Proof Fix an enumeration $\{A_{\beta }:{\beta }<{\mu }\}$ of $\mathcal {A}$ witnessing that $\mathcal {A}$ is nice.

We will define $\le _{A_{\beta }}$ by induction on ${\beta }<{\mu }$ .

Assume that $\le _{A_{\alpha }}$ is defined for ${\alpha }<{\beta }$ .

By (N2) we can fix a countable set $I_{\beta }=\{{\beta }_i:i<{\omega }\}\in \bigl [ {\beta } \bigr ]^ {\omega } $ such that for all ${\alpha }<{\beta }$ there is $n_{\alpha }<{\omega }$ such that

$$ \begin{align*} A_{\alpha}\cap A_{\beta}\subset \bigcup_{i<n_{\alpha}}A_{{\beta}_i}. \end{align*} $$

Choose an order $\le _{A_{\beta }}$ on $A_{\beta } $ such that

1. (i) for each $i<{\omega }$ writing $D_i=A_{{\beta }_i} \setminus \bigcup _{j<i}A_{{\beta }_j}$ we have

   $$ \begin{align*} \le_{A_{\beta}}\restriction (A_{\beta}\cap D_i)\quad=\quad \le_{A_{{\beta}_i}}\restriction (A_{\beta}\cap D_i); \end{align*} $$

2. (ii) $\textit{tp}(A_{\beta },\le _{A_{{\beta }}})={\omega }$ .

By induction on ${\beta }$ we show that (2) holds for $A_{\alpha }$ and $A_{\beta }$ for each ${\alpha }<{\beta }$ . Assume that this statement holds for each ${\beta }'<{\beta }$ . To check for ${\beta }$ fix ${\alpha }<{\beta }$ .

To define $\le _{\beta }$ we considered a set $I_{\beta }=\{{\beta }_i:i<{\omega }\}\in \bigl [ {\beta } \bigr ]^ {\omega } $ such that we had $n_{\alpha }<{\omega }$ with

$$ \begin{align*} A_{\alpha}\cap A_{\beta}\subset \bigcup_{i<n_{\alpha}}A_{{\beta}_i}. \end{align*} $$

For $i<n_{\alpha }$ let $C^{\prime }_i=A_{\alpha }\cap A_{\beta }\cap D_i$ , where $D_i=A_{{\beta }_i} \setminus \bigcup _{j<i}A_{{\beta }_j}$ . Then $\{C^{\prime }_i:i<n_{\alpha }\}$ is a partition of $A_{\alpha }\cap A_{\beta }$ and

$$ \begin{align*}\le _{A_{\beta}}\restriction C^{\prime}_i=\le_{A_{{\beta}_i}}\restriction C^{\prime}_i\end{align*} $$

by (i). By the inductive hypothesis, $A_{{\beta }_i}\cap A_{\alpha }$ has a partition into finitely many pieces $\{C_{i,j}:j<k_i\}$ such that $\le _{A_{\alpha }}\restriction C_{i,j}=\le _{A_{\beta _i}}\restriction C_{i,j}$ . Then the partition

$$ \begin{align*} \{C^{\prime}_i\cap C_{i,j}:i<n, j<k_i\} \end{align*} $$

of $A_{\alpha }\cap A_{\beta }$ works for ${\alpha }$ and ${\beta }$ . Indeed,

$$ \begin{align*} \le_{ A_{\alpha}}\restriction C^{\prime}_i\cap C_{i,j}\quad = \quad \le_{ A_{{\beta}_i}}\restriction C^{\prime}_i\cap C_{i,j}\quad = \quad \le_{A_{{\beta}}}\restriction C^{\prime}_i\cap C_{i,j}. \\[-35pt] \end{align*} $$

Proof For $A\in \mathcal {A}$ let

$$ \begin{align*} \mathcal G_A=\{{f}^+\in \operatorname{S}(\lambda): f\in \operatorname{S}(A) \land \text{there is a finite partition}\ \{C_i:i<n\}\ \text{of}\ A\nonumber\\ \qquad\qquad\qquad\qquad\qquad\text{ such that}\ f\restriction C_i\ \text{is}\ \le_A\text{-order preserving}\}. \end{align*} $$

Let G be the permutation group on ${\lambda }$ generated by

$$ \begin{align*} \bigcup\{\mathcal{G}_A:A\in \mathcal{A}\}. \end{align*} $$

Indeed, let $X,Y\in \bigl [ {\lambda } \bigr ]^{\omega }$ with $|{\lambda }\setminus X|=|{\lambda }\setminus Y|={\lambda }$ . Pick $A\in \mathcal {A}$ such that $X\cup Y\subset A$ and $|A\setminus X|=|A\setminus Y|={\omega }$ .

Let c be the unique $\le _A$ -monotone bijection between X and Y and d be the unique $\le _A$ -monotone bijection between $A\setminus X$ and $A\setminus Y$ . Then taking $g=c\cup d$ we have ${g}^+\in \mathcal {G}_A\subset G$ and ${g}^+[X]=Y$ .

Pick $A\in \mathcal {A}$ and choose $B\in \bigl [ {A} \bigr ]^ {\omega } $ such that $|A\setminus B|={\omega }$ .

Let $b_0,b_1,\dots $ be the $\le _A$ -increasing enumeration of B. Define a bijection $y:B\to {\omega }$ as follows: for $i<{\omega }$ and $j<2^i$ let

$$ \begin{align*}y(b_{2^i+j})=b_{2^{i+1}-j}.\end{align*} $$

Observe that if c is $\le _A$ -monotone then

$$ \begin{align*} |\{i<{\omega}:|\{j<2^i:c(b_{2^i+j})=r(b_{2^i+j})\}|\ge 2\}|\le 1. \end{align*} $$

Indeed, if $|\{j<2^i:c(b_{2^i+j})=y(b_{2^i+j})\}|\ge 2$ , then c should be $\le _A$ -decreasing, and if $|\{i:\{j<2^i:c(b_{2^i+j})=y(b_{2^i+j})\}\ne \emptyset \}|\ge 2$ , then y should be $\le _A$ -increasing.

So y cannot be covered by finitely many $\le _A$ -monotone functions. But for any $h\in G$ , $h\cap (A\times A)$ can be covered by finitely many $\le _A$ -monotone functions by (2) and by the construction of G.

Thus y is G-large.

Proof Let $A\in \bigl [ {Y} \bigr ]^ {\omega } $ . Then $\overline A$ is countable, so it is compact. Since a splendid space is zero-dimensional, A can be covered by finitely many compact open sets, and so A can be covered by an element of $\mathcal {U}$ . Thus $\mathcal {U}\lceil Y$ is cofinal in $\left \langle \bigl [ {Y} \bigr ]^ {\omega },\subset \right \rangle $ .

To check (N2) observe that every $U\in \mathcal {U}$ is a countable compact space, so it is homeomorphic to a countable successor ordinal. Thus U has only countably many compact open subsets. Hence $\mathcal {U}\lceil U$ is countable which implies (N2) in the following stronger form:

• (N2⁺) for each ${\beta }<{\mu }$ there is a set $I_{\beta }\in \bigl [ {\beta } \bigr ]^ {\omega } $ such that for all ${\alpha }<{\beta }$ there is ${\zeta }_{\alpha }\in I_{\beta }$ such that

  $$ \begin{align*} A_{\alpha}\cap A_{\beta} =A_{{\zeta}_\alpha}\cap A_{\beta}. \\[-35pt] \end{align*} $$

Proof Applying the Juhász–Nagy–Weiss theorem for ${\kappa }={\lambda }$ if $\operatorname {cf}({\lambda })>{\omega }$ , and for ${\kappa }={\lambda }^+$ if ${{\lambda }>\operatorname {cf}(\lambda })={\omega }$ , we obtain a splendid space on ${\kappa }\ge {\lambda }$ . So, by Lemma 2.4, we obtain a nice family $\mathcal {A}$ on ${\lambda }$ .

Thus, putting together Theorems 2.2 and 2.3 we obtained the desired permutation group on ${\lambda }$ .

Proof We prove only $\left \langle {\alpha },{\alpha }^* \right \rangle \notin \mathcal {H}[g_0]$ . The proof of the other statement is similar.

Assume on the contrary that $\left \langle {\alpha },{\alpha }^* \right \rangle \in \mathcal {H}[g_0]$ .

Pick the shortest term $t=\left \langle f_0,\dots , f_n\right \rangle $ from $\mathcal {H}$ such that $t[g_0]({\alpha })={\alpha }^*$ .

Write ${\alpha }_{n+1}={\alpha }$ and ${\alpha }_{i}=\left \langle f_{i},\dots ,f_n\right \rangle [g_0]({\alpha })$ for $0\le i\le n$ . Hence ${\alpha }_{0}={\alpha }^* $ .

Let i maximal such that ${\alpha }_i$ is $ {\zeta }_0$ or ${\eta }_0$ . Since $t[g]({\alpha })$ cannot be ${\alpha }^* $ by (3), i is defined.

Since ${\alpha }_i=\left \langle f_i,\dots ,f_{n}\right \rangle [g]({\alpha })$ , it follows that ${\alpha }_i\ne {\eta }_0$ by (5). So ${\alpha }_i={\zeta }_0$ .

Let j minimal such that ${\alpha }_j$ is $ {\zeta }_0$ or ${\eta }_0$ . Since

$$ \begin{align*}{\alpha}_j=(\left\langle f_0,\dots ,f_{j-1}\right\rangle [g])^{-1}({\alpha}^*),\end{align*} $$

it follows that ${\alpha }_j\ne {\eta }_0$ by (5). So ${\alpha }_j={\zeta }_0$ by (5). Thus ${\alpha }_i={\alpha }_j={\zeta }_0$ , and so

$$ \begin{align*} {\alpha}^* =\left\langle f_0,\dots, f_{j-1},f_i,\dots, f_n\right\rangle [g_0]({\alpha}). \end{align*} $$

Since $j< i$ , the term $t'=\left \langle f_0,\dots , f_{j-1},f_i,\dots , f_n\right \rangle $ is shorter than t and still ${\alpha }^* =t'[g_0]({\alpha })$ . So the length of t was not minimal. Contradiction.

Proof of Lemma 3.6 Write

$$ \begin{align*} \mathbb{TASK}_0= A\times \{\operatorname{dom},\operatorname{ran}\}\ \text{and}\ \mathbb{TASK}_1=\bigl[ {TERM({\mathcal{F}\cup\{x\}})} \bigr]^ {<{\omega}} \times {\kappa}. \end{align*} $$

Let $\{I_0,I_1\}\in \bigl [ {\bigl [ {\kappa } \bigr ]^ {\kappa } } \bigr ]^ {2} $ be a partition of ${\kappa }$ , and fix enumerations $\{T_i:i\in I_0\}$ of $\mathbb {TASK}_0$ , and $\{T_i:i\in I_1\}$ of $\mathbb {TASK}_1$ .

By transfinite induction, for $i<{\kappa }$ we will construct a function $g_i$ and if $i=j+1$ for some $j\in K_1$ then we also pick an ordinal ${\alpha }_{j+1}\in {\kappa }$ such that

1. (a) $g_i$ is an injective function, $\operatorname {dom}(g_i)\cup \operatorname {ran}(g_i)\subset A$ ;

2. (b) $g_i[B]\subset C$ and $g_i[A\setminus B]\subset A\setminus C$ ;

3. (c) $|g_i|\le i$ ;

4. (d) if $i=j+1$ , $j\in I_0$ , and $T_j=\left \langle {\zeta },\operatorname {dom}\right \rangle $ , then ${\zeta }\in \operatorname {dom}(g_i)$ ;

5. (e) if $i=j+1$ , $j\in I_0$ , and $T_j=\left \langle {\zeta },\operatorname {ran}\right \rangle $ , then ${\zeta }\in \operatorname {ran}(g_i)$ ;

6. (f) if $i=j+1$ , $j\in I_1$ , and $T_j=\left \langle \mathcal {H}_j,\chi _j\right \rangle $ , then

   1. (i) ${\alpha }_{j+1}\in {\kappa }\setminus \{{\alpha }_{j'+1}:j'\in I_1\cap j\}$ ; and

   2. (ii) $t[{g_i}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})$ is defined and $t[{g_i}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})\ne y({\alpha }_{j+1})$ for each $t\in \mathcal {H}_{j}$ .

Let $g_0=\emptyset $ .

If i is limit, then let $g_i=\bigcup _{j<i}g_j$ .

Assume that $i=j+1$ .

Claim 3.6.1.

(†) $$ \begin{align} |y\setminus \bigcup\mathcal{H}[g_j\cup \operatorname{id}_{{\lambda}\setminus A }]|={\kappa}, \end{align} $$

for each finite set $\mathcal {H}$ of $\mathcal {F}\cup \{x\}$ -terms.

Proof of the Claim Fix $\mathcal {H}$ . We can assume that $\mathcal {H}$ is closed for subterms. By (3) we have $|y\setminus \bigcup \mathcal {H}[\ ]|={\kappa }$ , and

(∘) $$ \begin{align} y\cap \bigcup \mathcal{H}[\ ]=y\cap \bigcup \mathcal{H}[\operatorname{id}_{{\lambda}\setminus A}], \end{align} $$

because $\mathcal {H}$ is closed for subterms. Since $|g_j|<{\kappa }$ , we have

(•) $$ \begin{align} |t[\operatorname{g_j\cup id}_{{\lambda}\setminus A}]\setminus t[\operatorname{id}_{{\lambda}\setminus A}]|<{\kappa}, \end{align} $$

for each $t\in \mathcal {H}$ . Putting together $|y\setminus \bigcup \mathcal {H}[\ ]|={\kappa }$ , $(\circ)$ , and $(\bullet)$ we obtain (†).

Case 1. $j\in I_0$ and so $T_j=\left \langle {\zeta }_j,x_j\right \rangle \in A\times \{\operatorname {dom},\operatorname {ran}\}$ .

Assume first that $x_j=\operatorname {dom}$ . If ${\zeta }_j\in \operatorname {dom} (g_j)$ , let $g_i=g_j$ . If ${\zeta }_j\notin \operatorname {dom} (g_j)$ , then pick ${\eta }\in C$ if ${\zeta }_i\in B$ , and pick ${\eta }\in A\setminus C$ if ${\zeta }_i\in A\setminus B$ such that and ${\eta }\notin \operatorname {ran}(g_j)$ .

Let $g_i=g_j\cup \left \langle {\zeta }_i,{\eta }\right \rangle $ . Then $g_i$ satisfies (a)–(f).

The case $x_j=\operatorname {ran}$ is similar.

Case 2. $j\in I_1$ and so $T_j=\left \langle \mathcal {H}_j,\chi _j\right \rangle \in \bigl [ {TERM({\mathcal {F}\cup \{x\}})} \bigr ]^ {<{\omega }} \times {\kappa }.$

We can assume that $\mathcal {H}_j$ is closed for subterms.

By Claim 3.6.1, we have

$$ \begin{align*} |y\setminus \bigcup \mathcal{H}_j[g_j\cup id_{({\lambda}\setminus A)} ]|={\kappa}. \end{align*} $$

So we can pick ${\alpha }_{j+1}\in {\kappa }\setminus \{{\alpha }_{j'+1}:j'\in I_1\cap j\}$ such that

• (*) for each $t\in \mathcal {H}_{j}$ either $t[{g_j}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})$ is undefined or $t[{g_j}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})\ne y({\alpha }_{j+1})$ .

Now in finitely many steps, using Lemma 3.5, we can extend the function $g_j$ to a function $g_i$ such that

• (*) $t[{g_i}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})$ is defined and $t[{g_i}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})\ne y({\alpha }_{j+1})$ for each $t\in \mathcal {H}_{j}$ .

Indeed, if $t[{g'}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})$ is not defined, where $t=\left \langle t_0,\dots , t_n\right \rangle $ then there is $i<n$ such that either

• ${\zeta }_i=\left \langle t_{i+1},\dots , t_n\right \rangle [{g'}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})$ is defined, $t_{i}=x$ , and ${\zeta }_i\in A\setminus \operatorname {dom}(g')$ ,

or

• ${\zeta }_i=\left \langle t_{i+1},\dots , t_n\right \rangle [{g'}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})$ is defined, $t_{i}=x^{-1}$ , and ${\zeta }_i\in A\setminus \operatorname {ran}(g')$ .

In both cases, using Lemma 3.5, we can extend $g'$ to $g"$ such that $\left \langle t_i,\dots , t_n\right \rangle [{g"}\cup \operatorname {id}_{{\lambda }\setminus A}]({\alpha }_{j+1})$ is defined and $\left \langle {\alpha }_{j+1},y({\alpha }_{j+1})\right \rangle \notin \bigcup \mathcal {H}_j[g"\cup id_{{\lambda }\setminus A}]$ .

After the inductive construction, the function $g=\bigcup _{i<{\kappa }}g_i$ meets the requirements.

Proof of Lemma 3.7 Fix a locally small, cofinal subfamily $\mathcal {C}\subset \bigl [ {\lambda } \bigr ]^ {\kappa } $ such that ${\mu }=|\mathcal {C}|$ is minimal. Then $|\{C\in \mathcal {C}:D\subset C\}|=|\mathcal {C}|$ for all $D\in \bigl [ {\lambda } \bigr ]^ {\kappa } $ .

Write $\mathcal {C}=\{C_{\alpha }:{\alpha }<{\mu }\}$ . Since $2^{\kappa }={\kappa }^+\le {\lambda }\le {\mu }$ there is a sequence $\left \langle B_{\alpha }:{\alpha }<{\mu }\right \rangle \subset {[{\lambda }]}^{{\kappa }}$ such that

1. (a) $\{B_{\alpha }:{\alpha }<{\kappa }^+\}\supset \{X\subset {\kappa }: |X|=|{\kappa }\setminus X|={\kappa }\}$ ,

2. (b) $\{B_{\alpha }:{\alpha }<{\mu }\}\supset \mathcal {C}$ .

Thus $\mathcal {B}=\{B_{\alpha }:{\alpha }<{\mu }\}$ is cofinal in ${[{\lambda }]}^{{\kappa }}$ . Now, for each ${\alpha }<{\mu }$ pick $A_{\alpha }\in \mathcal {C}$ such that $A_{\alpha }\supset C_{\alpha }\cup B_{\alpha }\cup {\kappa }$ and $|A_{\alpha }\setminus B_{\alpha }|={\kappa }$ .

Then $\mathcal {D}=\{\left \langle A_{\alpha },B_{\alpha }\right \rangle :{\alpha }<{\mu }\}$ satisfies the requirements.

Proof of Theorem 3.2 Fix $\mathcal {D}$ , $\mathcal {A}$ , and $\mathcal {B}$ as in Lemma 3.7.

For $\left \langle A,B\right \rangle \in \mathcal {D}$ consider the structure

$$ \begin{align*}\mathcal{M}_{\left\langle A,B\right\rangle }=\left\langle A,<,B,\{A\cap X:A\in \mathcal{A}\}\right\rangle .\end{align*} $$

Fix $\mathcal {D}'\in \bigl [ {\mathcal {D}} \bigr ]^ {{\kappa }^+} $ such that writing $\mathcal {A}'=\{A':\left \langle A',B'\right \rangle \in \mathcal {D}'\}$ and $\mathcal {B}'=\{B': \left \langle A',B'\right \rangle \in \mathcal {D}'\}$ we have

1. (a) $\forall \left \langle A,B\right \rangle \in \mathcal {D} \exists \left \langle A',B'\right \rangle \in \mathcal {D}'$ such that $\rho _{A,A'} $ is an isomorphism between $\mathcal {M}_{\left \langle A,B\right \rangle }$ and $\mathcal {M}_{\left \langle A',B'\right \rangle }$ .

2. (b) $\{X\subset {\kappa }: |X|=|{\kappa }\setminus X|={\kappa }\}\subset \mathcal {B}'$ .

Pick $K\in \bigl [ {\kappa } \bigr ]^ {\kappa } $ with $|{\kappa }\setminus K|={\kappa } $ . Choose $y \in S({\kappa })$ such that $y({\alpha })\ne {\alpha }$ for each ${\alpha }\in {\kappa }$ .

Lemma 3.8 (Key lemma)

There are functions $\mathcal {F}=\{f_{\left \langle A,B\right \rangle }:\left \langle A,B\right \rangle \in \mathcal {D}'\}$ such that

1. (a) $f_{\left \langle A,B\right \rangle }\in \operatorname {S}(A)$ ,

2. (b) $f_{\left \langle A,B\right \rangle }[B]=K$ ;

moreover, taking

$$ \begin{align*} & \mathcal{S}=\big\{\rho_{C_0,C_1}: \left\langle A_0,B_0\right\rangle ,\left\langle A_1,B_1\right\rangle \in \mathcal{D}', C_0\in \mathcal{A}\lceil^* A_0, C_1\in \mathcal{A}\lceil^* A_1,\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\rho_{C_0,C_1}[\mathcal{A}\lceil C_0]= \mathcal{A}\lceil C_1\}, \end{align*} $$

if $\mathcal {H}$ is a finite collection of $\ \mathcal {F}\cup \mathcal {S}$ -terms, then

$$ \begin{align*} |y\setminus \bigcup\mathcal{H}[\ ]|={\kappa}. \end{align*} $$

Before proving the Key lemma, we show how the Key Lemma completes the proof of Theorem 3.2.

So assume that the Key lemma holds.

For each $\left \langle A,B\right \rangle \in \mathcal {D}$ pick $\left \langle A',B'\right \rangle \in \mathcal {D}'$ such that $\rho _{A,A'} $ is an isomorphism between $\mathcal {M}_{\left \langle A,B\right \rangle }$ and $\mathcal {M}_{\left \langle A',B'\right \rangle }$ . We assume that $\left \langle A',B'\right \rangle =\left \langle A,B\right \rangle $ for $\left \langle A,B\right \rangle \in \mathcal {D}'$ .

Let

$$ \begin{align*} g_{\left\langle A,B\right\rangle }=\rho_{A',A}\circ f_{\left\langle A',B'\right\rangle }\circ \rho_{A,A'} \in S(A). \end{align*} $$

Let G be the permutation group on ${\lambda }$ generated by

$$ \begin{align*} \mathcal{G}=\{{g_{\left\langle A,B\right\rangle }}^+:\left\langle A,B\right\rangle \in \mathcal{D}\}. \end{align*} $$

Proof of Lemma 3.9 It is enough to show that for each $X\in \bigl [ {\lambda } \bigr ]^ {\kappa } $ there is $g\in G$ with $g[X]=K$ .

So fix $X\in \bigl [ {\lambda } \bigr ]^ {\kappa } $ . Pick $\left \langle A,B\right \rangle \in \mathcal {D}$ such that $X\subset B$ .

Then

$$ \begin{align*} Z=g_{\left\langle A,B\right\rangle }[X] \subset g_{\left\langle A,B\right\rangle }[B]=&(\rho_{A',A}\circ f_{\left\langle A',B'\right\rangle }\circ \rho_{A,A'})[B]\\ =&(\rho_{A',A}\circ f_{\left\langle A',B'\right\rangle })[B']=\rho_{A',A}[K]=K. \end{align*} $$

Since $|Z|=|{\kappa }\setminus Z|={\kappa }$ , there is C such that $\left \langle C,Z\right \rangle \in \mathcal {D}'$ . Then $f_{\left \langle C,Z\right \rangle }[Z]=K$ . Thus ${g_{\left \langle C,Z\right \rangle }}^+[Z]=K$ because $\left \langle C',Z'\right \rangle =\left \langle C,Z\right \rangle $ and so $f_{\left \langle C,Z\right \rangle }=g_{\left \langle C,Z\right \rangle }$ .

Thus $K=({g_{\left \langle C,Z\right \rangle }}^+\circ {g_{\left \langle A,B\right \rangle }}^+)[X]$ .

Proof of Lemma 3.10 We prove that $y\not \subset h$ for any $h\in G$ .

Assume that

$$ \begin{align*} h=(g_0^+)^{\ell_0}\circ(g_1^+)^{\ell_1}\circ \dots\circ (g_{n-1}^+)^{\ell_{n-1}}, \end{align*} $$

where $g_i=g_{\left \langle A_i,B_i\right \rangle }=\rho _{A^{\prime }_i,A_i}\circ f_{A^{\prime }_i, B^{\prime }_i}\circ \rho _{A_i,A^{\prime }_i}$ and $\ell _i\in \{-1,1\}$ for $i<n$ .

Since $g_i^+\setminus g_i$ is the identity function on ${\lambda }\setminus A_i$ , we have

$$ \begin{align*} &h\subset \bigcup \{(g_{i_0})^{\ell_{i_0}}\circ(g_{i_1})^{\ell_{i_1}}\circ \dots\circ (g_{i_{k-1}})^{\ell_{i_{k-1}}}:\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad k<n, i_0<i_1<\dots<i_{k-1}<n \}. \end{align*} $$

Fix $k\le n$ and $i_0<i_1<\dots <i_{k-1}<n$ .

Observe that if $\ell _i=-1$ then

$$ \begin{align*} (g_i)^{\ell_i}=(\rho_{A^{\prime}_i,A_i}\circ f_{A^{\prime}_i, B^{\prime}_i}\circ \rho _{A_i,A^{\prime}_i})^{-1} =\rho_{A^{\prime}_i,A_i}\circ (f_{A^{\prime}_i, B^{\prime}_i})^{-1}\circ \rho _{A_i,A^{\prime}_i}. \end{align*} $$

So

$$ \begin{align*} & (g_{i_0})^{\ell_{i_0}}\circ(g_{i_1})^{\ell_{i_1}}\circ \dots\circ (g_{i_{k-1}})^{\ell_{i_{k-1}}}\\ &\quad=\rho_{A^{\prime}_{i_0},A_{i_0}}\circ (f_{A^{\prime}_{i_0}, B^{\prime}_{i_0}})^{\ell_{i_0}}\circ \rho _{A_{i_0},A^{\prime}_{i_0}}\circ \rho_{A^{\prime}_{i_1},A_{i_1}}\circ (f_{A^{\prime}_{i_1}, B^{\prime}_{i_1}})^{\ell_{i_1}}\circ \rho _{A_{i_1},A^{\prime}_{i_1}}\circ. \end{align*} $$

For $j<k$ let

$$ \begin{align*} \rho^*_j=\rho _{A_{i_j},A^{\prime}_{i_j}}\circ \rho_{A^{\prime}_{i_{j+1}},A_{i_{j+1}}}. \end{align*} $$

Observe that writing

$$ \begin{align*} C_{j+1}=\rho_{A_{i_{j+1}},A^{\prime}_{i_{j+1}}}[A_{i_{j}}\cap A_{i_{j+1}}]\text{ and } C_j=\rho _{A_{i_{j}},A^{\prime}_{i_{j}}}[A_{i_{j}}\cap A_{i_{j+1}}], \end{align*} $$

we have

$$ \begin{align*} \rho^*_j=\rho_{C_{j+1},C_j}\in \mathcal{S} \end{align*} $$

(see Figure 1).

Figure 1

The function $\rho ^*_j$ .

Thus

$$ \begin{align*} &(g_{i_0})^{\ell_{i_0}}\circ(g_{i_1})^{\ell_{i_1}}\circ \dots\circ (g_{i_{k-1}})^{\ell_{i_{k-1}}}\\ &\qquad\qquad=\rho_{A_{i_0},A^{\prime}_{i_0}}\circ (f_{A^{\prime}_{i_0}, B^{\prime}_{i_0}})^{\ell_0}\circ \rho^*_0\circ (f_{A^{\prime}_{i_1}, B^{\prime}_{i_1}})^{\ell_1}\circ \rho^*_1\circ \dotsb \\ &\qquad\qquad\qquad\qquad\ \circ (f_{A^{\prime}_{i_{k-1}},B^{\prime}_{i_{k-1}}})^{\ell_{i_{k-1}}}\circ \rho_{A^{\prime}_{i_{k-1}},A_{i_{k-1}}}. \end{align*} $$

Since $\rho _{A_\ell ,A^{\prime }_\ell }\restriction {\kappa }=\operatorname {id}\restriction {\kappa }$ , we have

$$ \begin{align*} &\bigl( (g_{i_0})^{\ell_{i_0}}\circ(g_{i_1})^{\ell_{i_1}}\circ \dots\circ (g_{i_{k-1}})^{\ell_{i_{k-1}}}\bigr) \cap{\kappa}\times {\kappa}\\ &\qquad\qquad\qquad \subset (f_{A^{\prime}_{i_0}, B^{\prime}_{i_0}})^{\ell_0}\circ \rho^*_0\circ (f_{A^{\prime}_{i_1}, B^{\prime}_{i_1}})^{\ell_1}\circ \rho^*_1\circ \dotsb \\ &\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \ \, \circ (f_{A^{\prime}_{i_{k-1}},B^{\prime}_{i_{k-1}}})^{\ell_{i_{k-1}}}. \end{align*} $$

But $(f_{A^{\prime }_{i_0}, B^{\prime }_{i_0}})^{\ell _0}\circ \rho ^*_0\circ (f_{A^{\prime }_{i_1}, B^{\prime }_{i_1}})^{\ell _1}\circ \rho ^*_1\circ \dots \circ (f_{A^{\prime }_{i_{k-1}},B^{\prime }_{i_{k-1}}})^{\ell _{i_{k-1}}}=t[]$ for the $\mathcal {F}\cup \mathcal {S}$ -term $t=\left \langle (f_{A^{\prime }_{i_0}, B^{\prime }_{i_0}})^{\ell _0},\rho ^*_0, (f_{A^{\prime }_{i_1}, B^{\prime }_{i_1}})^{\ell _1},\rho ^*_1,\dots , (f_{A^{\prime }_{i_{k-1}},B^{\prime }_{i_{k-1}}})^{\ell _{i_{k-1}}}\right \rangle $ .

Since there are only finitely many sequences $i_0<\dots <i_{k-1}<n$ , we obtain that $h\cap {\kappa }\times {\kappa }$ is covered by the union of finitely many $\mathcal {F}\cup \mathcal {S}$ -terms.

But y is not covered by the union of finitely many $\mathcal {F}\cup \mathcal {S}$ -terms. So y witnesses that G is not ${\kappa }$ -transitive.

Proof of the Key Lemma 3.8 Write $\mathcal {D}'=\{\left \langle A_{\alpha },B_{\alpha }\right \rangle :{\alpha }<{\kappa }^+\}$ .

By transfinite induction, we define functions $\{f_{\alpha }:{\alpha }<{\kappa }^+\}$ such that taking

$$ \begin{align*} \mathcal{F}_{<{\beta}}=\{f_{\gamma}:{\gamma}<{\beta}\} \end{align*} $$

and

$$ \begin{align*} &\mathcal{S}_{<{\beta}}=\{\rho_{C_0,C_1}: {\delta},{\gamma}< {\beta}, C_0\in \mathcal{A}\lceil^* A_{\delta}, C_1\in \mathcal{A}\lceil^* A_{\gamma},\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\rho_{C_0,C_1}[\mathcal{A}\lceil C_0]=\mathcal{A}\lceil C_1 \}, \end{align*} $$

we have

1. (i) $f_{\alpha }\in \operatorname {S}(A_{\alpha })$ ,

2. (ii) $f_{\alpha }[B_{\alpha }]=K$ ,

3. (iii) if $\mathcal {H}$ is a finite collection of $\mathcal {F}_{<{\alpha }+1}\cup \mathcal {S}_{<\alpha +1}$ -terms, then

   $$ \begin{align*} |y\setminus \mathcal{H}[\ ]|={\kappa}. \end{align*} $$

Assume that we have constructed $f_{\beta }$ for ${\beta }<{\alpha }$ . Then we have:

(*) $$ \begin{align} \text{if}\ \mathcal{H}\ \text{is a finite collection of}\ \mathcal{F}_{<\alpha}\cup \mathcal{S}_{<\alpha}\text{-terms, then} |y\setminus \mathcal{H}[\ ]|={\kappa}. \end{align} $$

To continue the construction we need a bit more.

Claim 3.10.1. If $\mathcal {H}$ is a finite collection of $\mathcal {F}_{<{\alpha }}\cup \mathcal {S}_{<\alpha +1}$ -terms, then

$$ \begin{align*} |y\setminus \mathcal{H}[\ ]|={\kappa}. \end{align*} $$

Proof First observe that if ${\rho }_i={\rho }_{A_i,A^*_i}$ for $i<2$ , then

(†) $$ \begin{align} {\rho}_1\circ {\rho}_0={\rho}_{{\rho}_0^{-1}[A^*_0\cap A_1],{\rho}_1[A^*_0\cap A_1]}. \end{align} $$

Let

$$ \begin{align*} t=\left\langle t_0,t_1,\dots, t_n\right\rangle \end{align*} $$

be an element of $\mathcal {H}$ . Since ${\rho }_{C_0,C_1}\restriction {\kappa } =\operatorname {id}\restriction {\kappa }$ , if $t_0\in \mathcal {S}_{<\alpha +1}$ , then $t[\ ]\cap {\kappa }\times {\kappa }=\left \langle t_1,\dots , t_n\right \rangle [\ ]\cap {\kappa }\times {\kappa }$ . So we can assume that $t_0\in \mathcal {F}_{<{\alpha }}$ . Similar arguments give that we can assume that $t_n\in \mathcal {F}_{<{\alpha }}$ .

Now assume that

$$ \begin{align*} \left\langle t_i,\dots , t_j\right\rangle =\big\langle f_{\alpha_i},\rho_{C_{i+1},D_{i+1}},\rho_{C_{i+2},D_{i+2}}, \ldots, \rho_{C_{j-1},D_{j-1}},f_{{\alpha}_j}\big\rangle. \end{align*} $$

Then, by (†)

$$ \begin{align*} \rho_{C_{i+1},D_{i+1}}\circ\rho_{C_{i+2},D_{i+2}}\circ \cdots \circ \rho_{C_{j-1},D_{j-1}} = \rho_{E_i, E_j}, \end{align*} $$

for some $E_i\in \mathcal {A}\lceil C_{i+1}$ and $E_j\in \mathcal {A}\lceil D_{j-1}$ .

Thus we can assume that $j=i+2$ and

$$ \begin{align*} \left\langle t_i,t_{i+1}, t_{i+2}\right\rangle =\left\langle f_{{\alpha}_0},{\rho}_{E_0,E_1},f_{{\alpha}_1}\right\rangle. \end{align*} $$

Now

$$ \begin{align*} f_{{\alpha}_0}\circ {\rho}_{E_0,E_1}\circ f_{{\alpha}_1}= f_{{\alpha}_0}\circ {\rho}_{A_{{\alpha}_0}\cap E_0,A_{{\alpha}_1}\cap E_1}\circ f_{{\alpha}_1} \end{align*} $$

and ${\rho }_{A_{{\alpha }_0}\cap E_0,A_{{\alpha }_1}\cap E_1}\in \mathcal {S}_{<{\alpha }}$ .

Thus there is an $\mathcal {F}_{<{\alpha }}\cup \mathcal {S}_{<\alpha }$ -term $s_t$ such that

$$ \begin{align*} t[\ ]\cap ({\kappa}\times {\kappa})=s_t[\ ]\cap ({\kappa}\times {\kappa}). \end{align*} $$

Since $|y\setminus \bigcup \{s_t[\ ]:t\in \mathcal {H}\}|={\kappa }$ by (*), the Claim holds.

Since the claim holds, we can apply Lemma 3.6 for the family $\mathcal {F} =\mathcal {F}_{<{\alpha }}\cup \mathcal {S}_{{<\alpha }+1}$ to obtain $f_{\alpha }$ as g.

So we proved the Key Lemma 3.8.

So we proved Theorem 3.2.⊣

Proof of Theorem 4.1 If $\mathcal {F}$ is a set of functions, let

$$ \begin{align*} \left\langle\mathcal{F}\right\rangle _{gen}=\{f_0\circ\cdots \circ f_{n-1}:n\in {\omega}, f_i\in \mathcal{F} \text{ or } f_i^{-1}\in \mathcal{F} \text{ for}\ i<n\}. \end{align*} $$

Using Lemma 4.2 fix an H-large $r\in S({\omega })$ . Enumerate ${[{\omega }_1]}^{{\omega }}\times {[{\omega }_1]}^{{\omega }}$ as $\{\left \langle A_{\alpha },B_{\alpha }\right \rangle :{\alpha }<2^{\omega }\}$ . By transfinite recursion on ${\alpha }<2^{\omega }$ , we will construct permutations $f_{\alpha }\in \operatorname {S}_{{\omega }}({\omega }_1)$ such that $f_{\alpha }[A_{\alpha }]=B_{\alpha }$ and writing

$$ \begin{align*} \mathcal{F}_{\delta}=\{t[]:t\ \text{is a}\ H\cup \{f_{\zeta}:{\zeta}<{\delta}\}\text{-term}\}= \left\langle H\cup\{f_{\zeta}:{\zeta}<{\delta}\}\right\rangle _{gen}, \end{align*} $$

the permutation r is $\mathcal {F}_{\alpha +1}$ -large.

Since $\mathcal {F}_0=H$ , we know that $r\in S({\omega })$ is $\mathcal {F}_0$ -large.

Assume that we have constructed $\left \langle f_{\zeta }:{\zeta }<{\alpha }\right \rangle $ such that the function r is $\mathcal {F}_{{\zeta }+1}$ -large for ${\zeta }<{\alpha }$ . Then r is $\mathcal {F}_{\alpha }$ -large. Next we should construct $f_{\alpha }\in S({\omega }_1)$ such that $f_{\alpha }[A_{\alpha }]=B_{\alpha }$ and r is $\mathcal {F}_{{\alpha }+1}$ -large. We want to apply MA(countable) to construct $f_{\alpha }$ , but to do so we need some technical lemmas.

Fix first $C_{\alpha }\in {[{\omega }_1]}^{{\omega }}$ such that $A_{\alpha }\cup B_{\alpha }\subset C_{\alpha }$ and $C_{\alpha }\setminus (A_{\alpha }\cup B_{\alpha })={\omega }$ .

Definition 4.3. Given sets X and Y let us denote by $\operatorname {{Bij_p}}(X,Y)$ the set of all finite bijections between subsets of X and Y.

For $A,B,C\in {[{\omega }_1]}^{{\omega }}$ define the poset $\mathcal {P}_{C,A,B}=\left \langle P_{C,A,B},\le \right \rangle $ as follows. Let

$$ \begin{align*} P_{C,A,B}=\{p\in \operatorname{{Bij_p}}(C,C): p[A]\subset B, p[C\setminus A]\subset C\setminus B\}. \end{align*} $$

Write $p\le q$ iff $p\supseteq q$ .

We want to apply MA(countable) for the countable poset

$$ \begin{align*}\mathcal{P}=\mathcal{P}_{C_{\alpha},A_{\alpha},B_{\alpha}}.\end{align*} $$

Our plan is to define a family $\mathbb D$ of dense subsets in P with $|\mathbb D|<2^{\omega }$ such that if $\mathcal {K}$ is a $\mathbb D$ -generic filter in P, then $(\bigcup \mathcal {K})\cup \operatorname {id}_{{\omega }_1\setminus C_{\alpha }}$ works as $f_{\alpha }$ .

Lemma 4.5. If $M\in {\omega }$ and $\mathcal {H}$ is a finite set of $\mathcal {F}_{\alpha }\cup \{x\}$ -terms then

$$ \begin{align*} &E_{\mathcal{H},M}=\{p\in P: \exists m\in {\omega}\setminus M \\ &\qquad\qquad\quad t[p](m) \text{ is defined, but}\ t[p](m)\ne r(m)\ \text{for each}\ t\in \mathcal{H}\} \end{align*} $$

is dense in P.

Proof of the lemma Fix $q\in P$ . We can assume that $\mathcal {H}$ is closed for subterms.

We know that $|r\setminus \bigcup \mathcal {H}[\ ]|={\omega }$ because r is $\mathcal {F}_{\alpha }$ -large.

Since $\mathcal {H}$ is closed for subterms,

$$ \begin{align*}r\cap \bigcup \mathcal{H}[\ ]=r\cap \bigcup \mathcal{H}[ \operatorname{id}_{{\omega}_1\setminus C_{\alpha}} ].\end{align*} $$

Since $|q|<{\omega }$ , we have

$$ \begin{align*} |r\setminus \bigcup \mathcal{H}[q\cup \operatorname{id}_{{{\omega}_1}\setminus C_{\alpha}} ]|={\omega}. \end{align*} $$

So we can pick $m\in {\omega }\setminus M$ such that

• (*) for each $t\in \mathcal {H}$ either $t[q\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ is undefined or $t[q\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m) \ne r(m)$ .

Since $\mathcal {H}$ is finite, we can find $p\le q$ such that

• (*) for each $t\in \mathcal {H}$ either $t[p\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ is undefined or $t[p\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m) \ne r(m)$ ,

• (•) the cardinality of the finite set

  $$ \begin{align*}\{t\in \mathcal{H}\ :\ t[p\cup \operatorname{id}_{{{\omega}_1}\setminus C_{\alpha}}](m)\ \text{is undefined}\}\end{align*} $$
  is minimal.

To show that $p\in E_{\mathcal {H},M}$ we prove that

1. (∘) there is no $t\in \mathcal {H}$ such that $t[p\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ is undefined.

Assume on the contrary that this statement is not true.

Fix $t\in \mathcal {H}$ such that $t[{p}\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ is not defined, where $t=\left \langle t_0,\dots , t_n\right \rangle $ . Thus there is $i<n$ such that

1. (1) $\left \langle t_{i+1},\dots , t_n\right \rangle [{p}\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ is defined, but

2. (2) $\left \langle t_{i},\dots , t_n\right \rangle [{p}\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ is not defined.

Then $t'=\left \langle t_i,\dots , t_n\right \rangle \in \mathcal {H}$ . Let ${\zeta }_i=\left \langle t_{i+1},\dots , t_n\right \rangle [{p}\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ . Then either $t_{i}=x$ and ${\zeta }_i\notin \operatorname {dom}(p)$ or $t_{i}=x^{-1}$ and ${\zeta }_i\notin \operatorname {ran}(p).$

In both cases, using Lemma 3.5, we can extend p to $p'$ such that $\left \langle t_i,\dots , t_n\right \rangle [{p'}\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}](m)$ is defined and $\left \langle m,r(m)\right \rangle \notin \mathcal {H}[p'\cup \operatorname {id}_{{{\omega }_1}\setminus C_{\alpha }}]$ . Thus $p'\le q$ and

$$ \begin{align*} &\{t\in \mathcal{H}\ :\ t[p'\cup \operatorname{id}_{{{\omega}_1}\setminus C_{\alpha}}](m)\ \text{is undefined}\}\subsetneq\\ &\qquad\qquad\qquad\qquad\qquad\qquad\{t\in \mathcal{H}\ :\ t[p\cup \operatorname{id}_{{{\omega}_1}\setminus C_{\alpha}}](m)\ \text{is undefined}\}, \end{align*} $$

which contradicts $(\bullet)$ .

So we proved Lemma 4.5.

Let

$$ \begin{align*} &\mathbb D=\{D_{i},R_i:i\in C_{\alpha}\}\cup \\ &\qquad\qquad\qquad \{E_{\mathcal{F},M}:M\in {\omega},\ \mathcal{F}\ \text{is a finite set of}\ \mathcal{F}_{\alpha}\cup\{x\}\text{-terms.}\}. \end{align*} $$

Then $\mathbb D$ is a family of dense sets in $P_{C_{\alpha },A_{\alpha },B_{\alpha }}$ with cardinality $<2^{\omega }$ . So, by MA(countable), there is a $\mathbb D$ -generic filter $\mathcal {K} $ . Let $f_{\alpha }=(\bigcup \mathcal {K})\cup id_{{\omega }_1\setminus C_{\alpha }}$

The assumption $\{D_{i},R_j:i\in C_{\alpha }\}\subset \mathbb {D}$ yields $C_{\alpha }=\operatorname {dom}(\bigcup \mathcal {K})=\operatorname {ran}(\bigcup \mathcal {K})$ . Since $f_{\alpha }[A_{\alpha }]\subset B_{\alpha }$ and $f_{\alpha }[C_{\alpha }\setminus A_{\alpha }]\subset C_{\alpha }\setminus B_{\alpha }$ by the construction of $P_{C_{\alpha },A_{\alpha },B_{\alpha }}$ we have $f_{\alpha }[A_{\alpha }]=B_{\alpha }$ .

If $\mathcal {F}$ is a finite subset of ${\mathcal {F}_{\alpha +1}}$ , then there is a finite set $\mathcal {H}$ of $\mathcal {F}_{\alpha }\cup \{x\}$ -terms such that

$$ \begin{align*} \mathcal{F}=\{t[f_{\alpha}]:t\in \mathcal{H}\}. \end{align*} $$

Then $E_{\mathcal {H},M}\cap \mathcal {K}\ne \emptyset $ implies that there is $m>M$ such that $r(m)\notin \{t[f_{\alpha }](m):t\in \mathcal {H}\}=\{f(m):f\in \mathcal {F}\}$ . Thus r is $\mathcal {F}_{{\alpha }+1}$ -large. Hence $f_{\alpha }$ satisfies the requirements.

So we carried out the inductive construction, and so we have constructed $\left \langle f_{\alpha }:{\alpha }<2^{\omega }\right \rangle $ such that r is $\mathcal {F}_{2^{\omega }}$ -large. So the group $H^*=\mathcal {F}_{2^{\omega }}$ satisfies the requirements. This completes the proof of Theorem 4.1.

Proof of Theorem 4.6. First we show that $G^*$ is ${\omega }$ -homogeneous.

Let $X,Y\in \bigl [ {\lambda } \bigr ]^{\omega }$ be arbitrary. First, by (iii) we can pick $f,g\in G$ such that $f[{\omega }]\supset X$ and $g[{\omega }]\supset Y$ . Since $H^*$ is ${\omega }$ -homogeneous, there is $h\in H^*$ such that

$$ \begin{align*} h[f^{-1}(X)]= g^{-1}(Y). \end{align*} $$

Then $g\circ {h}^+\circ f^{-1}\in G^*$ and $(g\circ {h}^+\circ f^{-1})[X]=Y$ .

Next we show that $G^*$ is ${\omega }$ -intransitive. Fix a countable injective function r with $\operatorname {dom}(r)\cup \operatorname {ran}(r)\in {[{\omega }_1]}^{{\omega }}$ which is $H^*$ -large. Without loss of generality we can assume that $r\in S({\gamma })$ for some ${\gamma }<{\omega }_1$ . We will verify that

$$ \begin{align*} r\ \text{is}\ G^*\text{-large} \end{align*} $$

as well. It is enough to prove the next lemma.

Lemma 4.7. For each $g\in G^*$ there is a finite subset $H_g$ of $H^*$ such that

$$ \begin{align*}g\cap({\gamma}\times {\gamma})\subset \bigcup H_g.\end{align*} $$

Proof of the Lemma Since $G^*=\left \langle G\cup H^+\right \rangle _{gen}$ , where $H^+=\{h^+:h\in H^*\}$ and both G and $H^+$ are subgroups, we can assume that

$$ \begin{align*} g = e_0\circ g_0\circ \cdots \circ e_n\circ g_n, \end{align*} $$

where $g_i\in G$ and $e_i\in H^+$ .

For $e\in H^+$ , write $e^-=e\restriction {{\omega }_1}\in H^*$ .

By finite induction, define countable subsets $A_{n+1}, B_n,A_n,\dots , B_0, A_0$ of ${\lambda }$ as follows: let $A_{n+1}={\gamma }$ and $B_i=g_i[A_{i+1}]$ and $A_{i}=e_{i}[B_{i}]$ for $i=n,n-1,\dots ,0$ .

Pick ${\delta }<{\omega }_1$ with

$$ \begin{align*}\bigcup\{A_i,B_i:0\le i\le n+1\}\cap {\omega}_1\subset {\delta}.\end{align*} $$

For $0\le k<m\le n$ let

$$ \begin{align*}g_{k,m}=g_{k}\circ \cdots\circ g_{m-1}.\end{align*} $$

By (ii) we can pick $h_{k,m}\in H^*$ such that $h_{k,m}\supset g_{k,m}\cap ({\delta }\times {\delta })$ . Let

$$ \begin{align*} &\mathcal{H}_g=\{e^-_{i_0}\circ h_{i_0,i_1}\circ e^-_{i_1}\circ h_{i_1, i_2}\circ \cdots \circ e^-_{i_\ell} \circ h_{i_\ell,i_{\ell+1}}: \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad 0\le i_0<\dots <i_\ell<i_{\ell+1}= n \}. \end{align*} $$

Claim 4.7.1. $g\cap ({\gamma }\times {\gamma })\subset \bigcup \mathcal {H}_g$ .

Proof of the Claim Let ${\alpha }\in {\gamma }$ be arbitrary with $g({\alpha })\in {\gamma }$ . Write ${\alpha }_{n+1}={\alpha }$ , ${\beta }_i=g_i({\alpha }_{i+1})$ , and ${\alpha }_{i}=e_{i}({\beta }_i)$ for $i=n,n-1,\dots ,0$ . So ${\alpha }_0=g({\alpha })\in {\gamma }$ .

Let $i_0=0<\dots <i_s =n+1$ be the enumeration of the set $I=\{i\le n+1: {\alpha }_i\in {\omega }_1\}=\{i\le n+1: {\alpha }_i\in {\delta }\}$ .

Fix $\ell <s$ , and write $k=i_{\ell }$ and $m=i_{\ell +1}$ .

If $k+1=m$ , then ${\alpha }_k,{\beta }_k, {\alpha }_m\in {\delta }$ and so then

$$ \begin{align*}{\alpha}_k=e_k({\beta}_k)=e_k(g_k({\alpha}_m))= (e_k^-\circ h_{k,m})({\alpha}_m).\end{align*} $$

If $k+1<m$ , then

1. (i) ${\alpha }_k\in {\delta }$ , ${\beta }_m\in {\delta }$ , but

2. (ii) ${\alpha }_i, {\beta }_i\in {\lambda }\setminus {\omega }_1$ and so ${\alpha }_i={\beta }_i$ for $k<i<m$

(see Figure 2).

Figure 2

The function $h_{k,m}$ .

Thus

$$ \begin{align*} {\beta}_k&=(g_{k}\circ e_k\circ g_{k+1}\circ\dots\circ e_{m-1}\circ g_{m-1})({\alpha}_m) \\ &=(g_{k}\circ g_{k+1}\circ\dots\circ g_{m-1})({\alpha}_m)=g_{k,m}({\alpha}_{m})=h_{k,m}({\alpha}_m), \end{align*} $$

and so

$$ \begin{align*}{\alpha}_k=e_k({\beta}_k)= e_k(h_{k,m}({\alpha}_m))=(e_k^-\circ h_{k,m})({\alpha}_m).\end{align*} $$

Hence

$$ \begin{align*} g({\alpha})&= (e_0\circ g_0\circ \dots \circ e_n\circ g_n)({\alpha})\\ &=(e_{i_0}^-\circ h_{i_0,i_1}\circ \dots \circ e^-_{i_\ell}\circ h_{i_{s-1},i_{s}})({\alpha}) \end{align*} $$

and $(e_{i_0}^-\circ h_{i_0,i_1}\circ \dots \circ e^-_{i_\ell }\circ h_{i_{s-1},i_{s}})\in \mathcal {H}_g$ .

So we proved the Claim which completes the proof of the Lemma.

As we observed, the previous lemma implies that r is $G^*$ -large, and so $G^*$ is ${\omega }$ -intransitive which completes the proof of Theorem 4.6.⊣

Proof of Theorem 4.8 First observe that (2) implies that $|\{g\cap ({\omega }_1\times {\omega }_1):g\in G\}|\ge {\omega }_1$ , and so $2^{\omega }>{\omega }_1$ by (1).

For each countable injective function f with $\operatorname {dom}(f)\cup \operatorname {ran} (f)\subset {\omega }_1$ pick a permutation $h(f)\in \operatorname {S}_{{\omega }}({\omega }_1)$ with $h(f)\supset f$ .

Let

$$ \begin{align*} H=\left\langle\{h(g\cap ({\alpha}\times {\alpha})):g\in G, {\alpha}<{\omega}_1\}\right\rangle _{gen}. \end{align*} $$

Since $2^{\omega }>{\omega }_1$ , we have

1. (3) $|H|\le |\{g\cap ({\omega }_1\times {\omega }_1):g\in G\}|\cdot {\omega }_1<2^{\omega }$ , and

2. (4) $\forall g\in G \forall {\alpha }<{\omega }_1$ $\exists h\in H$ such that $g\cap ({\alpha }\times {\alpha })\subset h$ .

By (3) we can apply Theorem 4.1 and so there is an ${\omega }$ -homogeneous, but ${\omega }$ -intransitive permutation group $H^*\le \operatorname {S}_{\omega }({\omega }_1)$ with $H^*\supset H$ .

By (2) and (4) we can apply Theorem 4.6 for G and $H^*$ to show that the permutation group $G^*=\left \langle G\cup \{h^+:h\in H^+\}\right \rangle _{gen}\le \operatorname {S}_{\omega }({\lambda })$ is ${\omega }$ -homogeneous, but ${\omega }$ -intransitive.

Proof of Corollary 4.9 from Theorem 4.8 We work in $V^P$ . Let $G=\operatorname {S}_{{\omega }}({\lambda })^V$ . Then

$$ \begin{align*}|\{g\cap {\omega}_1\times {\omega}_1: g\in G\}|=|\operatorname{S}_{{\omega}}({\omega}_1)^V|= (2^{\omega})^V< ((2^{\omega})^+)^V= (2^{\omega})^{V^P}.\end{align*} $$

So (1) holds. Since P is c.c.c., $\{g[{\omega }]:g\in G\}={[{\lambda }]}^{{\omega }}\cap V$ is cofinal in $\left \langle {[{\lambda }]}^{{\omega }},\subset \right \rangle $ . Hence (2) also holds.

So we can apply Theorem 4.8 because it is known that MA(countable) holds after adding $(2^{\omega })^+$ -many Cohen reals to a ground model (e.g., $cov(\mathcal {M})=2^{\omega }$ in the Cohen model by [Reference Blass1, Table 4], and $cov(\mathcal {M})=2^{\omega }$ implies MA(countable) by [Reference Keremedis4, Theorem 1]).