Proof. (i) By hypothesis, let $\preceq $ be a linear order on X.

(a) $\Longleftrightarrow $ (b) This is well known and, moreover, is true for any infinite set Y in place of $\wp (X)$ .

(b) $\Rightarrow $ (c) $\Rightarrow $ (d) These are straightforward.

(d) $\Rightarrow $ (a) We first prove the following lemma.

Proof. Since $\preceq $ is a linear ordering of X, the family of all non-empty finite subsets of X has a choice function. So, it suffices to show that the family $\wp ^{\infty }(X)$ of all infinite subsets of X has a choice function. To this end, let $\mathscr {F}$ be the Fréchet filter on X, that is,

$$ \begin{align*}\mathscr{F}=\{Y\subseteq X:|X\setminus Y|<\aleph_{0}\}.\end{align*} $$

By hypothesis, $\mathscr {F}$ has a well-orderable base, say $\mathscr {B}=\{B_{\alpha }:\alpha <\kappa \}$ for some well-ordered cardinal $\kappa $ . We assert the following:

(2) $$ \begin{align} (\forall Y\in\wp^{\infty}(X))(\exists \alpha<\kappa)(|Y\setminus B_{\alpha}|>0). \end{align} $$

To see this, fix $Y\in \wp ^{\infty }(X)$ . Let $y\in Y$ . Then $X\setminus \{y\}\in \mathscr {F}$ , so since $\mathscr {B}$ is a base for $\mathscr {F}$ , there exists $\alpha <\kappa $ such that $B_{\alpha }\subseteq X\setminus \{y\}$ . As $y\in Y$ , we obtain that $|Y\setminus B_{\alpha }|>0$ . Therefore, (2) holds, as asserted.

By (2), for every $Y\in \wp ^{\infty }(X)$ , let

$$ \begin{align*}\alpha_{Y}=\min\{\alpha\in\kappa:|Y\setminus B_{\alpha}|>0\}.\end{align*} $$

Since $\mathscr {B}\subseteq \mathscr {F}$ , we have that

$$ \begin{align*}\forall Y\in\wp^{\infty}(X)(|Y\setminus B_{\alpha_{Y}}|<\aleph_{0}).\end{align*} $$

Then the function $h:\wp ^{\infty }(X)\rightarrow \bigcup \wp ^{\infty }(X)$ defined by

$$ \begin{align*}h(Y)=\mathrm{min}_{\preceq}(Y\setminus B_{\alpha_{Y}})\text{ for all }Y\in\wp^{\infty}(X),\end{align*} $$

is a choice function for $\wp ^{\infty }(X)$ . Thus, $\mathsf {AC}(X)$ is true, as required.

By (d) and Lemma 5.2, we conclude that X is well orderable. We recall here the following $\mathsf {ZF}$ -fact (see Rubin and Rubin [Reference Rubin and Rubin23, Theorem 5.5, p. 76]):

(*) The power set of a well-ordered set can be linearly ordered.

By (*) and the fact that X is well orderable, we infer that $\wp (X)$ is linearly orderable. By (the second paragraph of) Remark 1.10, let $\mathcal {\mathcal {I}}$ be an independent family in X of cardinality $|\wp (X)|$ . Since $\wp (X)$ is linearly orderable, so is its subset $\mathcal {I}$ . Our plan is to show that $\mathcal {I}$ is well orderable, which will give us that $\wp (X)$ is well orderable since $|\wp (X)|=|\mathcal {I}|$ . By way of contradiction, we assume that $\mathcal {I}$ is not well orderable. Let

$$ \begin{align*}\mathscr{B}=\{\bigcap\mathcal{Q}:\mathcal{Q}\in [\mathcal{I}]^{<\omega}\setminus\{\emptyset\}\}.\end{align*} $$

Since $\mathcal {I}$ is independent, $\mathscr {B}$ is a filter base. Let $\mathscr {F}_{\mathscr {B}}$ be the filter on X generated by $\mathscr {B}$ . By hypothesis, $\mathscr {F}_{\mathscr {B}}$ has a well-orderable base, say

$$ \begin{align*}\mathscr{C}=\{C_{\alpha}:\alpha<\kappa\},\end{align*} $$

for some well-ordered cardinal $\kappa $ . For every $\alpha <\kappa $ , we let

$$ \begin{align*}\mathcal{D}_{\alpha}=\{x\in \mathcal{I}:C_{\alpha}\subseteq x\}.\end{align*} $$

As $\mathcal {I}\subseteq \mathscr {F}_{\mathscr {B}}$ and $\mathscr {C}$ is a base for $\mathscr {F}_{\mathscr {B}}$ , we have

$$ \begin{align*}\mathcal{I}=\bigcup\{\mathcal{D}_{\alpha}:\alpha<\kappa\}.\end{align*} $$

Since $\mathcal {I}$ is not well orderable and the union of a well-ordered family of finite subsets of $\mathcal {I}$ is well orderable (because $\mathcal {I}$ is linearly orderable), it follows that, for some $\alpha <\kappa $ , $\mathcal {D}_{\alpha }$ is infinite. As $\mathscr {B}$ is a base for $\mathscr {F}_{\mathscr {B}}$ and $C_{\alpha }\in \mathscr {F}_{\mathscr {B}}$ , there exists $\mathcal {Q}\in [\mathcal {I}]^{<\omega }\setminus \{\emptyset \}$ such that $\bigcap \mathcal {Q}\subseteq C_{\alpha }$ . Since $\mathcal {D}_{\alpha }$ is infinite and $\mathcal {Q}$ is finite, there exists $z\in \mathcal {D}_{\alpha }\setminus \mathcal {Q}$ . It follows that $\mathcal {Q},\{z\}$ are two disjoint finite subsets of the independent family $\mathcal {I}$ such that

$$ \begin{align*}\bigcap\mathcal{Q}\subseteq C_{\alpha}\subseteq z.\end{align*} $$

But then, we have

$$ \begin{align*}\left(\bigcap\mathcal{Q}\right)\cap z^{c}=\emptyset,\end{align*} $$

contradicting the fact that $\mathcal {I}$ is independent. From the above arguments, we conclude that $\mathcal {I}$ is well orderable, and thus so is $\wp (X)$ since $|\wp (X)|=|\mathcal {I}|$ . Therefore, $\mathsf {AC}(\wp (X))$ holds, as required.

(ii) For the independence result, we will construct a new symmetric model of $\mathsf {ZF}$ and prove that it has the required properties. To this end, we start with a countable transitive model M of $\mathsf {ZFC}+\mathsf {CH}$ . As a forcing notion, we use the set

$$ \begin{align*}\mathbb{P}=\operatorname{\mathrm{{Fn}}}(\omega\times\omega_{1}\times\omega_{1},2,\omega_{1}),\end{align*} $$

of all countable partial functions from $\omega \times \omega _{1}\times \omega _{1}$ into $2=\{0,1\}$ equipped with the partial order of reverse inclusion, that is, for all $p,q\in \mathbb {P}$ , $p\leq q\Longleftrightarrow p\supseteq q$ . Note that the maximum element $\mathbf {1}_{\mathbb {P}}$ of $\mathbb {P}$ is the empty function. Let G be a $\mathbb {P}$ -generic filter over M, and also let $M[G]$ be the corresponding generic extension model of M. Since $\aleph _{1}$ is a regular cardinal, $(\mathbb {P},\leq )$ is $\aleph _{1}$ -closed, that is, for every $\gamma \in \aleph _{1}$ and every decreasing sequence $(p_{\xi })_{\xi <\gamma }$ of elements of $\mathbb {P}$ , there exists $q\in \mathbb {P}$ such that $q\leq p_{\xi }$ for all $\xi <\gamma $ . It follows from Kunen [Reference Kunen16, Theorem 6.14, p. 214] that no new functions $g:\omega \rightarrow 2$ , and thus no new reals, are added. Moreover, by [Reference Kunen16, Corollary 6.15, p. 215], we have that $(\mathbb {P},\leq )$ preserves cofinalities $\leq \aleph _{1}$ , and hence cardinals $\leq \aleph _{1}$ . Therefore, $\mathsf {CH}$ is true in $M[G]$ , as well as in every model N with $M\subseteq N\subseteq M[G]$ .

Forcing with $\mathbb {P}$ over M yields, in $M[G]$ , the following generic subsets of $\omega _{1}$ :

$$ \begin{align*}a_{n,k}=\{m\in\omega_{1}:\exists p\in G(p(n,k,m)=1)\},\ (n,k)\in\omega\times\omega_{1}.\end{align*} $$

Note that, via standard density arguments, the family

$$ \begin{align*}\mathcal{A}=\{a_{n,k}:(n,k)\in\omega\times\omega_{1}\}\end{align*} $$

is independent in $\omega _{1}$ . The sets $a_{n,k}$ , $(n,k)\in \omega \times \omega _{1}$ , and $\mathcal {A}$ , have respectively the following $\mathbb {P}$ -names:

$$ \begin{align*}\dot{a}_{n,k}=\{(\check{m},p):m\in\omega_{1}\wedge p\in \mathbb{P}\wedge p(n,k,m)=1\},\ (n,k)\in\omega\times\omega_{1}\end{align*} $$

and

$$ \begin{align*}\dot{\mathcal A}=\{(\dot{a}_{n,k},\mathbf{1}_{\mathbb{P}}):(n,k)\in\omega\times\omega_{1}\}.\end{align*} $$

Every permutation $\phi $ of $\omega \times \omega _{1}$ induces an order-automorphism of $(\mathbb {P},\leq )$ by requiring, for every $p\in \mathbb {P}$ , the following:

(3) $$ \begin{align} \begin{aligned} \operatorname{\mathrm{{dom}}}\phi(p)&=\{(\phi(i,j),k):(i,j,k)\in\operatorname{\mathrm{{dom}}}(p)\},\\ \phi(p)(\phi(i,j),k)&=p(i,j,k). \end{aligned} \end{align} $$

Let $\mathcal G$ be the group of all order-automorphisms of $(\mathbb {P},\le )$ which are induced by permutations $\phi $ of $\omega \times \omega _{1}$ , as in (3). Note that

(4) $$ \begin{align} (\forall\phi\in\mathcal{G})(\forall (n,m)\in\omega\times\omega_{1})(\phi(\dot{a}_{n,m})=\dot{a}_{\phi(n,m)}). \end{align} $$

By (4), we obtain that

(5) $$ \begin{align} \forall\phi\in\mathcal{G}(\phi(\dot{\mathcal A})=\dot{\mathcal A}). \end{align} $$

For every $E\in [\omega \times \omega _{1}]^{<\omega }$ , we let $\operatorname {\mathrm {{fix}}}_{\mathcal G}(E)=\{\phi \in \mathcal G:{\forall e\in E}(\phi (e)=e)\}$ , and we also let $\Gamma $ be the filter of subgroups of $\mathcal G$ generated by the filter base $\{\operatorname {\mathrm {{fix}}}_{\mathcal G}(E):E\in [\omega \times \omega _{1}]^{<\omega }\}$ . It is not hard to verify that $\Gamma $ is a normal filter on $\mathcal {G}$ . If $\dot {x}$ is a $\mathbb {P}$ -name and $E\in [\omega \times \omega _{1}]^{<\omega }$ is such that $\operatorname {\mathrm {{fix}}}_{\mathcal G}(E)\subseteq \operatorname {\mathrm {{sym}}}_{\mathcal {G}}(\dot {x})$ ( $=\{\phi \in \mathcal {G}:\phi (\dot {x})=\dot {x}\}$ ), then we call E a support of $\dot {x}$ . Let

$$ \begin{align*}N=\{\dot{x}_{G}:\dot{x}\in\operatorname{\mathrm{{HS}}}^{\Gamma}\},\end{align*} $$

be the symmetric extension model of M. By the definitions of $\Gamma $ and $\operatorname {\mathrm {{HS}}}^{\Gamma }$ , it follows that every $\dot {x}\in \operatorname {\mathrm {{HS}}}^{\Gamma }$ has a support in the above sense.

In view of the observations at the beginning of the proof (of part (ii)), we have:

$$ \begin{align*}N\models\mathsf{CH},\end{align*} $$

and thus

$$ \begin{align*}N\models\mathsf{AC}(\mathbb{R}).\end{align*} $$

Proof. Assume the contrary. Since $|\mathcal {A}|=\aleph _{1}$ in $M[G]$ , our assumption yields $|\mathcal {A}|=\aleph _{1}$ in N also. So, let $f:\omega _{1}\rightarrow \mathcal {A}$ be a bijection in N. Let $\dot {f}$ be an $\operatorname {\mathrm {{HS}}}^{\Gamma }$ -name for f. There exists $p\in G$ such that

(6) $$ \begin{align} p\Vdash"\dot{f}\text{ is a bijection from }\check{\omega_{1}}\text{ onto }\dot{\mathcal{A}}\text{"}. \end{align} $$

Let $E\in [\omega \times \omega _{1}]^{<\omega }$ be a support of $\dot {f}$ . Since f is a bijection and E is finite, there exist $i\in \omega _{1}$ and $(n_{0},m_{0})\in \omega \times \omega _{1}$ such that $E\cap (\{n_{0}\}\times \omega _{1})=\emptyset $ and $f(i)=a_{n_{0},m_{0}}$ . Let $q\in G$ be such that $q\leq p$ and

(7) $$ \begin{align} q\Vdash\dot{f}(\check{i})=\dot{a}_{n_{0},m_{0}}. \end{align} $$

Since $|q|<\aleph _{1}$ , there exists $k\in \omega _{1}$ such that $k>m_{0}$ and, for all $i\in \omega _{1}$ with $i\geq k$ , $(n_{0},i)\not \in \operatorname {\mathrm {{dom}}}(\operatorname {\mathrm {{dom}}}(q))$ . Let $\eta :[0,k)\rightarrow [k,2k)$ be an order-isomorphism. Then $\eta $ induces a permutation $\eta ^{*}$ of $\omega _{1}$ defined by

$$\begin{align*}\eta^{*}(i)= \begin{cases} \eta(i), & \text{if }i\in [0,k);\\ \eta^{-1}(i), & \text{if }i\in [k,2k);\\ i, & \text{if }2k\leq i. \end{cases} \end{align*}$$

Note that, as $m_{0}\in [0,k)$ , $\eta ^{*}(m_{0})>m_{0}$ , since $\eta ^{*}(m_{0})=\eta (m_{0})\in [k,2k)$ . We define a permutation $\phi $ of $\omega \times \omega _{1}$ by stipulating, for all $(n,m)\in \omega \times \omega _{1}$ ,

$$\begin{align*}\phi(n,m)= \begin{cases} (n_{0},\eta^{*}(m)), & \text{if }n=n_{0};\\ (n,m), &\text{if }n\ne n_{0}. \end{cases} \end{align*}$$

Denoting by $\phi $ the element of $\mathcal {G}$ induced by $\phi $ , as well as the extension of $\phi $ to an automorphism of $M^{\mathbb {P}}$ (the class of all $\mathbb {P}$ -names), we have that the following hold:

1. (a) $\phi \in \operatorname {\mathrm {{fix}}}_{\mathcal {G}}(E)$ , since $E\cap (\{n_{0}\}\times \omega _{1})=\emptyset $ and $\phi $ fixes $\{n\}\times \omega _{1}$ pointwise for all $n\in \omega \setminus \{n_{0}\}$ . It follows that $\phi (\dot {f})=\dot {f}$ , since E is a support of $\dot {f}$ and $\phi \in \operatorname {\mathrm {{fix}}}_{\mathcal {G}}(E)$ .

2. (b) $\phi (\dot {a}_{n_{0},m_{0}})=\dot {a}_{\phi (n_{0},m_{0})}=\dot {a}_{n_{0},\eta ^{*}(m_{0})}$ .

3. (c) $\phi (\dot {\mathcal {A}})=\dot {\mathcal {A}}$ (by (5)).

4. (d) q and $\phi (q)$ are compatible conditions. It follows that $q\cup \phi (q)$ is a well-defined extension of q, $\phi (q)$ , and p (recall that $q\leq p$ ).

By (7), (a), and (b), we obtain that

(8) $$ \begin{align} \phi(q)\Vdash\dot{f}(\check{i})=\dot{a}_{n_{0},\eta^{*}(m_{0})}. \end{align} $$

By (c) and (d), together with (6), (7) and (8), we conclude that

(9) $$ \begin{align} &q\cup\phi(q)\Vdash"\dot{f}\text{ is a bijection from }\check{\omega_{1}}\text{ onto }\dot{\mathcal{A}}\text{"}\wedge \dot{f}(\check{i})=\dot{a}_{n_{0},m_{0}}\nonumber\\ &\qquad\qquad\qquad\wedge\dot{f}(\check{i})=\dot{a}_{n_{0},\eta^{*}(m_{0})}. \end{align} $$

But then, (9) clearly yields a contradiction. Therefore, $\mathcal {A}$ is not well orderable in the model N, as required.

The above arguments complete the proof of (ii) and of the theorem.

Proof. Let $\preceq $ be a linear order on X. Let $\mathcal {B}=\{B_{\alpha }:\alpha <\kappa \}$ , where $\kappa $ is a well-ordered cardinal, be a family of well-orderable sets such that $X=\bigcup \mathcal {B}$ .

Let A be an infinite subset of X. If, for some $\alpha \in \kappa $ , $A\cap B_{\alpha }$ is infinite, then (since $B_{\alpha }$ is well orderable) $A\cap B_{\alpha }$ is well orderable. If $A\cap B_{\alpha }$ is finite for all $\alpha \in \kappa $ , then $A=\bigcup \{A\cap B_{\alpha }:\alpha <\kappa \}$ is well orderable as a union of a well-ordered family of finite linearly ordered sets.

Proof. First, we consider Solovay’s model $L(\mathbb {R})$ from [Reference Solovay24], labeled as ‘Model $\mathcal {M}5(\aleph )$ ’ in [Reference Howard and Rubin11], for which the following facts are well known (see [Reference Howard and Rubin11, Reference Solovay24]):

$$ \begin{align*}L(\mathbb{R})\models\mathsf{DC}\wedge\neg\mathsf{UF}(\omega)\wedge"\aleph_{1}\text{ is regular"}\wedge\aleph_{1}\not\leq 2^{\aleph_{0}}.\end{align*} $$

Our plan is to force over $L(\mathbb {R})$ with a suitable poset of conditions so that a free ultrafilter on $\omega $ is added, and in the generic extension it is still true that $\aleph _{1}\not \leq 2^{\aleph _{0}}$ .

For the reader’s convenience, we give a brief description of $L(\mathbb {R})$ . We first recall that for a well-ordered cardinal $\kappa $ , the Lévy collapsing order for $\kappa $ , Lv $(\kappa )$ (in the notation of Kunen [Reference Kunen16, Definition 8.6, p. 231]), is the set of all finite functions p with $\operatorname {\mathrm {{dom}}}(p)\subseteq \kappa \times \omega $ , $\operatorname {\mathrm {{ran}}}(p)\subseteq \kappa $ and for each $(\alpha ,n)\in p$ , $p(\alpha ,n)<\alpha $ . Lv $(\kappa )$ is partially ordered by reverse inclusion, i.e., $p\leq q$ if and only if $p\supseteq q$ .

$L(\mathbb {R})$ is constructed in two steps, starting with a model M of $\mathsf {ZFC}$ with a strongly inaccessible cardinal $\kappa $ . First, we force with Lv $(\kappa )$ over M, so that if G is an Lv $(\kappa )$ -generic filter over M, we obtain the generic extension model $M[G]$ in which all cardinals less than $\kappa $ have been collapsed to $\omega $ , i.e., $\kappa =(\omega _{1})^{M[G]}$ (see Kunen [Reference Kunen16, Theorem 8.8, p. 231]). Second, if $\mathbb {R}$ is the set of real numbers of $M[G]$ , then $L(\mathbb {R})$ is the inner model $\mathsf {HOD}(\mathbb {R})$ (defined within $M[G]$ ), i.e., the class of those sets of $M[G]$ which are hereditarily ordinal-definable from a finite number of elements of $\mathrm {On}\cup \mathbb {R}\cup \{\mathbb {R}\}$ , where $\mathrm {On}$ is the class of all ordinal numbers; for “Ordinal-definable sets”, see Jech [Reference Jech14, pp. 194–196] or Kunen [Reference Kunen16, Chapter V, Section 2, p. 157].

In $L(\mathbb {R})$ , we let

$$ \begin{align*}\mathbb{P}=\wp^{\infty}(\omega)/[\omega]^{<\omega}=\{[x]:x\in \wp^{\infty}(\omega)\},\end{align*} $$

where for $x\in \wp ^{\infty }(\omega )$ , $[x]=\{x\triangle y:y\in [\omega ]^{<\omega }\}$ (where “ $\triangle $ ” denotes the operation of symmetric difference between sets). We define a binary relation on $\mathbb {P}$ by stipulating, for all $p,q\in \mathbb {P}$ ,

$$ \begin{align*}p\leq q\Longleftrightarrow\forall x\in p\forall y\in q(x\setminus y\text{ is finite)}.\end{align*} $$

It is easy to verify that $\leq $ is a partial order on $\mathbb {P}$ .Footnote ⁵ Let G be a $\mathbb {P}$ -generic set over $L(\mathbb {R})$ and let $L(\mathbb {R})[G]$ be the generic extension model of $L(\mathbb {R})$ .

In $L(\mathbb {R})[G]$ , we let

$$ \begin{align*}\mathscr{U}=\bigcup G.\end{align*} $$

Claim 5.17. $\mathscr {U}$ is a free ultrafilter on $\omega $ in $L(\mathbb {R})[G]$ . Thus,

$$ \begin{align*}L(\mathbb{R})[G]\models\mathsf{UF}(\omega).\end{align*} $$

Proof. First, we show that $\mathscr {U}$ is a free filter on $\omega $ .

1. (1) $\emptyset \not \in \mathscr {U}$ . Otherwise, $[\emptyset ]=[\omega ]^{<\omega }\in \mathbb {P}$ , a contradiction.

2. (2) $\forall x\in \wp (\omega )(|\omega \setminus x|<\aleph _{0}\rightarrow x\in \mathscr {U})$ . Since G is a filter in $\mathbb {P}$ , $G\neq \emptyset $ . So, there exists $x\in \wp ^{\infty }(\omega )$ such that $[x]\in G$ . Since $[x]\leq [\omega ]$ and G is a filter, $[\omega ]\in G$ . As $[\omega ]=\{z\in \wp (\omega ):|\omega \setminus z|<\aleph _{0}\}$ , we conclude that every co-finite subset of $\omega $ is an element of $\mathscr {U}$ .

3. (3) $\forall x\in \mathscr {U}\forall y\in \mathscr {U}(x\cap y\in \mathscr {U})$ . Fix $x,y\in \mathscr {U}$ . Then $[x],[y]\in G$ and since G is a filter in $\mathbb {P}$ , there exists $z\in \wp ^{\infty }(\omega )$ such that $[z]\in G$ , $[z]\leq [x]$ , and $[z]\leq [y]$ . By definition of $\leq $ , we have that both $z\setminus x$ and $z\setminus y$ are finite. Therefore, $z\setminus (x\cap y)$ is finite and, since z is infinite, $x\cap y$ is infinite. It follows that $[x\cap y]\in \mathbb {P}$ and $[z]\leq [x\cap y]$ . Since $[z]\in G$ and G is a filter, we obtain that $[x\cap y]\in G$ . Thus, $x\cap y\in \mathscr {U}$ .

4. (4) $\forall x\in \mathscr {U}\forall y\in \wp (\omega )(x\subseteq y\rightarrow y\in \mathscr {U})$ . Fix $x\in \mathscr {U}$ and $y\in \wp (\omega )$ with $x\subseteq y$ . Then $[x]\in G$ , $[y]\in \mathbb {P}$ , and $[x]\leq [y]$ . Since $[x]\in G$ and G is a filter, we conclude that $[y]\in G$ , and so $y\in \mathscr {U}$ .

From (1)-(4), we infer that $\mathscr {U}$ is a free filter on $\omega $ .

Now, we show that $\mathscr {U}$ is an ultrafilter on $\omega $ in $L(\mathbb {R})[G]$ . Since, by (2), $\mathscr {U}$ contains every co-finite subset of $\omega $ , it suffices to show that, for every $x\in \wp ^{\infty }(\omega )$ with $x^{c}\in \wp ^{\infty }(\omega )$ , either $x\in \mathscr {U}$ or $x^{c}\in \mathscr {U}$ . So, fix such an $x\in \wp ^{\infty }(\omega )$ in $L(\mathbb {R})[G]$ . By Claim 5.16, we have that $x,x^{c}\in L(\mathbb {R})$ . Due to this fact, we may let within $L(\mathbb {R})$ ,

$$ \begin{align*}D=\{p\in\mathbb{P}:p\leq [x]\vee p\leq [x^{c}]\}.\end{align*} $$

Then D is a dense subset of $\mathbb {P}$ . To see this, fix $p\in \mathbb {P}$ . There exists $z\in \wp ^{\infty }(\omega )$ such that $p=[z]$ . Since z is infinite and $z=(z\cap x)\cup (z\cap x^{c})$ , it follows that $z\cap x$ is infinite or $z\cap x^{c}$ is infinite. If the first case holds, then $[z\cap x]\in D$ and $[z\cap x]\leq p$ , and, if the second case holds, then $[z\cap x^{c}]\in D$ and $[z\cap x^{c}]\leq p$ . Therefore, either of the above two cases yields an extension of p in D, i.e., D is dense in $\mathbb {P}$ , as required.

Since G is $\mathbb {P}$ -generic over $L(\mathbb {R})$ , $G\cap D\neq \emptyset $ . So, let $g\in G\cap D$ . It follows that $g\leq [x]$ or $g\leq [x^{c}]$ and, as G is a filter, we obtain that $[x]\in G$ or $[x^{c}]\in G$ , i.e., $x\in \mathscr {U}$ or $x^{c}\in \mathscr {U}$ . Therefore, $\mathscr {U}$ is an ultrafilter, finishing the proof of the claim.

The above arguments complete the proof of the theorem.

Proof. (i) Let $\mathcal {I}$ be an independent family in $\omega $ of size $2^{\aleph _{0}}$ , and also let

$$ \begin{align*}\mathcal{F}=\mathcal{I}\cup\left\{\bigcup_{A\in\mathcal{J}}A^{c}:\mathcal{J}\in\wp^{\infty}(\mathcal{I})\right\}.\end{align*} $$

Since $\mathcal {I}$ is independent, it is easy to see that $\mathcal {F}$ has the finite intersection property. Furthermore, $\bigcap \mathcal {F}=\emptyset $ since $\bigcap \mathcal {F}\subseteq (\bigcap \mathcal {I})\cap (\bigcup _{A\in \mathcal {I}}A^{c})=\emptyset $ .

By $\mathsf {BPI}(\omega )$ , there exists an ultrafilter $\mathscr {U}$ on $\omega $ which extends $\mathcal {F}$ . Since $\bigcap \mathcal {F}=\emptyset $ , $\mathscr {U}$ is free. We assert that $\mathscr {U}$ has no base of cardinality $<2^{\aleph _{0}}$ . Assume the contrary, and let $\mathscr {C}$ be a base for $\mathscr {U}$ such that $|\mathscr {C}|<2^{\aleph _{0}}$ . For each $C\in \mathscr {C}$ , let

$$ \begin{align*}\mathcal{S}_{C}=\{A\in\mathcal{I}:C\subseteq A\}.\end{align*} $$

Then

$$ \begin{align*}\mathcal{I}=\bigcup\{\mathcal{S}_{C}:C\in\mathscr{C}\}.\end{align*} $$

Since $|\mathscr {C}|< 2^{\aleph _{0}}$ and $|\mathcal {I}|=2^{\aleph _{0}}$ , it follows from the second principle of our hypothesis that for some $C\in \mathscr {C}$ , $S_{C}$ is an infinite set. By the definition of $S_{C}$ , we have $C\subseteq \bigcap S_{C}$ , and hence $\bigcap S_{C}\in \mathscr {U}$ since $C\in \mathscr {C}\subseteq \mathscr {U}$ and $\mathscr {U}$ is a filter. Furthermore, since $S_{C}\in \wp ^{\infty }(\mathcal {I})$ , it follows that $(\bigcap S_{C})^{c}=\bigcup _{A\in S_{C}}A^{c}\in \mathcal {F}\subseteq \mathscr {U}$ and thus $(\bigcap S_{C})^{c}\in \mathscr {U}$ . This, together with $\bigcap S_{C}\in \mathscr {U}$ , contradicts $\mathscr {U}$ ’s being a filter. Therefore, $\mathscr {U}$ has no base of cardinality $<2^{\aleph _{0}}$ .

(ii) Assuming $\mathsf {BPI}(\omega )$ and following the proof of (i), we have that, if $\mathscr {C}$ is a well-ordered base for the ultrafilter $\mathscr {U}$ which extends $\mathcal {F}$ and $|\mathscr {C}|<2^{\aleph _{0}}$ , then again for some $C\in \mathscr {C}$ , $\mathcal {S}_{C}$ is infinite; otherwise, $2^{\aleph _{0}}=|\mathcal {I}|=|\bigcup \{\mathcal {S}_{C}:C\in \mathscr {C}\}|=|\mathscr {C}|< 2^{\aleph _{0}}$ , a contradiction. We may now complete the proof as in (i), obtaining again a contradiction to $\mathscr {U}$ ’s being a filter.

(iii) This follows from (ii) and the fact that the Basic Cohen Model [Reference Howard and Rubin11, Model $1$ ] satisfies $\mathsf {BPI}+\neg \mathsf {AC}(\mathbb {R})$ (see [Reference Jech13], or [Reference Howard and Rubin11]).

(iv) The first assertion follows from Theorem 5.1(i) and part (iii), and the second assertion follows from the first and the fact that $\mathsf {AC}(\mathbb {R})$ (strictly) implies $\mathsf {BPI}(\omega )$ in  $\mathsf {ZF}$ .

Proof. (i) ( $\mathsf {AC}(\mathbb {R}) \Rightarrow \mathsf {BPI_{wob}}(\omega )$ ) This is an immediate consequence of the fact that $|\wp (\omega )|=|\mathbb {R}|$ and the fact that $\mathsf {AC}(\mathbb {R})$ is equivalent to “ $\mathbb {R}$ can be well ordered”.

( $\mathsf {BPI_{wob}}(\omega ) \Rightarrow \mathsf {AC}(\mathbb {R})$ ) Let $\mathcal {I}$ and $\mathcal {F}$ be defined as in the proof of Theorem 5.20(i). By $\mathsf {BPI_{wob}}(\omega )$ , $\mathcal {F}$ can be extended to an ultrafilter $\mathscr {U}$ with a well-orderable base, $\mathscr {C}$ say. Then $|\mathscr {C}|\leq 2^{\aleph _{0}}$ and, by the proof of Theorem 5.20(i), $|\mathscr {C}|\not < 2^{\aleph _{0}}$ . Therefore, $|\mathscr {C}|=2^{\aleph _{0}}$ , i.e., $\mathbb {R}$ can be well ordered. So, $\mathsf {AC}(\mathbb {R})$ holds true.

The second assertion of (i) follows from the first and the fact that $\mathsf {BPI}$ (and thus $\mathsf {BPI}(\omega )$ ) does not imply $\mathsf {AC}(\mathbb {R})$ in $\mathsf {ZF}$ (see [Reference Howard and Rubin11, Basic Cohen Model $1$ ]).

(ii) The first assertion follows from (i) (or Theorem 5.1(i)), and the second one follows from the fact that “every ultrafilter on $\omega $ has a well-orderable base” is true in Feferman’s model $\mathcal {M}2$ of [Reference Howard and Rubin11] since, in $\mathcal {M}2$ , every ultrafilter on $\omega $ is principal, but $\mathsf {AC}(\mathbb {R})$ is false in $\mathcal {M}2$ (see [Reference Feferman3] or [Reference Jech13, Problem 24, p. 82]).

Proof. (i) In Solovay’s model $L(\mathbb {R})$ [Reference Howard and Rubin11, Model $5\left(\mathrm{\aleph}\right)$ ], $\aleph _{1}\not \le 2^{\aleph _{0}}$ , and hence the only well-orderable subsets of $\mathbb {R}$ in $L(\mathbb {R})$ are the countable ones. It follows that in $L(\mathbb {R})$ , $\mathsf {MA}$ is equivalent to $\mathsf {MA}(\aleph _{0})$ . Since $\mathsf {DC}$ implies $\mathsf {MA}(\aleph _{0})$ (see Kunen [Reference Kunen16, Lemma 2.6(c), pp. 54–55]) and $\mathsf {DC}$ is true in $L(\mathbb {R})$ , we conclude that $\mathsf {MA}$ is true in $L(\mathbb {R})$ . However, $\mathsf {UF}(\omega )$ is false in $L(\mathbb {R})$ , and thus so is $\mathsf {UF_{wob}}(\omega )$ .

(ii) Assume $\mathsf {MA}$ . By way of contradiction, suppose that there exists a free ultrafilter $\mathscr {U}$ on $\omega $ with a well-orderable base $\mathscr {B}$ such that $|\mathscr {B}|<2^{\aleph _{0}}$ . We let

$$ \begin{align*}P=\{(x,y):x,y\in[\omega]^{<\omega},x\cap y=\emptyset\}\end{align*} $$

and define a binary relation $\preceq $ on P by stipulating, for all $(x,y),(z,w)\in P$ ,

$$ \begin{align*}(x,y)\preceq (z,w)\Longleftrightarrow z\subseteq x\wedge w\subseteq y.\end{align*} $$

It is easy to verify that $\preceq $ is a partial order on P. Furthermore, since $|[\omega ]^{<\omega }|=\aleph _{0}$ , it follows that $|P|=\aleph _{0}$ . Hence, the poset $(P,\preceq )$ has the c.c.c.

For each $B\in \mathscr {B}$ , we let

(10) $$ \begin{align} D_{B}=\{(x,y)\in P: B\cap x\ne\emptyset\text{ and } B\cap y\ne\emptyset\}. \end{align} $$

We let

$$ \begin{align*}\mathcal{D}=\{D_{B}:B\in\mathscr{B}\}.\end{align*} $$

Then we have:

$$ \begin{align*}|\mathcal{D}|\le|\mathscr{B}|<2^{\aleph_{0}},\end{align*} $$

since the mapping $B\mapsto D_{B}$ , $B\in \mathscr {B}$ , is onto, and thus $|\mathcal {D}|\le |\mathscr {B}|$ since $\mathscr {B}$ is well ordered. By $\mathsf {MA}$ (in particular, by $\mathsf {MA}(|\mathscr {B}|)$ ) applied to $(P,\preceq )$ and $\mathcal {D}$ , we obtain a $\mathcal {D}$ -generic filter in P, G say. Let

(11) $$ \begin{align} F_{1}=\bigcup\operatorname{\mathrm{{dom}}}(G),\ F_{2}=\bigcup\operatorname{\mathrm{{ran}}}(G). \end{align} $$

Now, we let

$$ \begin{align*}\mathscr{C}_{1}=\mathscr{B}\cup\{F_1\},\ \mathscr{C}_2=\mathscr{B}\cup\{F_2\}.\end{align*} $$

Since $\mathscr {B}$ is a base for $\mathscr {U}$ and (by Claim 5.26) $\mathscr {C}_{1}$ and $\mathscr {C}_{2}$ have the finite intersection property, it follows that

$$ \begin{align*}\forall U\in\mathscr{U}(F_{1}\cap U\ne\emptyset\text{ and }F_2 \cap U\ne\emptyset),\end{align*} $$

and thus, as $\mathscr {U}$ is an ultrafilter, we deduce that

$$ \begin{align*}F_1\in\mathscr{U}\text{ and }F_2\in\mathscr{U}.\end{align*} $$

This, together with Claim 5.25, yields a contradiction to the fact that $\mathscr {U}$ is a filter. Therefore, $\mathscr {U}$ cannot have a well-orderable base of cardinality $<2^{\aleph _{0}}$ , as required.

For the second assertion of (ii), we first note that Tachtsis [Reference Tachtsis26] had constructed a Fraenkel–Mostowski permutation model $\mathcal {N}$ in which $\mathsf {CH}$ is true, but $\mathsf {MA}(\aleph _{0})$ (and thus $\mathsf {MA}$ ) is false (see [Reference Tachtsis26, Theorem 2.15]). By Remark 4.3, it follows that “no free ultrafilter on $\omega $ has a well-orderable base of cardinality $< 2^{\aleph _{0}}$ ” is true in $\mathcal {N}$ . On the other hand, it is easy to verify that the latter statement and $\neg \mathsf {MA}(\aleph _{0})$ are both boundable (for the definition of the term “boundable statement”, see [Reference Howard and Rubin11, Note 103] or [Reference Jech13, Problem 1, p. 94]), and that this yields their conjunction is also boundable. Since this conjunction has a permutation model, it follows from the Jech–Sochor First Embedding Theorem [Reference Jech13, Theorem 6.1] that it also has a symmetric model of $\mathsf {ZF}$ . Therefore, “no free ultrafilter on $\omega $ has a well-orderable base of cardinality $< 2^{\aleph _{0}}$ ” does not imply $\mathsf {MA}$ in $\mathsf {ZF}$ , as required.

(iii) The first assertion follows immediately from the hypothesis and the fact that any base for a filter on $\omega $ has cardinality $\leq 2^{\aleph _{0}}$ .