Proof Suppose there exist $v\in V$ and a complete homomorphism $f\colon \mathbb {B}\to \mathbb {C}\mathbin {\upharpoonright } v$ such that $U=f^{-1}[V]$ . Let us define

and note that $c\le f(b)$ implies $g(c)\le b$ . Furthermore, observe that if $c\in V$ then, by completeness of the homomorphism f,

hence $f(g(c))\in V$ , which means $g(c)\in U$ . In conclusion, the maps f and g witness that $\langle U,\ge \rangle \le _{\textrm{T}}\langle V,\ge \rangle $ .

Proof (1 $\Longrightarrow $ 2) Assuming V is not $\sigma $ -complete, there exists a maximal antichain

in $\mathbb {B}$ such that for all $i<\omega $ , $a_i\notin V$ . The function

is a complete homomorphism, hence $U=f^{-1}[V]$ is the desired ultrafilter.

(2 $\Longrightarrow $ 3) By Proposition 2.7.

(3 $\Longrightarrow $ 1) Suppose U is a non-principal ultrafilter over $\omega $ such that $\langle U,\supseteq \rangle \le _{\textrm{T}} \langle V,\ge\rangle$ . By Proposition 2.3 we have $\operatorname {\textrm{add}}(V)\le \operatorname {\textrm{add}}(U)=\aleph _0$ , implying V is not $\sigma $ -complete.

Proof Let U be an ultrafilter on $\mathbb {B}(I,\kappa )$ . We define a function $x\colon \kappa \times 2\to \operatorname {\textrm{Clop}}(^\kappa 2)$ as follows: for every $\alpha <\kappa $ ,

As U is an ultrafilter, there exists a function $g\colon \kappa \to 2$ such that for all $\alpha <\kappa $ ,

. Let

it is sufficient to show that whenever $Y\subseteq X$ is infinite we have $\bigwedge Y\notin U$ , then conclude using Proposition 2.5.

Let $\Delta \colon \omega \to \kappa $ be an arbitrary injective function. Since we are assuming I contains all singletons, we have $\{g\circ \Delta \}\in I$ . Therefore, by definition,

Now using the fact that $I(\kappa )$ is a $\sigma $ -ideal,

as desired.

Proof The upper bound $\mathfrak {u}(\mathbb {C}_{\omega _1})+\mathfrak {u}(\mathbb {B}_{\omega _1})\le 2^{\aleph _0}$ follows directly from the observation that the Boolean algebras $\mathbb {C}_{\omega _1}$ and $\mathbb {B}_{\omega _1}$ have both cardinality $2^{\aleph _0}$ .

As for $\mathfrak {u}(\mathbb {C}_\omega )\le \mathfrak {u}(\mathbb {C}_{\omega _1})$ , Remark 2.11 gives a $\sigma $ -complete embedding $f\colon \mathbb {C}_\omega \to \mathbb {C}_{\omega _1}$ which, due to the fact that $\mathbb {C}_\omega $ is c.c.c., is in fact a complete embedding. Consequently, for every ultrafilter U on $\mathbb {C}_{\omega _1}$ we have $\operatorname {\textrm{cof}}\bigl (f^{-1}[U]\bigr )\le \operatorname {\textrm{cof}}(U)$ by Proposition 2.3 and Proposition 2.7. The inequality $\mathfrak {u}(\mathbb {B}_\omega )\le \mathfrak {u}(\mathbb {B}_{\omega _1})$ is proved analogously.

To prove $\mathfrak {u}\le \mathfrak {u}(\mathbb {C}_\omega )$ , it is sufficient to note that no ultrafilter on $\mathbb {C}_\omega $ is $\sigma $ -complete, then apply Proposition 2.3 and Corollary 2.8. Similarly for $\mathfrak {u}\le \mathfrak {u}(\mathbb {B}_\omega )$ .

We now prove the inequality $\mathfrak {d}\le \mathfrak {u}(\mathbb {C}_\omega )$ . Let U be an ultrafilter on $\mathbb {C}_\omega $ ; since U is not $\sigma $ -complete, there exists a maximal antichain

such that for all $i<\omega $ , $a_i\notin U$ . Let $X\subset U$ be such that ${\lvert X \rvert }<\mathfrak {d}$ , we aim to prove that X is not cofinal in $\langle U,\ge \rangle $ . Whenever $s\in {^{<\omega }2}$ , we let

be the corresponding element of $\operatorname {\textrm{Clop}}(^\omega 2)$ . By point (1) of Proposition 3.2, for every

there exists some $s\in {^{<\omega }2}$ such that

. Using this fact, for each $x\in X$ we may define a function $f_x\colon \omega \to \omega $ as follows. Let $i<\omega $ ; as every element of U meets infinitely many members of

, let $j_i$ be the least $j\ge i$ such that

. Then we define

Since ${\lvert X \rvert }<\mathfrak {d}$ , there exists an increasing function $g\colon \omega \to \omega $ such that for all $x\in X$ we have $g\nleq f_x$ . Now define

and assume $u\in U$ ; the other case when $\neg u\in U$ is completely analogous. Towards a contradiction suppose now X is cofinal in $\langle U,\ge \rangle $ , this implies the existence of some $x\in X$ such that $x\le u$ . Since g is not dominated by $f_x$ , we can find some $i<\omega $ such that $f_x(i)<g(i)$ ; then, by definition of $f_x$ there exists some $s\colon f_x(i)\to 2$ such that

. Putting everything together, we deduce

which is a contradiction because $f_x(i)<g(j_i)$ . This shows that X cannot be cofinal in the ultrafilter and completes the proof of $\mathfrak {d}\le \mathfrak {u}(\mathbb {C}_\omega )$ .

The fact that $\operatorname {\textrm{cof}}(\mathcal {N})\le \mathfrak {u}(\mathbb {B}_\omega )$ is a consequence of the work of Burke [Reference Burke3]. More precisely, in Case 1 of [Reference Burke3, Theorem 1], he showed the following: if has the property that for all $b\in \mathbb {B}_\omega $ there exists $x\in X$ such that either $x\le b$ or , then there exists a cofinal subset $Y\subseteq \mathcal {N}$ such that ${\lvert X \rvert }={\lvert Y \rvert }$ . Therefore, the inequality $\operatorname {\textrm{cof}}(\mathcal {N})\le \mathfrak {u}(\mathbb {B}_\omega )$ follows from the observation that, whenever U is an ultrafilter on $\mathbb {B}_\omega $ and $X\subseteq U$ is cofinal, clearly X satisfies the assumption of Burke’s result.

The other inequalities in the diagram are well known.

Proof As the subalgebra $\mathbb {D}_\alpha $ is countable, there are only countably many $s\in \mathbb {C}_\alpha $ such that and the set is finite. For every such s, let , so that $\mathbb {M}(U,\mathbb {C}_\alpha )$ is the countable union of the $M_s$ . Furthermore, finitely many conditions $p_1,\dots ,p_n\in M_s$ are compatible, as witnessed by the stronger condition $\langle s,S_{p_1}\wedge \dots \wedge S_{p_n}\rangle $ .

Proof Let us assume V satisfies $\aleph _1<2^{\aleph _0}$ . We proceed with a finite-support iteration of the forcing introduced above; alongside the iteration, for each $\alpha <\omega _1$ we construct a $\mathbb {P}_\alpha $ -name $\dot {U}_\alpha $ for an ultrafilter on $\mathbb {C}_{\omega +\alpha }$ . As the iteration is $\sigma $ -centered, in particular the c.c.c. will guarantee that cardinals are preserved in the forcing extension.

First of all, let $\mathbb {P}_0=\{\emptyset \}$ and let $\dot {U}_0$ be the name for any ground-model ultrafilter $U_0$ on $\mathbb {C}_\omega $ with the property that for all $i<\omega $ , $a_i\notin U_0$ . For $\alpha <\omega _1$ , suppose $\mathbb {P}_\alpha $ and $\dot {U}_\alpha $ are given; we let

$$ \begin{align*} \mathbb{P}_{\alpha+1}=\mathbb{P}_\alpha*\dot{\mathbb{M}}(U_\alpha,\mathbb{C}_{\omega+\alpha}) \end{align*} $$

and $\dot {g}_\alpha $ be a $\mathbb {P}_{\alpha +1}$ -name for the element of $\mathbb {C}_{\omega +\alpha }$ added generically at this stage. We then take $\dot {U}_{\alpha +1}$ to be the name for an ultrafilter on $\mathbb {C}_{\omega +\alpha +1}$ extending $\dot {U}_\alpha \cup \{\dot {g}_\alpha \}$ . Finally, at limit ordinals $\delta <\omega _1$ , we simply let $\dot {U}_\delta $ be a name for an ultrafilter on $\mathbb {C}_{\omega +\delta }$ extending $\bigcup _{\alpha <\delta }\dot {U}_\alpha $ .

In the resulting forcing extension $V[G]$ , let $U=\bigcup _{\alpha <\omega _1}U_\alpha $ . We claim that U is an ultrafilter on $\mathbb {C}_{\omega _1}$ ; indeed, if $b\in \mathbb {C}_{\omega _1}$ then there exists some $\alpha <\omega _1$ such that $b\in \mathbb {C}_\alpha $ . Hence, at some later stage of the iteration, either b or $\neg b$ is decided to be in U. In conclusion, U is an ultrafilter on $\mathbb {C}_{\omega _1}$ such that

$$ \begin{align*} \Big\{g_\alpha\wedge\bigvee\{a_i \mid i\ge n\} \mid \alpha<\omega_1,\ n<\omega \Big\} \end{align*} $$

is cofinal in $\langle U,\ge \rangle $ , thus witnessing the fact that $\aleph _1=\mathfrak {u}(\mathbb {C}_{\omega _1})<2^{\aleph _0}$ in $V[G]$ .

Proof of Theorem 3.5 Let us assume V satisfies Martin’s Axiom [Reference Martin and Solovay17] and consider the forcing of Lemma 3.7, which resulted in a model of $\mathfrak {u}(\mathbb {C}_{\omega _1})<2^{\aleph _0}$ . The forcing is $\sigma $ -centered, being an iteration of $\sigma $ -centered forcings, and it is known that such extension will also satisfy $\operatorname {\textrm{non}}(\mathcal {N})=2^{\aleph _0}$ . Indeed, this is explained in detail in the proof of [Reference Blass, Brendle, Brian, Hamkins, Hardy and Larson1, Corollary 41], which is enough to conclude the proof of our theorem.

Proof of Theorem 3.8 Let us assume V satisfies $\aleph _1<2^{\aleph _0}$ . As in the proof of Theorem 3.5, we proceed with a finite-support iteration of the c.c.c. forcing just introduced; alongside the iteration, for each $\alpha <\omega _1$ we construct a $\mathbb {P}_\alpha $ -name $\dot {U}_\alpha $ for an ultrafilter on $\mathbb {B}_{\omega +\alpha }$ .

The base case of the iteration presents no complications. For $\alpha <\omega _1$ , suppose $\mathbb {P}_\alpha $ and $\dot {U}_\alpha $ are given; we let

$$ \begin{align*} \mathbb{P}_{\alpha+1}=\mathbb{P}_\alpha*\dot{\mathbb{K}}(U_\alpha,\mathbb{B}_{\omega+\alpha}) \end{align*} $$

and $\dot {g}_\alpha $ be a $\mathbb {P}_{\alpha +1}$ -name for the function $\omega \longrightarrow \mathbb {B}_{\omega +\alpha }$ added generically at this stage. We then take $\dot {U}_{\alpha +1}$ to be the name for an ultrafilter on $\mathbb {B}_{\omega +\alpha +1}$ extending . Finally, at countable limit ordinals, we proceed as usual by taking unions.

In the resulting forcing extension $V[G]$ , let $U=\bigcup _{\alpha <\omega _1}U_\alpha $ . Then, U is an ultrafilter on $\mathbb {B}_{\omega _1}$ such that

is cofinal in $\langle U,\ge \rangle $ , thus witnessing the fact that $\aleph _1=\mathfrak {u}(\mathbb {B}_{\omega _1})<2^{\aleph _0}$ in $V[G]$ .

Question 3.10. Is it consistent that $\mathfrak {u}(\mathbb {B}_\omega )<\mathfrak {u}(\mathbb {C}_\omega )$ ?

Proof (1 $\Longrightarrow $ 2) Suppose U is a coherent P-ultrafilter; let a maximal antichain

and a subset

be given. We aim to find first a maximal antichain

such that

• • for all $i<\omega $ there exists $j<\omega $ such that $p_i\le a_j$ and

• • for every $n<\omega $ the set is finite.

To do so, it is convenient to use the following notation: for $n<\omega $ , let

$$ \begin{align*} x^1_n=x_n\quad\text{ and }x^0_n=\neg x_n. \end{align*} $$

Now, for $s\in {^{<\omega }2}$ we define

First, it is clear that for all $s,t\in {^{<\omega }2}$ , if

then $s=t$ . Secondly, by finite distributivity we have for each $i<\omega $

and consequently

Thirdly, we observe that for every $n<\omega $

and the set on the right-hand side is finite. In conclusion, if we enumerate

then it is clear that

is a maximal antichain in $\mathbb {B}$ satisfying the two desired properties.

For every $n<\omega $ , the set

is such that

hence, by the assumption that U is a coherent P-ultrafilter, we can find $Y\subseteq \omega $ such that

and for each $n<\omega $ ,

is finite. Now we observe that for $n<\omega $

which is finite. From this and the fact that

refines

, it follows that for each $n<\omega $ also

is finite.

(2 $\Longrightarrow $ 1) Let be a maximal antichain in $\mathbb {B}$ . Assume that for each $n<\omega $ , $X_n\subseteq \omega $ is such that . By our hypothesis, there exists $y\in U$ such that for each $n<\omega $ the set is finite. Letting , it is obvious that and for each $n<\omega $ , is finite.

Proof Let $u\in U$ be arbitrary; by density of $\mathbb {D}$ , there exists a maximal antichain such that:

• • for all $i<\omega $ there exists $a\in A$ such that $p_i\le a$ and

• • for all $i<\omega $ , either $p_i\le u$ or .

Using the assumption $A\cap U=\emptyset $ , it follows that for all $a\in A$

Now, by the fact that

is a P-point over $\omega $ , there exists $Y\subseteq \omega $ , with

, such that for every $a\in A$ the set

is finite.

We claim that the element defined as

has the desired properties. Clearly $v\in U$ and $v\le u$ . Furthermore, for each $a\in A$ we have

but there are only finitely many $i\in Y$ such that

. This means that $a\wedge v$ is a finite (possibly empty) disjunction of elements of the subalgebra $\mathbb {D}$ , and therefore $a\wedge v\in \mathbb {D}$ .

Proof Let $X\subseteq U$ be such that $\pi (\mathbb {B})<{\lvert X \rvert }$ ; we aim to find some infinite $Y\subseteq X$ such that $\bigwedge Y\in U$ . If U is $\sigma $ -complete then this is immediate, so let us assume U is not $\sigma $ -complete, which means there exists a maximal antichain such that for all $i<\omega $ , $a_i\notin U$ .

Let $\mathbb {D}\subseteq \mathbb {B}$ be a dense subalgebra such that ${\lvert {\mathbb {D}} \rvert }=\pi (\mathbb {B})$ . Without loss of generality, we may assume that for all $u\in X$ , . Indeed, by Lemma 4.5, for all $u\in X$ there exists $v_u\in U$ such that $v_u\le u$ and . In case we have chosen the same $v_u$ for infinitely many $u\in X$ , we have the thesis already. On the other hand, if the map $u\mapsto v_u$ is finite-to-one, without loss of generality we may replace X with , which also has cardinality greater than $\pi (\mathbb {B})$ .

Now, we claim that there is some $x\in X$ such that for all $n<\omega $

is infinite. If not, then for each $x\in X$ we could find some $n_x<\omega $ such that the set

(2)

is finite. Clearly there exists $X'\subseteq X$ , with $\pi (\mathbb {B})<{\lvert {X'} \rvert }$ , such that for all $x\in X'$ , $n_x=m$ for some fixed $m<\omega $ . Now, the function

cannot be injective, hence there must be some infinite $X''\subseteq X'$ such that $x,x'\in X''$ implies $(\forall i<m)(a_i\wedge x=a_i\wedge x')$ , contradicting the assumption that the set in (2) is finite. This completes the proof of the claim.

Since for every $n<\omega $ the set $X_n$ is infinite, recursively we can choose

such that for each $n<\omega $ ,

. By Lemma 4.3, there exists $y\in U$ such that for each $n<\omega $ the set

is finite. From now on, we follow the steps of Kanamori [Reference Kanamori13, Theorem 1.10]. For $n<\omega $ we define two natural numbers

and

We also define the corresponding interval of natural numbers

$$ \begin{align*} I_n=[\gamma_n,\delta_n), \end{align*} $$

possibly empty.

Note that for each $n<\omega $ we have $n\le \gamma _n$ , hence we can easily find by recursion an infinite set $W\subseteq \omega $ such that for all $n,m\in W$ , if $I_n\cap I_m\neq \emptyset $ then $n=m$ . Now there is clearly an infinite subset $Z\subseteq W$ with the property that

Since U is an ultrafilter, we have $x\wedge y\wedge \neg z\in U$ ; our goal is now to show that for every $n\in Z$

(3) $$ \begin{align} x\wedge y\wedge\neg z\le x_n. \end{align} $$

Let $n\in Z$ be given. As

is a maximal antichain, to establish (3) it is sufficient to prove that for any $i<\omega $

$$ \begin{align*} a_i\wedge x\wedge y\wedge\neg z\le x_n. \end{align*} $$

If i is such that $a_i\le z$ then we are done, so let us assume

. This implies that $i\notin I_n$ , so there are now two possibilities. If $i<\gamma _n$ then by definition $a_i\wedge x\le x_n$ . Otherwise, if $\delta _n\le i$ then again by definition $a_i\wedge y\le x_n$ . This establishes (3) for each $n\in Z$ .

In conclusion, by taking , it follows that Y is an infinite subset of X such that $x\wedge y\wedge \neg z\le \bigwedge Y$ and therefore $\bigwedge Y\in U$ . The thesis now follows from Proposition 2.5.

Proof By Proposition 4.4 and Theorem 4.6.

Proof By $\diamondsuit $ , let us fix a sequence such that $S_\alpha \subseteq \alpha $ and, for every $X\subseteq \omega _1$ , the set is stationary. As noted before, the existence of such a sequence implies $2^{\aleph _0}=\aleph _1$ , hence we may enumerate . We proceed to construct recursively a sequence such that, at each stage $\alpha <\omega _1$ , the set has the finite intersection property.

Let . For $\alpha <\omega _1$ , inductively the set has the finite intersection property, so we may choose $U_{\alpha +1}$ to be either $B_\alpha $ or $\neg B_\alpha $ in such a way that still has the finite intersection property. Suppose now $\delta <\omega _1$ is a limit ordinal; we distinguish two cases:

1. (1) There exists an infinite $Y\subseteq S_\delta $ such that has the finite intersection property. Then choose such a Y and define $U_\delta =\bigcap _{\alpha \in Y}U_\alpha $ .

2. (2) There is no such a Y. Then define .

This completes the recursive construction, whence it follows that is an ultrafilter on $\mathbb {B}_\omega $ .

Now let $X\subseteq \omega _1$ be uncountable; we aim to find an infinite $Y\subseteq X$ such that $\bigcap _{\alpha \in Y}U_\alpha \in U$ , then conclude by Proposition 2.5 that U is not Tukey maximal. By stationarity there exists, for some sufficiently large $\kappa $ , a countable elementary substructure $M\preceq H_\kappa $ such that $\{X,U\}\subset M$ , and a limit ordinal $\delta <\omega _1$ such that $X\cap M=S_\delta $ . Now we only need to find an infinite $Y\subseteq X\cap M$ such that has the finite intersection property, as this will imply that, at stage $\delta $ , we are in Case (1) of the construction of U.

To simplify notation, as $\delta $ is countable, we may enumerate all the possible finite intersections from as . By recursion, we shall construct for all $n<\omega $ an uncountable $X_n\in \mathcal {P}(X)\cap M$ , a countable ordinal $\alpha _n\in X_n\cap M$ , and non-empty clopen sets . At each step n, the following crucial property will hold true:

(⋆_n) $$\begin{align} (\forall m\le n)(\forall\alpha\in X_n)\ \mu\Biggl(F_m\cap\bigcap_{k<n} U_{\alpha_k}\cap U_\alpha\cap C_{n,m}\Biggr)\ge\biggl(1-\frac{1}{2^{n+2}}\biggr)\mu(C_{n,m}). \end{align}$$

Once the recursive construction is complete, we shall conclude the proof by showing that has the desired property.

For the base case of the recursion, we first observe that for every $\alpha \in X$ the set $F_0\cap U_\alpha $ has positive measure. But X is uncountable, hence by Proposition 3.2 there exists an uncountable $X_0\subseteq X$ and a non-empty clopen $C_{0,0}$ such that

(⋆₀) $$\begin{align} (\forall\alpha\in X_0)\ \mu(F_0\cap U_\alpha\cap C_{0,0})\ge\frac{3}{4}\mu(C_{0,0}), \end{align}$$

as we wanted. Now choose any $\alpha _0\in X_0$ . At this point, we observe that the construction which has just been described can be performed within the model M, hence we can further assume that $X_0\in M$ and $\alpha _0\in M$ .

For the general case we proceed similarly, but taking into account the decreasing measure of the sets $C_{n,m}$ . More precisely, let $n>0$ and suppose inductively we have the countable ordinals , an uncountable $X_{n-1}\subseteq X$ , and clopen sets . As in particular $\alpha _{n-1}\in X_{n-1}$ , by $(\star _{n-1})$ we have for all $m<n$

(4) $$ \begin{align} \mu\Biggl(F_m\cap\bigcap_{k<n} U_{\alpha_k}\cap C_{n-1,m}\Biggr)\ge\biggl(1-\frac{1}{2^{n+1}}\biggr)\mu(C_{n-1,m}). \end{align} $$

Therefore, if $m<n$ then for every $\alpha \in X_{n-1}$

$$ \begin{align*} &\mu\Biggl(F_m\cap\bigcap_{k<n} U_{\alpha_k}\cap U_\alpha\cap C_{n-1,m}\Biggr)\\&\quad \ge\mu\Biggl(F_m\cap\bigcap_{k<n} U_{\alpha_k}\cap C_{n-1,m}\Biggr)+\mu\Biggl(F_m\cap\bigcap_{k<n-1} U_{\alpha_k}\cap U_\alpha\cap C_{n-1,m}\Biggr)-\mu(C_{n-1,m})\\ &\quad\ge\biggl(1-\frac{1}{2^{n+1}}\biggr)\mu(C_{n-1,m})+\biggl(1-\frac{1}{2^{n+1}}\biggr)\mu(C_{n-1,m})-\mu(C_{n-1,m})\\ &\quad=\biggl(1-\frac{1}{2^n}\biggr)\mu(C_{n-1,m}), \end{align*} $$

where the second line follows from finite additivity of the measure $\mu $ , whereas the third is a consequence of (4) and $(\star _{n-1})$ . Now, since $X_{n-1}$ is uncountable, by Proposition 3.2 again there exists an uncountable $X_n\subseteq X_{n-1}$ and non-empty clopens such that holds and furthermore:

$$ \begin{align*} \text{if }m<n\text{, then }C_{n,m}\subseteq C_{n-1,m}\text{ and }\mu(C_{n,m})\ge\biggl(1-\frac{1}{2^n}\biggr)\mu(C_{n-1,m}). \end{align*} $$

Now choose any . Again, we observe that each step of the recursive construction can be performed within the model M, hence we can further assume that $X_n\in M$ and $\alpha _n\in M$ .

We have thus obtained , so the proof is complete once we show that, for each fixed $m<\omega $ ,

$$ \begin{align*} \mu\Biggl(F_m\cap\bigcap_{n<\omega}U_{\alpha_n}\Biggr)>0. \end{align*} $$

To do so, consider $\varepsilon =\prod _{\ell =1}^{\infty }\bigl (1 - \frac {1}{2^\ell }\bigr )>0$ and observe that, by our choice of the clopen sets, for every $n\ge m$ we have

$$ \begin{align*} \mu(C_{n,m})\ge\prod_{\ell=1}^{n}\biggl(1 - \frac{1}{2^\ell}\biggr)\mu(C_{m,m})\ge\varepsilon\,\mu(C_{m,m}). \end{align*} $$

Consequently, for every $n\ge m$

$$ \begin{align*} \mu\Biggl(F_m\cap\bigcap_{k<n} U_{\alpha_k}\Biggr)\ge\frac{3}{4}\mu(C_{n,m})\ge\varepsilon\frac{3}{4}\mu(C_{m,m}) \end{align*} $$

and finally, by countable additivity of the measure $\mu $ ,

$$ \begin{align*} \mu\Biggl(F_m\cap\bigcap_{n<\omega}U_{\alpha_n}\Biggr)=\lim_{n\to\infty}\mu\Biggl(F_m\cap\bigcap_{k<n} U_{\alpha_k}\Biggr)\ge\varepsilon\frac{3}{4}\mu(C_{m,m})>0, \end{align*} $$

as desired. This shows that has the required properties and completes the proof.

Question 5.3. Does $2^{\aleph _0}=\aleph _1$ imply the existence of an ultrafilter on $\mathbb {B}_\omega $ which is not Tukey maximal?