$ . We may now follow the proof of “(i) $\Rightarrow $ (ii)” of Theorem 4.1 in order to establish that

$$ \begin{align*}Z=\langle\{u_{n}:n\in\omega\}\rangle,\end{align*} $$

is an infinite-dimensional subspace of V such that $Z\not \in \mathcal {A}$ and $\mathcal {A}\cup \{Z\}$ is AD2.⊣

Proof. Assume the hypothesis. We will use the following lemma, which has been established by Howard and Tachtsis [Reference Howard and Tachtsis5]. For the reader’s convenience, and in the interest of making our paper self-contained, we include its proof here.⊣

Proof. Let F be any field and let V be an infinite-dimensional vector space over F. For each $n\in \omega \setminus \{0\}$ , let

$$ \begin{align*}A_n=\{(v_0,v_1,\ldots,v_n)\in V^{n+1}:v_0\ne 0,\text{ and for all}\ 1\le i\le n, v_i\not\in\langle v_0,v_1,\ldots,v_{i-1}\rangle\}.\end{align*} $$

Since V is infinite-dimensional, it follows that $A_n\ne \emptyset $ for all $n\in \omega \setminus \{0\}$ . Let $\mathcal A=\{A_n:n\in \omega \setminus \{0\}\}$ and let, by $\mathsf {AC}^{\aleph _0}$ , $f=\{(n,(v_0^{(n)},v_1^{(n)},\ldots ,v_n^{(n)})):n\in \omega \setminus \{0\}\}$ be a choice function for $\mathcal A$ . Note that by definition of $A_n$ , $\operatorname {\mathrm {{ran}}}(f(n))$ is an $(n+1)$ -sized set of linearly independent vectors of V.

Let $A=\bigcup \{\operatorname {\mathrm {{ran}}}(f(n)):n\in \omega \setminus \{0\}\}$ . It is clear that A is denumerable. Furthermore, since A has finite sequences of linearly independent vectors of arbitrary finite length, we may construct via mathematical induction a denumerable linearly independent subset of V. Indeed, let $w_0=v_0^{(1)}$ . Then $w_0$ is linearly independent, since $w_0\ne 0$ (see the definition of $A_n$ ). Assume that for some $n\in \omega \setminus \{0\}$ we have chosen linearly independent vectors $w_0,w_1,\ldots ,w_n\in A$ . Since $\dim (\langle w_0,w_1,\ldots ,w_n\rangle )= n+1$ and $\operatorname {\mathrm {{ran}}}(f(n+1))$ consists of $n+2$ linearly independent vectors, there exists $v\in \operatorname {\mathrm {{ran}}}(f(n+1))\setminus \langle w_0,w_1,\ldots ,w_n\rangle $ . Let $j_{n+1}=\min \{j:j< n+2$ and $v_j^{(n+1)}\in \operatorname {\mathrm {{ran}}}(f(n+1))\setminus \langle w_0,w_1,\ldots ,w_n\rangle \}$ . Put $w_{n+1}=v_{j_{n+1}}^{(n+1)}$ . This completes the inductive step.

From the above construction, we conclude that $\{w_n:n\in \omega \}$ is a denumerable linearly independent subset of V.⊣

Now we turn to the proof of the theorem. Fix a field F and an infinite-dimensional vector space V over F. By Lemma 4.7, let $I=\{v_{n}:n\in \omega \}$ be a denumerable linearly independent subset of V (the mapping $n\mapsto v_{n}$ is a bijection). We also let $W=\langle I\rangle $ . Then W is an infinite-dimensional subspace of V (and I is a denumerable basis for W). By Theorem 2.9(i), let $\mathcal {A}$ be an almost disjoint family in $\omega $ of cardinality $2^{\aleph _{0}}$ . We let

$$ \begin{align*}W_{X}={\langle}\{v_{n}:n\in X\}{\rangle}\ (X\in\mathcal{A}).\end{align*} $$

Then $\mathcal {W}=\{W_{X}:X\in \mathcal {A}\}$ is an AD2 family in V with cardinality $2^{\aleph _{0}}$ (since $|\mathcal {W}|=|\mathcal {A}|=2^{\aleph _{0}}$ ). To see that $\mathcal {W}$ is AD2, note that if $\mathcal {C}$ is a finite subset of $\mathcal {W}$ , then for every $C\in \mathcal {C}$ we have

$$ \begin{align*}C\kern-1pt\cap\kern-1pt \left(\kern-1pt\sum_{D\in\mathcal{C}\setminus\{C\}}D\kern-2pt\right)\kern-1.5pt\subseteq \left\langle \bigcup\{\kern-1pt\{v_{n}:n\in X\kern-1pt\cap\kern-1pt Y\}: X, Y\in\mathcal{A},\ X\ne Y, \text{and } W_{X},W_{Y}\in\mathcal{C}\}\kern-1pt \right\rangle .\end{align*} $$

Since the set $\bigcup \{\{v_{n}:n\in X\cap Y\}: X, Y\in \mathcal {A}, X\ne Y$ , and $W_{X},W_{Y}\in \mathcal {C}\}$ is finite (for $\mathcal {C}$ is finite and $\mathcal {A}$ is almost disjoint in $\omega $ ), we conclude that $C\cap (\sum _{D\in \mathcal {C}\setminus \{C\}}D)$ is finite-dimensional.⊣

Proof. We will first establish the independence of the above statement from $\mathsf {BPI}$ in $\mathsf {ZFA}$ and then we will transfer the result to $\mathsf {ZF}$ by using a suitable transfer theorem of Pincus [Reference Pincus13].

We consider the Mostowski Linearly Ordered Model of $\mathsf {ZFA}$ , which is labeled as ‘Model $\mathcal {N}3$ ’ in Howard–Rubin [Reference Howard and Rubin3]: The set A of atoms is denumerable and is equipped with an ordering $\leq $ chosen so that $(A,\le )$ is order-isomorphic to the set $\mathbb {Q}$ of rational numbers with the usual ordering; G is the group of all order automorphisms of $(A,\le )$ ; and $\mathcal {F}$ is the finite support (normal) filter on G, i.e., $\mathcal {F}$ is the filter on G which is generated by the subgroups $\operatorname {\mathrm {{fix}}}_{G}(E)=\{\phi \in G:\forall e\in E(\phi (e)=e)\}$ , $E\in [A]^{<\omega }$ . $\mathcal {N}3$ is the permutation model determined by A, G and $\mathcal {F}$ .

Let us recall that every $x\in \mathcal {N}3$ has a least (finite) support, which we shall denote by $E_{x}$ (see Jech [Reference Jech7, Lemma 4.5]). Furthermore, it is a renowned result of Halpern [Reference Halpern2] that $\mathsf {BPI}$ is true in the model $\mathcal {N}3$ . Thus we only need to show that there exists an infinite-dimensional vector space in $\mathcal {N}3$ which has no infinite AD1 family in $\mathcal {N}3$ .

To this end, we consider the set

$$ \begin{align*}V=\{f\in\mathbb{Z}_{2}^{A}:|\operatorname{\mathrm{{supp}}}(f)|<\aleph_{0}\},\end{align*} $$

equipped with pointwise operations of addition and multiplication with scalars from $\mathbb {Z}_{2}$ . Then $(V,+,\cdot )$ is an infinite-dimensional vector space over $\mathbb {Z}_{2}$ , and $(V,+,\cdot )\in \mathcal {N}3$ since, for every $\phi \in G$ , $\phi ((V,+,\cdot ))=(V,+,\cdot )$ .

We assert that V has no infinite AD1 family in $\mathcal {N}3$ . Assume not, and let $\mathcal {W}\in \mathcal {N}3$ be an infinite AD1 family of infinite-dimensional subspaces of V. Let $E=\{e_{1},e_{2},\ldots ,e_{n}\}\subset A$ , where $e_{1}<e_{2}<\cdots <e_{n}$ , be a finite support of $\mathcal {W}$ . Then E determines $n+1$ pairwise disjoint, open intervals in the ordering of A, namely $(-\infty ,e_{1})$ , $(e_{i},e_{i+1})$ ( $1\leq i<n$ ), $(e_{n},+\infty )$ . Let

$$ \begin{align*}\mathcal{Z}=\{(-\infty,e_{1})\}\cup\{(e_{i},e_{i+1}):1\leq i<n\}\cup\{(e_{n},+\infty)\}.\end{align*} $$

A couple of observations are in order:

1. (a) Since $\mathcal {W}$ is AD1 and the field of scalars (i.e. $\mathbb {Z}_{2}$ ) is finite, $\mathcal {W}$ is almost disjoint in V in the sense of Definition 2.3, i.e., for every $W,W'\in \mathcal {W}$ , if $W\ne W'$ then $W\cap W'$ is finite.

2. (b) Since $\mathcal {W}$ is infinite and $\mathcal {Z}\cup \{E\}$ is a finite partition of A, a straightforward pigeonhole-type argument yields that there exist distinct $W,W'\in \mathcal {W}$ and an $I\in \mathcal {Z}$ such that:

   (3) $$ \begin{align} \forall X\in\{W,W'\}(\{I\cap\operatorname{\mathrm{{supp}}}(f):f\in X\} \text{ is infinite}). \end{align} $$

Fix $W,W'\in \mathcal {W}$ and $I\in \mathcal {Z}$ as in (b). Since (by (b)) $W\neq W'$ , (a) yields that $W\cap W'$ is finite. This, together with (3) and the fact that $E_{W}\cup E_{W'}$ is finite, readily implies that there exist distinct $f_{W}\in W$ , $f_{W'}\in W'$ such that $(I\cap \operatorname {\mathrm {{supp}}}(f_{W}))\setminus E_{W}\ne \emptyset $ and $(I\cap \operatorname {\mathrm {{supp}}}(f_{W'}))\setminus E_{W'}\ne \emptyset $ .

Let $z=\min ((I\cap \operatorname {\mathrm {{supp}}}(f_{W}))\setminus E_{W})$ and also let $z'\in I\setminus E_{W}$ such that $z'<z$ and $(z',z)\cap E_{W}=\emptyset $ . Construct a $\phi \in \operatorname {\mathrm {{fix}}}_{G}(E\cup E_{W})$ so that $\phi (z)=z'$ and $f_{W},\phi (f_{W})$ agree on $A\setminus \{z',z\}$ . Since $E_{W}$ is a support of W and $\phi \in \operatorname {\mathrm {{fix}}}_{G}(E\cup E_{W})$ , we have $\phi (W)=W$ . Thus $\phi (f_{W})\in \phi (W)=W$ , and since W is a subspace of V, we have $f_{W}+\phi (f_{W})\in W$ . Furthermore, it is clear that $\operatorname {\mathrm {{supp}}}(f_{W}+\phi (f_{W}))=\{z',z\}\subset I\setminus E_{W}$ .

Using $f_{W}+\phi (f_{W})$ and suitable elements of $\operatorname {\mathrm {{fix}}}_{G}(E\cup E_{W})$ , it is not hard to verify that there exist infinitely many $g_{W}\in W$ such that $\operatorname {\mathrm {{supp}}}(g_{W})\in [I\setminus E_{W}]^{2}$ and $\max (\operatorname {\mathrm {{supp}}}(g_{W}))<z'$ , and infinitely many $h_{W}\in W$ such that $\operatorname {\mathrm {{supp}}}(h_{W})\in [I\setminus E_{W}]^{2}$ and $z<\min (\operatorname {\mathrm {{supp}}}(h_{W}))$ . In particular, for every $a,b\in I\setminus E_{W}$ such that $a<b<z'$ and $(a,z')\cap E_{W}=\emptyset $ , we may consider a $\pi \in \operatorname {\mathrm {{fix}}}_{G}(E\cup E_{W})$ such that $\operatorname {\mathrm {{supp}}}(\pi (f_{W}+\phi (f_{W})))=\{a,b\}$ . Letting $g_{W}=\pi (f_{W}+\phi (f_{W}))$ , we have $g_W\in W$ (for $f_{W}+\phi (f_{W})\in W$ , $\pi \in \operatorname {\mathrm {{fix}}}_{G}(E\cup E_{W})$ and $E_{W}$ is a support of W), $\operatorname {\mathrm {{supp}}}(g_{W})=\{a,b\}\in [I\setminus E_{W}]^{2}$ and $\max (\operatorname {\mathrm {{supp}}}(g_{W}))=b<z'$ . We may work in a similar manner for the $h_{W}$ ’s, and thus we leave this to the interested reader. By the above considerations, we have:

(4) $$ \begin{align} \forall g_{W}\forall h_{W}(\max(\operatorname{\mathrm{{supp}}}(g_{W}))<z'<z<\min(\operatorname{\mathrm{{supp}}}(h_{W}))). \end{align} $$

As with $f_{W}$ , we may let $t=\min ((I\cap \operatorname {\mathrm {{supp}}}(f_{W'}))\setminus E_{W'})$ , $t'\in I\setminus E_{W'}$ such that $t'<t$ and $(t',t)\cap E_{W'}=\emptyset $ , and a $\psi \in \operatorname {\mathrm {{fix}}}_{G}(E\cup E_{W'})$ such that $f_{W'}+\psi (f_{W'})\in W'$ and $\operatorname {\mathrm {{supp}}}(f_{W'}+\psi (f_{W'}))=\{t',t\}\subset I\setminus E_{W'}$ . Furthermore, similarly to the arguments of the previous paragraph, we may conclude that there exist infinitely many $g_{W'}\in W'$ such that $\operatorname {\mathrm {{supp}}}(g_{W'})\in [I\setminus E_{W'}]^{2}$ and $\max (\operatorname {\mathrm {{supp}}}(g_{W'}))<t'$ , and infinitely many $h_{W'}\in W'$ such that $\operatorname {\mathrm {{supp}}}(h_{W'})\in [I\setminus E_{W'}]^{2}$ and $t<\min (\operatorname {\mathrm {{supp}}}(h_{W'}))$ . Hence, we have:

(5) $$ \begin{align} \forall g_{W'}\forall h_{W'}(\max(\operatorname{\mathrm{{supp}}}(g_{W'}))<t'<t<\min(\operatorname{\mathrm{{supp}}}(h_{W'}))). \end{align} $$

Since $W\cap W'$ is finite and the sets of functions $h_{W}$ and of $h_{W'}$ are infinite subsets of W and $W'$ , respectively, we may pick an $h_{W}\in W\setminus W'$ and an $h_{W'}\in W'\setminus W$ (and thus $h_{W}\neq h_{W'}$ ).

There are the following cases:

1. (i) $\min (\operatorname {\mathrm {{supp}}}(h_{W}))\leq \min (\operatorname {\mathrm {{supp}}}(h_{W'}))$ .Footnote ⁴ In view of (4), we may construct an $\eta \in \operatorname {\mathrm {{fix}}}_{G}(E)$ such that $\eta (h_{W})=h_{W'}$ and $\eta $ fixes each of the elements $g_{W}$ of W, which are infinitely many. Since $h_{W'}=\eta (h_{W})\in \eta (W)$ and $h_{W'}\not \in W$ , we obtain that $\eta (W)\neq W$ . Furthermore, as $W\in \mathcal {W}$ , E is a support of $\mathcal {W}$ and $\eta \in \operatorname {\mathrm {{fix}}}_{G}(E)$ , we have $\eta (W)\in \eta (\mathcal {W})=\mathcal {W}$ . Since $\eta $ fixes an infinite subset of W pointwise, namely the set of the functions $g_{W}$ , we deduce that $\eta (W)\cap W$ is infinite. But this contradicts the fact that $\mathcal {W}$ is almost disjoint in V.

2. (ii) $\min (\operatorname {\mathrm {{supp}}}(h_{W'}))<\min (\operatorname {\mathrm {{supp}}}(h_{W}))$ . In view of (5), we may construct a $\sigma \in \operatorname {\mathrm {{fix}}}_{G}(E)$ such that $\sigma (h_{W'})=h_{W}$ and $\sigma $ fixes each of the elements $g_{W'}$ of $W'$ , which are infinitely many. Working similarly to (i), we infer that $\sigma (W')\neq W'$ , $\sigma (W')\in \mathcal {W}$ , and $\sigma (W')\cap W'$ is infinite; thus contradicting $\mathcal {W}$ ’s being almost disjoint in V.

By the above arguments, we conclude that the vector space V has no infinite AD1 family in the model $\mathcal {N}3$ , as asserted.

Now, we refer the reader to Pincus [Reference Pincus12] or [Reference Pincus13] (or to Howard–Rubin [Reference Howard and Rubin3, Note 103, pp. 284–285]) for the definition of the terms “boundable statement” and “injectively boundable statement.” To transfer the above result to $\mathsf {ZF}$ , we will apply the following transfer theorem of Pincus (see [Reference Pincus13, Theorem 4 and note added in proof]): “If a conjunction of injectively boundable statements and $\mathsf {BPI}$ has a Fraenkel–Mostowski model, then it also has a $\mathsf {ZF}$ -model.”

It is easy to see that $\Psi =$ “There exists an infinite-dimensional vector space which has no infinite AD1 family” is a boundable statement, and thus $\Psi $ is injectively boundable (boundable statements are (up to equivalence) injectively boundable, see Pincus [Reference Pincus12, p. 722]). Letting $\Phi =\mathsf {BPI}\wedge \Psi $ , we have that $\Phi $ satisfies the hypotheses of [Reference Pincus13, Theorem 4] stated above, and thus $\Phi $ has a $\mathsf {ZF}$ -model. This completes the proof of the theorem.⊣

Proof. Assume the hypothesis. Let V be an infinite-dimensional vector space over a field F, and let $\mathcal {A}$ be an AD1 family in V. Assume that $\mathcal {A}$ is not MAD1 (otherwise, there is nothing to show). By $\mathsf {MC}$ , let f be a multiple choice function for $\wp (\wp (V))\setminus \{\emptyset \}$ .

By transfinite recursion, we will construct a MAD1 family in V which contains $\mathcal {A}$ . Let

$$ \begin{align*}\mathcal{R}_{0}=\{Z:Z\leq V, Z\ \text{is infinite-dimensional,}\ Z\not\in\mathcal{A},\ \text{and}\ \mathcal{A}\cup\{Z\}\ \text{is}\ AD1\}.\end{align*} $$

Since $\mathcal {A}$ is not maximal, $\mathcal {R}_{0}\neq \emptyset $ . We let

$$ \begin{align*}\mathcal{S}_{0}\kern-1pt =\kern-1pt\{\mathcal{U}\kern-1pt :\kern-1pt \mathcal{U}\text{ is a} \subseteq\text{-maximal subset of}\, f(\mathcal{R}_{0})\, \text{such that}\, \bigcap\mathcal{U}\ \text{is infinite-dimensional}\},\end{align*} $$

$$ \begin{align*}\mathcal{T}_{0}=\left\{\bigcap\mathcal{U}:\mathcal{U}\in\mathcal{S}_{0}\right\},\end{align*} $$

and

$$ \begin{align*}\mathcal{A}_{0}=\mathcal{A}\cup\mathcal{T}_{0}.\end{align*} $$

The family $\mathcal {A}_{0}$ has the following properties:

(a) $\mathcal {A}\subsetneq \mathcal {A}_{0}$ . Indeed, pick any element T of $\mathcal {T}_{0}$ . Then $T=\bigcap \mathcal {U}$ for some $\mathcal {U}\in \mathcal {S}_{0}$ . We assert that $T\not \in \mathcal {A}$ . If not (i.e., if $T\in \mathcal {A}$ ), then we choose any element Z of $\mathcal {U}$ . Since $Z\in \mathcal {R}_{0}$ (for $Z\in f(\mathcal {R}_{0})\subseteq \mathcal {R}_{0}$ ), we have that $\mathcal {A}\cup \{Z\}$ is AD1. However, $T\cap Z=(\bigcap \mathcal {U})\cap Z=\bigcap \mathcal {U}$ (for $Z\in \mathcal {U}$ ). Since $\bigcap \mathcal {U}$ is infinite-dimensional, we have obtained a contradiction to the fact that $\mathcal {A}\cup \{Z\}$ is AD1. Therefore, $T\in \mathcal {A}_{0}\setminus \mathcal {A}$ , and thus $\mathcal {A}\subsetneq \mathcal {A}_{0}$ .

(b) $\mathcal {A}_{0}$ is AD1. Let $U,W\in \mathcal {A}_{0}$ with $U\ne W$ . There are the following cases:

(b1) $U,W\in \mathcal {A}$ . Then $U\cap W$ is finite-dimensional since $\mathcal {A}$ is AD1.

(b2) $U,W\in \mathcal {T}_{0}$ . Then $U=\bigcap \mathcal {U}$ and $W=\bigcap \mathcal {U}'$ for some $\mathcal {U},\mathcal {U}'\in \mathcal {S}_{0}$ . Since $U\neq W$ , we have $\mathcal {U}\neq \mathcal {U}'$ so $\mathcal {U}\not \subseteq \mathcal {U}'$ and $\mathcal {U}'\not \subseteq \mathcal {U}$ , for $\mathcal {U}$ and $\mathcal {U}'$ are maximal subsets of $f(\mathcal {R}_{0})$ such that $\bigcap \mathcal {U}$ and $\bigcap \mathcal {U}'$ are infinite-dimensional. It follows that $\mathcal {U},\mathcal {U}'\subsetneq \mathcal {U}\cup \mathcal {U}'$ , so $\bigcap (\mathcal {U}\cup \mathcal {U}')=(\bigcap \mathcal {U})\cap (\bigcap \mathcal {U}')=U\cap W$ is finite-dimensional.

(b3) $U\in \mathcal {A}$ , $W\in \mathcal {T}_{0}$ . There exists $\mathcal {U}\in \mathcal {S}_{0}$ such that $W=\bigcap \mathcal {U}$ . Pick any element Z of $\mathcal {U}$ . Then $U\cap Z$ is finite-dimensional, for $U\in \mathcal {A}$ and $\mathcal {A}\cup \{Z\}$ is AD1. Since $U\cap W\subseteq U\cap Z$ , we have that $U\cap W$ is finite-dimensional.

(b4) $U\in \mathcal {T}_{0}$ , $W\in \mathcal {A}$ . Similarly to (b3), we conclude that $U\cap W$ is finite-dimensional.

Hence, $\mathcal {A}_{0}$ is AD1.

We assume that for some ordinal number $\alpha $ we have constructed an $\subseteq $ -increasing sequence $(\mathcal {A}_{\beta })_{\beta <\alpha }$ of AD1 families in V such that $\mathcal {A}\subsetneq \mathcal {A}_{0}$ . Let

$$ \begin{align*}\mathcal{B}_{\alpha}=\bigcup\{\mathcal{A}_{\beta}:\beta<\alpha\},\end{align*} $$

and also let

$$ \begin{align*}\mathcal{R}_{\alpha}=\{Z:Z\leq V, Z\ \text{is infinite-dimensional}, Z\not\in\mathcal{B}_{\alpha},\ \text{and}\ \mathcal{B}_{\alpha}\cup\{Z\}\ \text{is AD1}\}.\end{align*} $$

Since $(\mathcal {A}_{\beta })_{\beta <\alpha }$ is a chain of AD1 families, $\mathcal {B}_{\alpha }$ is AD1. If $\mathcal {R}_{\alpha }=\emptyset $ , then $\mathcal {B}_{\alpha }$ is a MAD1 family in V which contains $\mathcal {A}$ , and we are done. Otherwise, (that is, if $\mathcal {R}_{\alpha }\ne \emptyset $ ), we define

$$ \begin{align*}\mathcal{A}_{\alpha}=\mathcal{B}_{\alpha}\cup & \Big\{\bigcap\mathcal{U}:\mathcal{U}\text{ is a}\ \subseteq\text{-maximal subset of}\ f(\mathcal{R}_{\alpha})\ \text{such that}\\ & \quad \bigcap\mathcal{U}\ \text{is infinite-dimensional}\Big\}. \end{align*} $$

Since $\mathsf {Ord}$ (i.e., the class of all ordinal numbers) is a proper class, the recursion must terminate at some ordinal stage. By the above construction, the ending of the recursion yields a MAD1 family in V which contains $\mathcal {A}$ .

The second assertion of the theorem follows from the first and the fact that $\mathsf {MC}$ does not imply $\mathsf {AC}$ in $\mathsf {ZFA}$ (see [Reference Jech7, Theorem 9.2(i)]). This completes the proof of the theorem.⊣

Proof. We will use a permutation model which was constructed by Howard and Tachtsis [Reference Howard and Tachtsis4], and whose description is as follows: We start with a model M of $\mathsf {ZFA}$ + $\mathsf {AC}$ with an $\aleph _1$ -sized set A of atoms, which is a disjoint, denumerable union of $\aleph _1$ -sized sets so that

$$ \begin{align*}A=\bigcup\{A_i:i\in\omega\},\ |A_{i}|=\aleph_{1}.\end{align*} $$

Let G be the group of all permutations of A, which fix $A_i$ for all $i\in \omega $ . Note that any element of G also fixes the bijection $i\mapsto A_{i}$ , and hence fixes the ordered partition $(A_{i})_{i\in \omega }$ of A. (And also note that for every $i\in \omega $ and every permutation $\phi $ of A (not necessarily in G), $\phi (i)=i$ since the natural numbers are pure sets (i.e., their transitive closure contains no atoms) and pure sets are fixed by any permutation of A.) Let $\mathcal {F}$ be the (normal) filter of subgroups of G generated by the subgroups $\operatorname {\mathrm {{fix}}}_G(E)$ , where $E\subset A$ and $|E|<\aleph _1$ . Let $\mathcal {N}$ be the Fraenkel–Mostowski model determined by M, G and $\mathcal {F}$ .

In [Reference Howard and Tachtsis4, proof B of Theorem 2], it was shown that each of $\mathsf {LW}$ and $\mathsf {AC^{LO}}$ (and hence $\mathsf {AC^{WO}}$ ) are true in $\mathcal {N}$ . Therefore, it suffices to show that in $\mathcal {N}$ , there exists an infinite-dimensional vector space over some field which has an AD1 family in $\mathcal {N}$ that cannot be extended to a MAD1 family in $\mathcal {N}$ .

To this end, we consider the following infinite-dimensional vector space V over $\mathbb {Z}_{2}$ ,

$$ \begin{align*}V=\{f\in \mathbb{Z}_{2}^{A}:|\operatorname{\mathrm{{supp}}}(f)|<\aleph_{0}\},\end{align*} $$

where V is equipped with pointwise operations. Then $(V,+,\cdot )\in \mathcal {N}$ (for $(V,+,\cdot )$ is fixed by every permutation of A in G). For each $n\in \omega $ , consider the infinite-dimensional vector subspace of V,

$$ \begin{align*}V_{n}=\{f\in V:\operatorname{\mathrm{{supp}}}(f)\subset A_{n}\}.\end{align*} $$

Let

$$ \begin{align*}\mathcal{V}=\{V_{n}:n\in\omega\}.\end{align*} $$

$\mathcal {V}$ is an AD1 family in V which belongs to the model $\mathcal {N}$ (any element of G fixes $\mathcal {V}$ pointwise, so $\mathcal {V}$ is also denumerable in $\mathcal {N}$ ).

$\mathcal {V}$ is not MAD1 in $\mathcal {N}$ . Indeed, let H be a choice function for $\mathcal {A}=\{A_{n}:n\in \omega \}$ . (Note that by definition of the filter $\mathcal {F}$ , $\operatorname {\mathrm {{ran}}}(H)$ is a support of (every element of) H, and hence $H\in \mathcal {N}$ .) Let $W={\langle }\{F_{n}:n\in \omega \}{\rangle }$ , where for $n\in \omega $ , $F_{n}\upharpoonright A_{m}=0$ (the zero function) for $m\in \omega \setminus \{n\}$ , and $F_{n}\upharpoonright A_{n}=\chi _{\{H(n)\}}$ . Then W is an infinite-dimensional subspace of V such that $W\not \in \mathcal {V}$ and $\mathcal {V}\cup \{W\}$ is AD1 in $\mathcal {N}$ .⊣

Proof. We prove the claim by contradiction. Let $\mathcal {M}$ be a MAD1 family in $\mathcal {N}$ which contains $\mathcal {V}$ (and thus, by the observations in the previous paragraph, $\mathcal {V}\subsetneq \mathcal {M}$ ), and also let $E\subset A$ be a support of $\mathcal {M}$ . We assert the following:

(6) $$ \begin{align} (\forall Z\in\mathcal{M}\setminus\mathcal{V})(\forall f\in Z)(\forall n\in\omega)(\operatorname{\mathrm{{supp}}}(f\upharpoonright A_{n})\subseteq E\cap A_{n}). \end{align} $$

If not, then there exist $Z\in \mathcal {M}\setminus \mathcal {V}$ , $f\in Z$ , $n\in \omega $ , and $a\in \operatorname {\mathrm {{supp}}}(f\upharpoonright A_{n})\setminus (E\cap A_{n})$ .

Since $Z\cap V_{n}$ is finite-dimensional (actually, $Z\cap V_{n}$ is finite since it is a finite-dimensional space over $\mathbb {Z}_{2}$ ) and any support of Z meets $A_n$ in a countable set, there exists $b\in A_{n}\setminus \{a\}$ such that for every $g\in Z$ , $b\not \in \operatorname {\mathrm {{supp}}}(g\upharpoonright A_{n})$ . If not, i.e., if for every $b\in A_{n}\setminus \{a\}$ there exists $g\in Z$ with $b\in \operatorname {\mathrm {{supp}}}(g\upharpoonright A_{n})$ , then let $E_{Z}$ be a support of Z. (So $E_{Z}\cap A_{n}$ is countable.) Fix $c\in R$ , where $R=A_{n}\setminus [(\operatorname {\mathrm {{supp}}}({f\upharpoonright A_{n}})\cup E\cup E_{Z}]$ , and let $g\in Z$ such that $c\in \operatorname {\mathrm {{supp}}}(g\upharpoonright A_{n})$ . For every $r\in R\setminus (\{c\}\cup \operatorname {\mathrm {{supp}}}(g)\})$ , let $\phi _{r}=(c,r)$ (i.e., $\phi _{r}$ interchanges the atoms c and r and leaves all the other atoms of A fixed). Then $S=\{\phi _{r}(g):r\in R\setminus (\{c\}\cup \operatorname {\mathrm {{supp}}}(g)\})\}$ is an infinite, linearly independent subset of Z (since $\phi _r\in \operatorname {\mathrm {{fix}}}_{G}(E_Z)$ and $g\in Z$ ), and thus $\{\sum _{x\in F}x:F\in [S]^{<\omega }\}$ is an infinite, linearly independent subset of $Z\cap V_{n}$ . But this contradicts $Z\cap V_{n}$ ’s being finite-dimensional.

Let $\pi =(a,b)$ , where (by the above argument) $b\in A_{n}\setminus (\{a\}\cup \operatorname {\mathrm {{supp}}}(g\upharpoonright A_{n}))$ for all $g\in Z$ . Then $\pi \in \operatorname {\mathrm {{fix}}}_{G}(E)$ , and hence $\pi (\mathcal {M})=\mathcal {M}$ . Thus $\pi (Z)\in \mathcal {M}$ (and also $\pi (Z)\not \in \mathcal {V}$ ), and furthermore, $\pi (Z)\ne Z$ (for $\pi (f)\in \pi (Z)\setminus Z$ , since $f\in Z$ and $b\in \operatorname {\mathrm {{supp}}}(\pi (f)\upharpoonright A_{n})$ ).

Since Z is not included in any finite sum of the $V_{i}$ ’s, and $Z\cap V_{i}$ is finite-dimensional for all $i\in \omega $ , it is not hard to verify that $\pi (Z)\cap Z$ contains an infinite linearly independent subset. But this contradicts the fact that $\mathcal {M}$ is AD1. Thus (6) is true as asserted.

On the basis of (6), and working similarly to the argument that $\mathcal {V}$ is not MAD1 in $\mathcal {N}$ , we may show that neither $\mathcal {M}$ is MAD1 in $\mathcal {N}$ . But this contradicts our assumption on $\mathcal {M}$ .

Thus the AD1 family $\mathcal {V}$ cannot be extended to a MAD1 family in the model $\mathcal {N}$ , finishing the proof of the claim.⊣

For the last assertion of the theorem (i.e., nonprovability of the algebraic statement in $\mathsf {ZF}$ ), note that since the negation of “In every infinite-dimensional vector space V, every AD1 family in V can be extended to a MAD1 family” is a boundable statement, and has a $\mathsf {ZFA}$ -model, it follows from the Jech–Sochor First Embedding Theorem (see [Reference Jech7, Theorem 6.1 and Problem 1 (p. 94)]) that it has a (symmetric) $\mathsf {ZF}$ -model. This completes the proof of the theorem.⊣