Proof. First note that if $\mu <\lambda $ and P is a $\lambda $ -directed poset of cardinality at most $\mu $ , then P has a greatest element and the theorem is trivially true. Hence, let us assume that $\lambda \leq \mu $ . Fix any one-to-one function $g:P\rightarrow \mu $ , and let us prove that the function $f(p)=\{g(p)\}$ is an unbounded function from P to $[\mu ]^{<\lambda }$ . Suppose that X is unbounded. Then $|X|\geq \lambda $ (by $\lambda $ -directedness) and therefore $g"X$ has size at least  $\lambda $ . This implies that $\cup f"X=g"X$ has size $\lambda $ . If follows that $f"X$ is unbounded in  $[\mu ]^{<\lambda }$ .

Proof. Suppose otherwise that there is an unbounded function $f:P\rightarrow Q$ . Since P is not $\lambda $ -directed closed, there is a family $B:=\{p_i\mid i<\theta \}$ for $\theta <\lambda $ which is unbounded. Thus $f"B\subseteq Q$ is unbounded. Since Q is $\lambda $ -directed, and $f"B$ is a family of size less than $\lambda $ , there is $q\in Q$ which bounds $f"B$ , contradiction.

Fact 3.3. Suppose $(P,\leq _P)$ is not $\lambda $ -directed closed, then for every $(Q,\leq _Q)$ , $(P,\leq _P)\times (Q,\leq _Q)$ is not $\lambda $ -directed closed.

Proof. Since $(U,\supseteq )$ is $\lambda $ -directed whenever U is $\lambda $ -complete, and since $|U|=2^\kappa $ , Theorem 3.1 applies ( $\lambda $ being itself and $\mu $ being $2^\kappa $ ) and yields $(U,\supseteq )\leq _T ([2^\kappa ]^{<\lambda },\subseteq )$ . So the “In particular” part of the theorem follows from the first part.

Suppose that ${\langle } A_i\mid i<\kappa {\rangle }$ witnesses that $\neg \operatorname {\mathrm {Gal}}(U,\lambda ,\kappa )$ . To show that $(U,\supseteq )\ge _T ([\kappa ]^{<\lambda },\subseteq )$ , define $f:[\kappa ]^{<\lambda }\rightarrow U$ by $f(I)=\bigcap _{i\in I}A_i$ , and let us prove that f is an unbounded map: Suppose that $X\subseteq [\kappa ]^{<\lambda }$ is unbounded. Then $|\cup X|\geq \lambda $ . Now

$$ \begin{align*}\bigcap f"X=\bigcap\{\cap_{i\in I}A_i\mid I\in X\}=\bigcap_{i\in\cup X}A_i.\end{align*} $$

By the assumption that the sequence witnesses the failure of Galvin’s property, it follows that $\bigcap f"X\notin U$ and therefore $f"X$ is unbounded in U.

For the other direction, suppose $g:[\kappa ]^{<\lambda }\rightarrow U$ is unbounded, and let us show that $\neg \operatorname {\mathrm {Gal}}(U,\lambda ,\kappa )$ . The witnessing sequence will be ${\langle } X_\alpha \mid \alpha <\kappa {\rangle }$ , where $X_\alpha =g(\{\alpha \})$ . Let us prove that $X_\alpha $ is as wanted. Fix any $I\in [\kappa ]^{\lambda }$ . Then the set $\hat {I}=\{\{i\}\mid i\in I\}\subseteq [\kappa ]^{<\lambda }$ is unbounded. By our assumption, $g"\hat {I}=\{X_i\mid i\in I\}$ is also unbounded, it follows that $\bigcap _{i\in I}X_i\notin U$ .

Proof. Let $\{b_i\mid i<\operatorname {\mathrm {Ch}}(U)\}$ be a filter base for U. Then $B=\{b_i\setminus \xi \mid i<\operatorname {\mathrm {Ch}}(U),\xi <\kappa \}$ is cofinal in U. Let $f:U\rightarrow W$ be a cofinal map. Then $f"B$ is a filter base for W, and therefore $\operatorname {\mathrm {Ch}}(W)\leq |f"B|\leq |B|=\operatorname {\mathrm {Ch}}(U)$ .

Proof. To see that $\operatorname {\mathrm {Ch}}(U)=2^\kappa $ , let ${\langle } A_i\mid i<2^\kappa {\rangle }$ witness that $\neg \operatorname {\mathrm {Gal}}(U,\mu ,2^\kappa )$ , and suppose toward a contradiction that there is cofinal set $\mathcal {B}\subseteq U$ such that $|\mathcal {B}|<2^\kappa $ . In particular, for each $\alpha <2^\kappa $ there is $b_\alpha \in \mathcal {B}$ such that $b_\alpha \subseteq A_\alpha $ . This defines a function from $2^\kappa $ to $\mathcal {B}$ , and by the pigeonhole principle there is $b^*\in \mathcal {B}$ such that $I:=\{\alpha <2^\kappa \mid b_\alpha =b^*\}$ has size at least $\kappa $ . It follows that $b^*\subseteq \cap _{\alpha \in I}A_\alpha $ , and therefore $\cap _{\alpha \in I}A_\alpha \in U$ , contradicting the fact that ${\langle } A_\alpha \mid \alpha <2^\kappa {\rangle }$ witnesses the failure of $\operatorname {\mathrm {Gal}}(U,\mu ,2^\kappa )$ .

Question 4.5. Suppose that U is a $\kappa $ -complete ultrafilter over $\kappa $ such that $\neg \operatorname {\mathrm {Gal}}(U,\kappa ,\lambda )$ . Is there a $\lambda $ -independent family ${\langle } A_i\mid i<\lambda {\rangle }\subseteq U$ which generates U?

Proof. By Corollary 3.17, such an ultrafilter U always exists, and by the previous corollary, $2^\kappa =\operatorname {\mathrm {Ch}}(U)$ .

Proof. Let $f:U\rightarrow W$ be the map sending $f(X^U_i)=X^W_i$ and for $b\in U$ , let $f(b)=f(X^U_i)$ for the minimal i such that $X^U_i\subseteq ^* b$ ., Given a $\subseteq ^*$ -cofinal set B in U, then for every $b\in B$ , let $i_b$ be the minimal index such that $X^U_{i_b}\subseteq ^* b$ (hence $f(g)=X^W_{i_b}$ ). We claim that $\{i_b\mid b\in B\}$ must be unbounded in $\theta $ . Otherwise, there is $i^*$ which is a bound. In particular, $X^U_{i^*}\subseteq ^* b$ for every $b\in B$ . Take any set $Y\in U$ such that $X^U_{i^*}\setminus Y$ is unbounded in $\kappa $ (just partition $X^U_{i^*}$ and take Y to be the piece which belongs to U). Now there is $b\in B$ such that $b\subseteq ^* Y$ and thus

$$ \begin{align*}b\subseteq^* Y\subseteq^* X^U_{i^*}\subseteq b.\end{align*} $$

Hence $b=^*Y=^*X^U_{i^*}$ which contradicts the fact that $X^U_{i^*}\setminus Y$ is unbounded in $\kappa $ . So we conclude that $\{i_b\mid b\in B\}$ is unbounded in $\theta $ . By definition of f, we have that $f"B\subseteq \{X^W_i\mid i<\theta \}$ then is cofinal in W. It follows that f is cofinal and $(U,\supseteq ^*)\geq _T (W,\supseteq ^*)$ . The other direction is symmetric.

Proof. Indeed, the identity map witnesses that $\leq _T$ , and equality cannot hold since $(U,\supseteq ^*)$ is $\kappa ^+$ -directed (as it is p-point) while $(U,\supseteq )$ is not $\kappa ^+$ -directed.

Proof. Take any sequence ${\langle } A_i\mid i<\kappa {\rangle }$ which has no pseudo intersection. Since U is $\kappa $ -complete we may assume that $A_i$ is decreasing in $\subseteq $ . This assumption guarantees that there is no pseudo intersection for ${\langle } A_i\mid i\in I{\rangle }$ where $I\in [\kappa ]^\kappa $ . By $2^\kappa $ -OK, there is a sequence ${\langle } B_i\mid i<2^\kappa {\rangle }$ such that for every $a\in [2^\kappa ]^{<\kappa }$ , $\cap _{i\in a}B_i\subseteq ^* A_{|a|}$ . It follows that ${\langle } B_i\mid i<2^\kappa {\rangle }$ witnesses $\neg \operatorname {\mathrm {Gal}}(U,\kappa ,2^\kappa )$ .

Proof. Suppose ${\langle } B_\alpha \mid \alpha <2^\kappa {\rangle }\in [U]^{2^\kappa }$ witnesses that $\neg \operatorname {\mathrm {Gal}}^*(U,\kappa ,2^\kappa )$ , and let $g:[2^\kappa ]^{<\kappa }\rightarrow \{B_\alpha \mid \alpha <2^\kappa \}$ be any enumeration. Let us prove that g is unbounded. Suppose that $\mathcal {A}\subseteq [2^\kappa ]^{<\kappa }$ is unbounded. Then $|\mathcal {A}|\geq \kappa $ , so $g[\mathcal {A}]$ is a subset of $\{B_\alpha \mid \alpha <2^\kappa \}$ of cardinality $\kappa $ and therefore has no pseudointersection; hence $g[\mathcal {A}]$ is unbounded. In the other direction, suppose that $\operatorname {\mathrm {Gal}}^*(U,\kappa ,2^\kappa )$ holds, and let ${g:[2^\kappa ]^{<\kappa }\rightarrow U}$ . We will prove that g is not unbounded. If $|\operatorname {\mathrm {Im}}(g)|<2^\kappa $ , then there are $\kappa $ -many elements mapped to the same set in U. Any $\kappa $ -many elements in $[2^\kappa ]^{<\kappa }$ are unbounded (since $\kappa $ is a strong limit) so g is not unbounded. If $|\operatorname {\mathrm {Im}}(g)|=2^\kappa $ , there is $Y\subseteq \operatorname {\mathrm {Im}}(g)$ , such that $|Y|=\kappa $ which has a pseudointersection. Again $g^{-1}[Y]$ is unbounded in $[2^\kappa ]^{<\kappa }$ but it is mapped to a bounded set.

Question 4.19. It is consistent from just a measurable cardinal that there is a $\kappa $ -complete ultrafilter such that $\neg \operatorname {\mathrm {Gal}}^*(U,\kappa ,2^\kappa )$ ?

Question 4.20. Is it consistent that U is a $\kappa $ -complete ultrafilter over $\kappa $ such that $(U,\supseteq ^*)$ is not Tukey-top but $(U,\supseteq )$ is? Namely, is it consistent that $\neg \operatorname {\mathrm {Gal}}(U,\kappa ,2^\kappa )$ but $\operatorname {\mathrm {Gal}}^*(U,\kappa ,2^\kappa )$ ?

Proof. Let U be a p-point on $\kappa $ . Suppose V is a uniform ultrafilter such that $U \ge _T V $ , and let $f: U \rightarrow V $ be a monotone cofinal map and $\langle \delta _{\alpha }\mid \alpha <\kappa \rangle $ be a strictly increasing sequence. Construct a $\subseteq $ -decreasing sequence $\langle X_{\alpha }:\alpha <\kappa \rangle $ (i.e., $\alpha \leq \beta \Rightarrow X_\beta \subseteq X_\alpha $ ) of members of $ U $ such that for each $\alpha <\kappa $ , $X_{\alpha }\cap \rho _\alpha =\emptyset $ and the following holds:

• $(*)$ For each pair $s\subseteq \rho _\alpha $ and $\delta \le \delta _\alpha $ , if there is a $Y\in U $ such that $s=Y\cap \rho _\alpha $ and $\delta \not \in f(Y)$ , then $\delta \not \in f(s\cup X_\alpha )$ (i.e., $f(s\cup X_\alpha )\cap \delta _\alpha \subseteq f(Y)\cap \delta _\alpha $ ).

Take $X_0\in U $ such that for each $\delta \le \delta _0$ , $\delta \not \in f(X_0)$ . Such an $X_0$ exists since $ V $ is uniform. Suppose $1\le \alpha <\kappa $ and $\langle X_{\beta }\mid \beta <\alpha \rangle $ have been chosen satisfying $(*)$ . For each $s\subseteq \rho _\alpha $ and $\delta \le \delta _\alpha $ , if there is a $Y\in U$ such that $s=Y\cap \rho _\alpha $ and $\delta \not \in f(Y)$ , then take $Y_{s,\delta }$ so that $Y_{s,\delta }\cap \rho _\alpha =s$ and $\delta \not \in f(Y_{s,\delta })$ ; otherwise, let $Y_{s,\delta }=[\rho _\alpha ,\kappa )$ . Define

(1) $$ \begin{align} X_\alpha=\bigcap\{Y_{s,\delta}:s\subseteq\rho_\alpha,\ \delta\le\delta_\alpha\} \cap \bigcap\{X_{\beta}:\beta<\alpha\}. \end{align} $$

Then $X_{\alpha }\cap \rho _\alpha =\emptyset $ and our construction ensures that $X_{\alpha }$ satisfies property $(*)$ : Given $s\subseteq \rho _\alpha $ and $\delta \le \delta _\alpha $ , if there is a $Y\in U $ such that $Y\cap \rho _\alpha =s$ and $\delta \not \in f(Y)$ then $\delta \not \in f(Y_{s,\delta })$ so by monotonicity of f, $\delta \not \in f(s\cup X_{\alpha })$ .

By definition of the $\rho _\alpha $ ’s and the fact that U is a p-point, there is an $X\in U$ such that $X\setminus \rho _\alpha \subseteq X_\alpha $ , for each $\alpha <\kappa $ . We claim that f is continuous when restricted to $ U \upharpoonright X$ .

Let $W\in U\upharpoonright X$ and $\alpha <\kappa $ . It follows that

(2) $$ \begin{align} W\setminus \rho_{\alpha} \subseteq X\setminus \rho_{\alpha} \subseteq X_{\alpha}. \end{align} $$

Let $s=W\cap \rho _\alpha $ , and let $\delta \le \delta _\alpha $ be given. If $\delta \not \in f(W)$ , then by $(*)$ it follows that $\delta \not \in f(s\cup X_{\alpha })$ . Since $X\setminus \rho _\alpha \subseteq X_{\alpha }$ , by monotonicity also $\delta \not \in f(s\cup (X\setminus \rho _\alpha ))$ . On the other hand, if $\delta \in f(W)$ , then by monotonicity $\delta \in f(s\cup X\setminus \rho _\alpha )$ . Thus,

(3) $$ \begin{align} f(W)\cap \delta_\alpha= f((W\cap\rho_\alpha)\cup (X\setminus \rho_\alpha))\cap \delta_\alpha. \end{align} $$

Define the function $\hat {f}:[\kappa ]^{<\kappa }\rightarrow [\kappa ]^{<\kappa }$ as follows: For $s\in [\kappa ]^{<\kappa }$ , letting $\alpha $ be the least such that $s\subseteq \rho _{\alpha }$ , define

(4) $$ \begin{align} \hat{f}(s)= f(s\cup (X\setminus \rho_\alpha))\cap \delta_\alpha. \end{align} $$

This $\hat {f}$ is monotone and uniformly continuous and for each $A\in U\upharpoonright X$ ,

(5) $$ \begin{align} f(A)=\bigcup_{\alpha<\kappa}\hat{f}(A\cap\rho_\alpha) \end{align} $$

since there are club many $\alpha $ , namely those in $\lim (A)\cap C_{\alpha \mapsto \rho _\alpha }$ , for which $f(A)\cap \delta _{\alpha }=\hat {f}(A\cap \rho _{\alpha })$ .

Defining g as in the statement of the theorem, it is routine to check that g is as claimed.

Proof. As before, let us construct a $\supseteq $ -decreasing sequence for each $1\leq i\leq n$ , $\langle X_{\alpha ,i}:\alpha <\kappa \rangle $ of members of $ U _i$ such that for all $\alpha <\kappa $ , $X_{\alpha ,i}\cap \rho ^i_\alpha =\emptyset $ and the following holds:

• $(*)$ For each pair $s_i\subseteq \rho ^i_\alpha $ and $\delta \le \delta _\alpha $ , if there is a $\vec {Y}={\langle } Y_1,\dots , Y_n{\rangle }\in U_1\times \dots \times U_n$ such that $s_i=Y_i\cap \rho ^i_\alpha $ and $\delta \not \in f(\vec {Y})$ , then $\delta \not \in f({\langle } s_1\cup X_{\alpha ,1},\dots ,s_n\cup X_{\alpha ,n}{\rangle })$ .

Take $X_{0,i}\in U_i$ such that $0\not \in f({\langle } X_{0,1},\dots ,X_{0,n}{\rangle })$ . Since V is uniform, and f is cofinal, such an $X_{0,i}$ exists. Suppose $1\le \alpha <\kappa $ and $\langle X_{\beta ,i}:\beta <\alpha \rangle $ have been constructed. For each $s_i\subseteq \rho ^i_\alpha $ and $\delta \le \delta _\alpha $ , if there are a sets $Y_i\in U _i$ such that $s_i=Y_i\cap \rho ^i_\alpha $ and $\delta \not \in f({\langle } Y_1,\dots , Y_n{\rangle })$ , then let $Y_{s_1,\dots ,s_n,i,\delta }$ be such a $Y_i$ ; otherwise, let $Y_{s_1,\dots ,s_n,i,\delta }=\kappa $ . Define

(6) $$ \begin{align} X_{\alpha,i}=\bigcap\{Y_{s_1,\dots,s_n,i,\delta}:s_1\subseteq \rho^1_\alpha,\dots,s_n\subseteq\rho^n_\alpha,\ \delta\le\delta_\alpha\} \cap \bigcap\{X_{\beta,i}:\beta<\alpha\}. \end{align} $$

Note that $X_{\alpha ,i}\cap \rho ^i_\alpha =\emptyset $ . Moreover, given $s_i\subseteq \rho ^i_\alpha $ , if there is a $Y_i\in U_i$ such that $Y_i\cap \rho ^i_\alpha =s_i$ then

$$ \begin{align*}f({\langle} s_1\cup X_{\alpha,1},\dots,s_n\cup X_{\alpha,n}{\rangle})\cap \delta_\alpha\subseteq f({\langle} Y_1,\dots, Y_n{\rangle})\cap \delta_\alpha,\end{align*} $$

by choice of $X_{\alpha ,i}$ and monotonicity of f.

Since each $U_i$ is a normal ultrafilter, we let $X_i^*:=\Delta ^*_{i<\kappa }X_{\alpha ,i}\in U_i$ , then $X_i^*\setminus \rho ^i_\alpha \subseteq X_{\alpha ,i}$ . We claim that f is continuous when restricted to $U_1\times \dots \times U_n\upharpoonright {\langle } X^*_1,\dots ,X^*_n{\rangle }$ . Toward this, let $W_i\in U_i$ be such that $W_i\subseteq X_i^*$ , let $\alpha <\kappa $ and let $s_i=W_i\cap \rho ^i_\alpha $ . It follows that

(7) $$ \begin{align} W_i\setminus \rho^i_\alpha \subseteq X^*_i\setminus \rho^i_\alpha \subseteq X_{\alpha,i}. \end{align} $$

Thus,

$$ \begin{align*}f({\langle} W_1,\dots, W_n{\rangle})\cap\delta_\alpha\subseteq f({\langle} s_1\cup X^*_{1}\setminus\rho^1_\alpha,\dots,s_n\cup X^*_{n}\setminus\rho^n_\alpha)\cap \delta_\alpha\subseteq\end{align*} $$

$$ \begin{align*}\subseteq f(s_1\cup X_{\alpha,1},\dots,s_n\cup X_{\alpha,n})\cap\delta_\alpha\subseteq f(W_1,\dots ,W_n)\cap\delta_\alpha.\end{align*} $$

Then $f({\langle } W_1,\dots , W_n{\rangle })\cap \delta _\alpha =f({\langle } s_1\cup X^*_{1}\setminus \rho ^1_\alpha ,\dots ,s_n\cup X^*_{n}\setminus \rho ^n_\alpha )\cap \delta _\alpha $ is continuous on $U_1\times \dots \times U_n\upharpoonright {\langle } X^*_1,\dots ,X^*_n{\rangle }$ .

Proof. The proof of $(1)$ is a straightforward generalization of [Reference Dobrinen and Todorcevic16, Theorem 32]: Given any $A\in \sum _{U}U_\alpha $ , define $(A)_\alpha =\{\beta <\kappa \mid {\langle }\alpha ,\beta {\rangle }\in A\}$ and let $\mathcal {B}$ be the set of those $A\in \sum _{U} U_\alpha $ such that for each $\alpha <\kappa $ , either $(A)_\alpha \in U_\alpha $ or else $(A)_\alpha =\emptyset $ . Then $\mathcal {B}$ is a filter base for $\sum _{ U} U_\alpha $ . Given $A\in \mathcal {B}$ define $\pi _0(A)$ to be the set of $\alpha <\kappa $ such that $(A)_\alpha \in U_\alpha $ , and for each $\alpha <\kappa $ , define $u_\alpha (A)=(A)_\alpha $ if $\alpha \in \pi _0(A)$ and $\kappa $ otherwise. Define $f(A)$ to be $\langle \pi _0(A)\rangle ^{\frown }\langle u_\alpha (A):\alpha <\kappa \rangle $ . Then f is a Tukey map. $(2)$ and $(3)$ are immediate from $(1)$ .

Fact 6.2. For any sequence ${\langle } A_i\mid i<\kappa {\rangle }\in U$ , where U is $\sigma $ -complete, $\Delta ^*_{i<\kappa }A_i\in U$ .

Proof. Just otherwise, the complement $B:=\kappa \setminus \Delta ^*_{i<\kappa }A_i$ would be in U, and for every $\beta \in B$ , there is $i_\beta <\pi (\beta )$ such that $\beta \notin A_{i_\beta }$ , but then $[\beta \mapsto i_\beta ]_U<[\pi ]_U$ , and since $[\pi ]_U$ is the least non-constant function, then there is a set $B'\in U\restriction B$ and $i^*$ such that for every $\beta \in B'$ , $i_\beta =i^*$ . This implies that $B'\cap A_{i^*}=\emptyset $ , contradicting that both $B'$ and $A_{i^*}$ are in U.

Proof. Recall that $\pi :\kappa \rightarrow \kappa $ denotes the least non-constant function. Since $ U $ is a p-point, for each $\alpha <\kappa $ , $\pi ^{-1}[\alpha ]:=\{\gamma <\kappa \mid \pi (\gamma )<\alpha \}$ is bounded in $\kappa $ ; let $h(\alpha )=(\sup \pi ^{-1}[\alpha +1])+1$ . Let ${\langle } A_i\mid i<\kappa {\rangle }\in \prod _{i<\kappa } U $ be given. Then for any $\alpha <\kappa $ , and any $\nu \in (\Delta ^*_{i<\kappa }A_i)\setminus h(\alpha )$ , $\nu \notin \pi ^{-1}[\alpha +1]$ and therefore $\pi (\nu )>\alpha $ which implies by the definition of the modified diagonal intersection that $\nu \in A_\alpha $ .

Proof. Let $\pi :\kappa \rightarrow \kappa $ be as before, and define $B_\alpha =\pi ^{-1}[\alpha +1]$ for each $\alpha <\kappa $ . Since $\pi $ is not constant mod U and U is $\kappa $ -complete, each $B_\alpha \notin U$ . Let ${\langle } A_i\mid i<\kappa {\rangle }\in \prod _{i<\kappa } U $ be any sequence. To see that $(\Delta ^*_{i<\kappa }A_i)\setminus B_\alpha \subseteq A_\alpha $ , let $\nu \in (\Delta ^*_{i<\kappa }A_i)\setminus \pi ^{-1}[\alpha +1]$ . Then $\pi (\nu )\geq \alpha +1>\alpha $ , so $\nu \in A_\alpha $ by the definition of $\Delta ^*_{i<\kappa }A_i$ .

Proof. Let us define the map $F:\prod _{i<\kappa }U\rightarrow U$ by

$$ \begin{align*}F({\langle} A_i\mid i<\kappa{\rangle})=\Delta^*_{i<\kappa}A_i.\end{align*} $$

To see that F is unbounded, let $\{{\langle } A^j_i\mid i<\kappa {\rangle }\mid j\in I\}$ be unbounded in $\prod _{i<\kappa }U$ and suppose toward contradiction that there is $B\in U$ such that $B\subseteq \Delta ^*_{i<\kappa }A^j_i$ for every $j\in I$ . Let ${\langle } B_i\mid i<\kappa {\rangle }$ be the sequence from the previous lemma, and note that ${\langle } B\setminus B_i\mid i<\kappa {\rangle }\in \prod _{i<\kappa }U$ . Also note that by the property of the sequence ${\langle } B_i\mid i<\kappa {\rangle }$ , for each $i<\kappa $ and each $j\in I$ ,

$$ \begin{align*}B\setminus B_i\subseteq (\Delta^*_{i<\kappa}A^j_i)\setminus B_i\subseteq A^j_i.\end{align*} $$

We conclude that in the everywhere domination order of the product $\prod _{i<\kappa }U$ , we have ${\langle } B\setminus B_i\mid i<\kappa {\rangle }\geq {\langle } A^j_i\mid i<\kappa {\rangle }$ . Therefore, the sequence ${\langle } B\setminus B_i\mid i<\kappa {\rangle }$ bounds the set $\{{\langle } A^j_i\mid i<\kappa {\rangle }\mid j\in I\}$ , contradicting our initial assumption that this set should be unbounded.

Proof. $U\times V\leq _{T} U\cdot V$ by Proposition 2.5. For the other direction, $U\cdot V\leq _T U\times \prod _{\alpha <\kappa }V$ by Lemma 6.1 item $(2)$ , and by Lemma 6.5, $ U\times \prod _{\alpha <\kappa }V\leq _T U\times V$ . The moreover part follows by a straightforward induction.

Proof. $U^n\equiv _T U\times U\times \dots \times U\equiv _T U$ .

Proof. U is Galvin iff U is not Tukey-top, and $U^n$ is Galvin iff $U^n$ is not Tukey-top. Since $U\equiv _T U^n$ , we see that U is Galvin iff $U^n$ is Galvin.

Question 6.9. Suppose that U, $V_\alpha $ are all Galvin. Does it follow that $\sum _UV_\alpha $ is Galvin? Is the Fubini product of two Galvin ultrafilters always Galvin?

Proof. Suppose that U is a p-point and let $\pi $ be the least non-constant function mod U. Since U is a p-point, there is $Y\in U$ such that for every $\gamma <\kappa $ , $\pi ^{-1}[\gamma ]\cap Y$ is bounded in $\kappa $ . For each $\alpha <\kappa $ , pick $\rho _\alpha :=\sup (\pi ^{-1}[\alpha +1]\cap Y)+1$ . To see that U is uniformly basic, let ${\langle } A_i\mid i<\kappa {\rangle }$ be a sequence converging to some set ${A\in U}$ . By definition of convergence, there is a sequence $\delta _\alpha $ such that for any $i\geq \delta _\alpha $ , $A_i\cap \rho _{\alpha }=A\cap \rho _{\alpha }$ ; define $f(\alpha )=\rho _{\alpha }$ . Let us prove that for any $f\le _{bd} g$ we have that $\cap _{\alpha <\kappa }A_{g(\alpha )}\in U$ . Towards this, let $\theta <\kappa $ be such that for every $\theta \leq \alpha <\kappa $ , $\delta _\alpha \leq g(\alpha )$ . It suffices to prove that $(\cap _{\beta <\theta }A_{g(\beta )})\cap A\cap Y\cap \Delta ^*_{\alpha <\kappa }A_{\delta _\alpha }\subseteq \cap _{\alpha <\kappa }A_{g(\alpha )}$ , whereFootnote ⁷

$$ \begin{align*}\Delta^*_{\alpha<\kappa}A_{\delta_\alpha}:=\{\nu<\kappa\mid \forall \alpha<\pi(\nu),\ \nu\in A_{\delta_{\alpha}}\}.\end{align*} $$

Indeed, suppose $\xi \in (\cap _{\beta <\theta }A_{g(\beta )})\cap A\cap Y\cap \Delta ^*_{\alpha <\kappa }A_{\delta _\alpha }$ . Let $\alpha _0<\kappa $ and split into cases:

1. (1) If $\alpha _0<\theta $ , then clearly $\xi \in A_{g(\alpha _0)}$ .

2. (2) If $\alpha _0<\pi (\xi )$ , then by the definition of $\Delta ^*_{\alpha <\kappa }A_{g(\alpha )}$ , $\xi \in A_{g(\alpha _0)}$ .

3. (3) If $\alpha _0\geq \max (\pi (\xi ),\theta )$ then $\xi \in \pi ^{-1}[\alpha _0+1]\cap Y$ , which means that $\xi <\rho _{\alpha _0}$ , so $\xi \in A\cap \rho _{\alpha _0}$ . Also we conclude that $\delta _{\alpha _0}\leq g(\alpha _0)$ , and by the choice of $\delta _{\alpha _0}$ , it follows that $A_{g(\alpha _0)}\cap \rho _{\alpha _0}=A\cap \rho _{\alpha _0}$ . Thus $\xi \in A_{g(\alpha _0)}$ .

Proof. Suppose that U is basic and uniform. Let ${\langle } X_i\mid i<\kappa {\rangle }\in [U]^\kappa $ be a sequence of sets in U. We want to prove that this sequence has a pseudointersection. By $\kappa $ -completeness we may assume that the sequence of $X_i$ is decreasing. Define the sets $Y_i=X_i\cup i$ . Clearly, $Y_i\supseteq X_i$ and $\lim _{i\rightarrow \kappa }Y_i=\kappa $ . Hence, since U is basic, there is a subsequence ${\langle } i_\alpha \mid \alpha <\kappa {\rangle }$ such that $X^*:=\cap _{\alpha <\kappa } Y_{i_\alpha }\in U$ . Note that for each $i<\kappa $ , there is $\alpha <\kappa $ such that $i<i_\alpha $ and therefore $X_i\supseteq X_{i_\alpha }=^*Y_{i_\alpha }\supseteq X^*$ . Hence U is a p-point.

Proof. Let ${\langle } A_i\mid i<\kappa ^+{\rangle }$ be any sequence of sets in U, and for each $\alpha <\kappa $ , $i<\kappa ^+$ define

$$ \begin{align*}H_{\alpha,i}:=\{j<\kappa^+\mid A_i\cap\alpha=A_j\cap\alpha\}.\end{align*} $$

Galvin observed that :

The proof can be found in [Reference Benhamou and Gitik9, Proposition 5.13]. Then we can find a sequence of distinct $i_\alpha $ ’s such that $A_{i_\alpha }\cap \alpha =A_{i^*}\cap \alpha $ . In particular, $\text {lim}_{\alpha \rightarrow \kappa } A_{i_\alpha }=A_{i^*}\in U $ , and since U is basic, there is a subsequence ${\langle } \alpha _j \mid j<\kappa {\rangle }$ such that $\cap _{j<\kappa } A_{i_{\alpha _j}}\in U$ .

Proof. Let $B\subseteq U$ be a filter base witnessing that U is basically generated. Fix a subset $C\subseteq U$ of cardinality $2^{\kappa }$ ; for each $X\in C$ fix one $Y_X\in B$ such that $Y_X\subseteq X$ . If there is a subset $D\subseteq C$ of cardinality $\kappa $ for which all $Y_X$ , $X\in D$ , are equal, then there is a subset of C of size $\kappa $ with intersection in U.

Otherwise, $|\{Y_X \mid X\in C\}|\ge \kappa ^+$ . Without loss of generality, assume that for each $X\ne X'$ in C, $Y_X\ne Y_{X'}$ . For each $s\in 2^{<\kappa }$ , let $C_s=\{X\in C \mid Y_X\cap \operatorname {\mathrm {dom}}(s)= s^{-1}[\{1\}]\}$ . Note that for each $\alpha <\kappa $ , there is at least one $s\in 2^\alpha $ for which $|C_s|\ge \kappa ^+$ .

Thus, fix $X\in C$ and the sequence $\langle s_{\alpha } \mid \alpha <\kappa \rangle $ from the claim such that each $|C_{s_\alpha }|\ge \kappa ^+$ . Now we build a convergent subsequence of $\{Y_X \mid X\in C\}$ quite simply: For $\alpha <\kappa $ , take any $X_\alpha \in C_{s_\alpha }$ . Then $\langle Y_{X_\alpha } \mid \alpha <\kappa \rangle $ is a sequence of members of B which converges to $Y_X$ which is a member of B. Since B witnesses that U is basically generated, there is a subsequence $\langle \alpha _i \mid i<\kappa \rangle $ such that $Y^*:=\bigcap _{i<\kappa } Y_{X_{\alpha _i}}$ is in U. Finally, let $D=\{X\in C \mid Y^*\subseteq X\}$ , and note that $\{X_{\alpha _i}\mid i<\kappa \}\subseteq D$ ; in particular, $|D|\ge \kappa $ . Then $\bigcap D\supseteq Y^*$ which is in U. Hence, U is not Tukey-top.