$\beta<2^{\aleph_{0}}$, have cardinality continuum, and be such that:

  1. $(*)_{I_{\beta}}$

For every $f:I_{\beta}\to\theta$, $\theta<2^{\aleph_{0}}$, for some $\eta\in
P^{I_{\beta}}_{\omega}$, for every $n<\omega$,

$\big{(}\forall\ell<h(n)\big{)}\big{[}f(\mathrm{Res}^{0}_{n}(\eta))=f(\mathrm{Res}^{\ell}_{n}(\eta))\big{]}$

(i.e.
$\eta(m)=\big{\langle}\alpha_{\ell}:\ell<h(n)\big{\rangle}\Rightarrow\big{|}\big{\\{}f(\eta\restriction
n{\char 94\relax}\langle\alpha_{\ell}\rangle):\ell<h(n)\big{\\}}\big{|}=1$.)

[Why do such $I$-s exist? The full tree will serve; that is, we let

$\displaystyle
I_{\beta}=\big{\\{}\langle\bar{\alpha}^{\ell}:\ell<\gamma\rangle:$
$\displaystyle\ \gamma\leq\omega,\ \bar{\alpha}^{\ell}\mbox{ an increasing
sequence of length }h(\ell)$ $\displaystyle\mbox{ from }2^{\aleph_{0}},\mbox{
except in the case }0<\gamma<\omega\wedge\ell=\gamma-1;$ $\displaystyle\mbox{
then we demand }\bar{\alpha}^{\ell}\mbox{ is just an
ordinal}<2^{\aleph_{0}}\big{\\}}.$

This is as required, as for any $f:I_{\beta}\to\theta$ we can choose a
sequence
$\bar{\alpha}_{\ell}=\langle\beta_{\ell,0},\ldots,\beta_{\ell,h(\ell)-1}\rangle$
by induction on $\ell<\omega$, where
$\beta_{\ell,0}<\ldots<\beta_{\ell,h(\ell)-1}<2^{\aleph_{0}}$ and
$f(\langle\bar{\alpha}^{0},\ldots,\bar{\alpha}^{\ell-1},\beta_{\ell,i}\rangle)$
does not depend on $i$. This is possible as
$2^{\aleph_{0}}>|\mathrm{rang}(f)|$. So $I_{\beta}$-s as required in
$(*)_{I_{\beta}}$ indeed exist.]

We shall now construct $\mathbf{B}_{\alpha}$ (for $\alpha\leq 2^{\aleph_{0}}$)
and $\bar{a}^{\alpha}=\langle a^{\alpha}_{n}:n<\omega\rangle$ such that:

  1. (I)

    1. (a)

$\mathbf{B}_{0}$ is a subalgebra of $\mathcal{P}(\omega)$ of cardinality $\mu$
with a chain of cardinality $\mu$ satisfying the ccc (even the strong
$\aleph_{1}$-cc).

[E.g. let $A$ be a set of $\mu$ reals, let $h$ be a one to one function from
$\omega$ onto the rationals and $\mathbf{B}$ is the Boolean algebra of subset
of $\omega$ generated by $\big{\\{}\\{n:h(n)<a\\}:a\in A\big{\\}}$. Clearly
$\mathfrak{B}$ has a linearly ordered subset of cardinality $\mu$, e.g. its
set of generators. Of course, its length is not $>\mu$ as its cardinality is
$\mu$.]

    2. (b)

$\mathbf{B}_{\alpha}$ is increasing continuous, of cardinality
$2^{\aleph_{0}}$ if $\alpha>0$.

    3. (c)

$\bar{a}^{\alpha}$ is an $\omega$-sequence of pairwise disjoint non-zero
elements of $\mathbf{B}_{\alpha}$.

    4. (d)

If $\alpha<2^{\aleph_{0}}$,
$a_{n}\in\mathbf{B}_{\alpha}\setminus\\{0_{\mathbf{B}_{\alpha}}\\}$, and
$\bigwedge\limits_{n\neq m}a_{n}\cap a_{m}=0$ then for $2^{\aleph_{0}}$ many
ordinals $\alpha$, we have $\bigwedge\limits_{n<\omega}a_{n}=a^{\alpha}_{n}$.

[You can demand that $\\{a^{\alpha}_{n}:n<\omega\\}$ is a maximal antichain;
it does not matter.]

    5. (e)

$\mathbf{B}_{\alpha+1}=\mathrm{ba}[\mathbf{B}_{\alpha},\bar{a}^{\alpha},I_{\alpha}]$
(We denote the $x_{\eta}$ by $x^{\alpha}_{\eta}$ for $\eta\in I_{\alpha}$.)

There is no problem to do the bookkeeping, and
$\mathbf{B}_{\alpha}\subseteq\mathbf{B}_{\alpha+1}$ by 3.4(1). We shall show
that
$\mathbf{B}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=\mathbf{B}_{2^{\aleph_{0}}}$
is as required. Obviously $\mathbf{B}$ has cardinality $2^{\aleph_{0}}$.

By 3.4(3) clearly $\mathbf{B}_{\alpha}\lessdot\mathbf{B}_{\alpha+1}$, so we
can prove by induction on $\alpha$ that $\beta<\alpha\ \Rightarrow\
\mathbf{B}_{\beta}\lessdot\mathbf{B}_{\alpha}$, by 2.9, 2.10. We can also
prove by induction on $\alpha$ that $\mathbf{B}_{\alpha}$ satisfies the
$\aleph_{1}$-cc (even the strong $\lambda$-cc when
$\lambda=\mathrm{cf}(\lambda)>\aleph_{0}$): the successor stage is proved by
3.5, the limits steps by 2.10. So demand (A) from 3.6 holds. If $\mathbf{f}$
is a homomorphism from $\mathbf{B}$ onto some $\mathbf{B}^{\prime}$ with
$\aleph_{0}\leq\|\mathbf{B}^{\prime}\|<2^{\aleph_{0}}$ then there are
$b_{n}\in\mathbf{B}^{\prime}\setminus\\{0\\}$ pairwise disjoint. Now for some
$a_{n}\in\mathbf{B}$, $\mathbf{f}(a_{n})=b_{n}$ and without loss of generality
$\bigwedge\limits_{n\neq m}a_{n}\cap a_{m}=0$ (otherwise use
$a^{\prime}_{n}=a_{n}\setminus\bigcup\limits_{m<n}a_{m}$). Hence for every
infinite co-infinite $Y\subseteq\omega$. for some $\alpha=\alpha_{Y}$:

$\\{a^{\alpha}_{2n}:n<\omega\\}=\\{a_{n}:n\in
Y\\}\quad\mbox{and}\quad\\{a^{\alpha}_{2n+1}:n<\omega\\}=\\{a_{n}:n\in\omega\setminus
Y\\}.$

Now define $g:I_{\alpha}\to\mathbf{B}^{\prime}$ by
$g(\eta)=\mathbf{f}(x^{\alpha}_{\eta})$, so by the choice of the
$I_{\alpha}$-s (i.e. by $(*)_{I_{\beta}}$) for some $\eta=\eta_{Y}\in
P^{I_{\alpha}}_{\omega}$, for every $n$, letting
$\eta(n)=\langle\alpha_{0},\alpha_{i},\ldots,\alpha_{h(n)-1}\rangle$, we have

$\bigwedge\limits_{\ell<h(n)}f(x^{\alpha}_{\eta\restriction n{\char
94\relax}\langle\alpha_{\ell}\rangle})=f(x^{\alpha}_{\eta\restriction n{\char
94\relax}\langle\alpha_{0}\rangle}).$

Hence $\mathbf{f}\bigl{(}x^{\alpha}_{\eta\restriction n{\char
94\relax}\langle\alpha_{\ell}\rangle}-x^{\alpha}_{\eta\restriction n{\char
94\relax}\langle\alpha_{\ell+1}\rangle}\bigr{)}=0_{\mathbf{B}^{\prime}}$ for
$\ell<h(n)-1$ and hence

$\mathbf{f}\bigl{(}a^{\alpha}_{n}\cap\bigcap\limits_{\ell<\frac{h(n)-1}{2}}\bigl{(}1-(x^{\alpha}_{\eta\restriction
n{\char 94\relax}\langle\alpha_{2\ell}\rangle}-x^{\alpha}_{\eta\restriction
n{\char
94\relax}\langle\alpha_{2\ell+1}\rangle})\bigr{)}\bigr{)}=\mathbf{f}(a^{\alpha}_{n})$

and

$\mathbf{f}\bigl{(}a^{\alpha}_{n}-\bigcup\limits_{\ell<\frac{h(n)-1}{2}}\bigl{(}x^{\alpha}_{\eta\restriction
n{\char 94\relax}\langle\alpha_{2\ell}\rangle}-x^{\alpha}_{\eta\restriction
n{\char
94\relax}\langle\alpha_{2\ell+1}\rangle}\bigr{)}\bigr{)}=\mathbf{f}(a^{\alpha}_{n}).$

Hence (see Definition 3.3)

$\begin{array}[]{rcl}n\mbox{ is
even}&\Rightarrow&\mathbf{B}^{\prime}\models\mathbf{f}(a^{\alpha}_{n})\leq\mathbf{f}(x^{\alpha}_{\eta}),\\\
n\mbox{ is
odd}&\Rightarrow&\mathbf{B}^{\prime}\models\mathbf{f}(a^{\alpha}_{n})\cap\mathbf{f}(x^{\alpha}_{\eta})=0.\end{array}$

Therefore,

$\begin{array}[]{rcccl}m\in Y&\Rightarrow&\mbox{ for some even $n$,
}a^{\alpha}_{n}=a_{m}&\Rightarrow&\mathbf{B}^{\prime}\models
b_{m}\leq\mathbf{f}(x^{\alpha}_{\eta}),\\\ m\in\omega\setminus
Y&\Rightarrow&\mbox{ for some odd $n$,
}a^{\alpha}_{n}=a_{m}&\Rightarrow&\mathbf{B}^{\prime}\models
b_{m}\cap\mathbf{f}(x^{\alpha}_{\eta})=0.\end{array}$

As this occurs for every infinite co-infinite $Y\subseteq\omega$, for some
$\alpha=\alpha_{Y}$, and $\eta=\eta_{Y}$, clearly we get $2^{\aleph_{0}}$ many
distinct members of $\mathbf{B}^{\prime}$ (simply, the
$\mathbf{f}(x_{\eta_{Y}})$), a contradiction. So demand (C) of 3.6 holds.

What about the length, i.e, clauses (B) and (B)+? For (B), first note that
$\mathbf{B}_{0}$ has a chain of cardinality $\mu$ and hence so does
$\mathbf{B}$. If $J\subseteq\mathbf{B}$ is a chain, $|J|=\mu^{+}$, then
$(\mathbf{B})^{+}$ gives a contradiction and even the “weakly indiscernible
sequence” version does because as $\mathbf{B}\models$ ccc, it has no subset of
order type $\mu^{+}$ or $(\mu^{+})^{*}$; but the variant of $(\mathbf{B})^{+}$
implies just this ($m=1$ suffices).

So it suffices to prove that clause $\textbf{(B)}^{+}$ holds for
$\mathbf{B}_{\alpha}$ by induction on $\alpha$.

Case 1: $\alpha=0$.

Trivial (can get $\bar{c}^{\zeta}$ constant on $Y\in[\mu^{+}]^{\mu^{+}}$).

Case 2: $\alpha$ is limit, $\mathrm{cf}(\alpha)\neq\mu^{+}$.

For some $\beta<\alpha$,

$Y_{1}=\\{\zeta<\mu^{+}:\bar{c}^{\zeta}\subseteq\mathbf{B}_{\beta}\\}\in[\mu^{+}]^{\mu^{+}},$

(note that if $\mathrm{cf}(\alpha)<\mu^{+}$, then we can get $Y_{1}=\mu^{+}$)
and use the induction hypothesis.

Case 3: $\mathrm{cf}(\alpha)=\mu^{+}$.

Let $\langle\beta_{\varepsilon}:\varepsilon<\mu^{+}\rangle$ be an increasing
continuous sequence with limit $\alpha$. Let
$n,m,\langle\bar{c}^{\zeta}:\zeta<\mu^{+}\rangle$ be given. Without loss of
generality $\bar{c}^{\zeta}=\langle c^{\zeta}_{\ell}:\ell<m\rangle$ is a
partition of $1_{\mathbf{B}_{\alpha}}$ (i.e., $\ell_{1}\neq\ell_{2}\Rightarrow
c^{\zeta}_{\ell_{1}}\cap c^{\zeta}_{\ell_{2}}=0$ and
$1_{\mathbf{B}_{\alpha}}=\bigcup\limits_{\ell<m}c^{\zeta}_{\ell}$). For each
$\zeta<\mu^{+}$, we can find
$a^{\zeta}_{\ell},b^{\zeta}_{\ell}\in\mathbf{B}_{\beta_{\zeta}}$ such that:

  1. (A)

$a^{\zeta}_{\ell}\leq c^{\zeta}_{\ell}\leq b^{\zeta}_{\ell}$

  2. (B)

$(0<x\leq b^{\zeta}_{\ell}-a^{\zeta}_{\ell})\wedge
x\in\mathbf{B}_{\beta_{\zeta}}\Rightarrow(x\cap
c^{\zeta}_{\ell}-a^{\zeta}_{\ell}\neq 0)\wedge(x-c_{\ell}^{\zeta}\neq 0)$.

[Why? By use of 3.4(5)). If $\zeta$ is limit then for some $f(\zeta)<\zeta$ we
have
$\\{a^{\zeta}_{\ell},b^{\zeta}_{\ell}:\ell<m\\}\subseteq\mathbf{B}_{f(\beta_{\zeta})}$.
By Fodor lemma for some $\varepsilon(*)<\mu^{+}$ and a stationary set
$S\subseteq\mu^{+}$, we have $\bigwedge\limits_{\zeta\in
S}f(\zeta)=\varepsilon(*)$.

So

  1. (c)

$\varepsilon\in
S\Rightarrow\\{a^{\varepsilon}_{\ell},b^{\varepsilon}_{\ell}:\ell<m\\}\subseteq\mathbf{B}_{\beta_{\dot{\varepsilon}(*)}}$.

Also without loss of generality

  1. (d)

If $\varepsilon<\zeta\in S$ then
$\\{c^{\varepsilon}_{\ell}:\ell\\}\subseteq\mathbf{B}_{\beta_{\zeta}}$.

Now apply the induction hypothesis on
$\mathbf{B}_{\beta_{\dot{\varepsilon}(*)}}$ and $\langle\bar{a}^{\zeta}{\char
94\relax}\bar{b}^{\zeta}:\zeta<\mu^{+}\rangle$, where $\bar{a}^{\zeta}=\langle
a^{\zeta}_{\ell}:\ell<m\rangle$,
$\bar{b}^{\zeta}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=\langle
b^{\zeta}_{\ell}:\ell<m\rangle$.

So there is $Y\in[S]^{\mu^{+}}$ such that $\langle\bar{b}^{\zeta}:\zeta\in
Y\rangle$ is an $(n,\mathrm{qf})$-indiscernible set. So let
$\zeta_{0}<\ldots<\zeta_{n-1}$ be from $S$ and for $k\leq n$ let
$\mathbf{B}^{\prime}_{k}$ be the subalgebra of $\mathbf{B}$ generated by
$X_{k}=\\{\bar{b}^{\zeta_{i}}_{\ell},b^{\zeta_{i}}_{\ell}:i<n,\
\ell<m\\}\cup\\{\bar{c}^{\zeta_{i}}_{\ell}:i<k,\ \ell<m\\}$. We understand
$\mathbf{B}^{\prime}_{0}$ by the choice of $Y$ and we can understand
$\mathbf{B}^{\prime}_{n}$ by clauses (c) + (d) above.

Case 4: $\alpha=\beta+1$.

Let $n,m<\omega$ and $\bar{c}^{\zeta}\in{}^{m}(\mathbf{B}_{\beta+1})$ for
$\zeta<\mu^{+}$ be given, $\bar{c}^{\zeta}=\langle
c^{\zeta}_{\ell}:\ell<m\rangle$. So there are $k_{\zeta,0}=k(\zeta,0)<\omega$,
$k_{\zeta,1}=k(\zeta,1)<\omega$ and
$b^{\zeta}_{0},\ldots,b^{\zeta}_{k_{\zeta,0}-1}\in\mathbf{B}_{\beta}$,
$\eta^{\zeta}_{0},\ldots,\eta^{\zeta}_{k_{\zeta,1}-1}\in I_{\beta}$ and
Boolean terms $\sigma^{\zeta}_{\ell}$ (for $\ell<m)$ such that

$c^{\zeta}_{\ell}=\sigma^{\zeta}_{\ell}\bigl{(}b^{\zeta}_{0},\ldots,b^{\zeta}_{k_{\zeta,0}-1},x_{\eta^{\zeta}_{0}},\ldots,x_{\eta^{\zeta}_{k(\zeta,1)}}\bigr{)}.$

Without loss of generality
$\langle\eta^{\zeta}_{\ell}:\ell<k_{\zeta,1}\rangle$ is a $\Delta$-system.

Without loss of generality $k_{\zeta,0}=k_{0}$, $k_{\zeta,1}=k_{1}$,
$\sigma^{\zeta}_{\ell}=\sigma_{\ell}$ and
${\ell\kern-1.07639ptg}(\eta^{\zeta}_{\ell})=m_{\ell}\leq\omega$ for every
$\zeta<\mu^{+}$.

Also, there is $k_{\zeta,2}<\omega$ such that:

  1. (II)

    1. $(\alpha)$

${\ell\kern-1.07639ptg}(\eta^{\zeta}_{\ell})<\omega\Rightarrow{\ell\kern-1.07639ptg}(\eta^{\zeta}_{\ell})<k_{\zeta,2}$

    2. $(\beta)$

$\eta^{\zeta}_{\ell_{1}}\neq\eta^{\zeta}_{\ell_{2}}\Rightarrow\eta^{\zeta}_{\ell_{1}}\restriction
k_{\zeta,2}\neq\eta^{\zeta}_{\ell_{2}}\restriction k_{\zeta,2}$

    3. $(\gamma)$

$2n+2<k_{\zeta,2}$.

Without loss of generality $\bigwedge\limits_{\zeta}k_{\zeta,2}=k_{2}$.

Without loss of generality the statement $(*)$ (with $k_{2}$ here for $k^{*}$
there and is $>n$) from the proof of 3.5 holds (essentially being a
$\Delta$-system), i.e.

  1. $(*)$

If $i\neq j<\lambda$, $k\leq k_{2}+1$, and
$m^{\prime},m^{\prime\prime}<m_{1}$, $\ell^{\prime},\ell^{\prime\prime}<h(k)$
and
$\mathrm{Res}^{\ell^{\prime}}_{k}(\eta_{m^{\prime},i})=\mathrm{Res}^{\ell^{\prime\prime}}_{k}(\eta_{m^{\prime\prime},j})$,
then for every $\alpha,\beta<\lambda$ we have:

$\mathrm{Res}^{\ell^{\prime}}_{k}(\eta_{m^{\prime},\alpha})=\mathrm{Res}^{\ell^{\prime\prime}}_{k}(\eta_{m^{\prime\prime},\alpha})=\mathrm{Res}^{\ell^{\prime}}_{k}(\eta_{m^{\prime},\beta})=\mathrm{Res}^{\ell^{\prime\prime}}_{k}(\eta_{m^{\prime\prime},\beta}).$

Let $\bar{b}^{\zeta}=\langle b^{\zeta}_{\ell}:\ell<k_{0}\rangle$. By the
induction hypothesis, without loss of generality
$\big{\langle}\bar{b}^{\zeta}{\char 94\relax}\langle a^{\beta}_{\ell}:\ell\leq
k_{2}\rangle:\zeta<\mu^{+}\big{\rangle}$ is $(\mathrm{qf},n)$-indiscernible
and without loss of generality the sequence
$\big{\langle}\langle\eta^{\zeta}_{\ell}\restriction(k_{2}+1):\ell<k_{1}\rangle:\zeta<\mu^{+}\rangle$
is indiscernible (sequence of finite sequences of ordinals).

To finish the proof of 3.6 it suffices to observe 3.8 below. ∎

######  Observation 3.8.

If $\mathbf{B}^{*}=\mathrm{ba}_{2}[\mathbf{B},\bar{a},I]$, $n^{*}<\omega$,
$I^{0}=\\{\eta\in I:{\ell\kern-1.07639ptg}(\eta)\leq n^{*}\\}$, $Z\subseteq
P^{I}_{\omega}$, and for every $\nu\in Z$ and $n\geq n^{*}$ the set

$\\{\nu^{\prime}\restriction(n+1):\nu^{\prime}\in Z,\ \nu^{\prime}\restriction
n=\nu\restriction n\\}$

has $<\lfloor h(n)/2\rfloor$ elements, then $\\{x_{\eta}:\eta\in Z\\}$ is
independent in $\mathbf{B}^{*}$ over
$\mathbf{B}^{0}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=\mathrm{ba}_{2}[\mathbf{B},\bar{a},I^{0}]$,
except the equations $c^{+}_{\eta}\leq x_{\eta}\wedge c^{-}_{\eta}\cap
x_{\eta}=0$ for $\eta\in Z$, where

$\begin{array}[]{lcr}c^{+}_{\eta}&=&\bigcup\Big{\\{}a_{2n}-\bigcup\\{x_{\mathrm{Res}^{2\ell}_{2n}(\eta)}-x_{\mathrm{Res}^{2\ell+1}_{2n}(\eta)}:\ell<h(n)/2\\}:2n<n^{*}\Big{\\}},\\\
c^{-}_{\eta}&=&\bigcup\Big{\\{}a_{2n+1}-\bigcup\\{x_{\mathrm{Res}^{2\ell}_{2n+1}(\eta)}-x_{\mathrm{Res}^{2\ell+1}_{2n+1}(\eta)}:2\ell+1<h(2n+1)\\}:\\\
&&2n+1<n^{*}\Big{\\}}.\end{array}$

(Note: $\eta_{1}\restriction n^{*}=\eta_{2}\restriction n^{*}\ \Rightarrow\
(c^{+}_{\eta_{1}},c^{-}_{\eta_{1}})=(c^{+}_{\eta_{2}},c^{-}_{\eta_{2}})$.)

###### Proof..

Let $\mathbf{f}_{0}$ be any function with domain $X=\\{x_{\eta}:\eta\in Z\\}$
such that
$\mathbf{f}_{0}(x_{\eta})\in\\{c^{+}_{\eta},1_{\mathbf{B}}-c^{-}_{\eta}\\}$,
and let

$J^{1}=I^{0}\cup X\cup\\{\mathrm{Res}^{\ell}_{n}(\nu):\nu\in Z,\ \ell<h(n),\
n<\omega\\}.$

Clearly, by 3.4(2) it suffices to find a homomorphism from
$\mathbf{B}_{1}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=\mathrm{ba}[\mathbf{B},\bar{a},J^{1}]$
into $\mathbf{B}^{0}$ extending
$\mathrm{id}_{\mathbf{B}^{0}}\cup\mathbf{f}_{0}$. For this it suffices to find
a mapping $\mathbf{f}$ from $\mathbf{B}\cup\\{x_{\eta}:\eta\in J^{1}\\}$ into
$\mathbf{B}^{0}$ extending $\mathrm{id}_{\mathbf{B}^{0}},\mathbf{f}_{0}$, and
$\mathrm{id}_{\\{x_{\eta}:\eta\in I^{0}\\}}$, and preserving the equations
defining $\mathrm{ba}[\mathbf{B},\bar{a},J^{1}]$. As
$\mathbf{f}\restriction\mathbf{B}$, $\mathbf{f}\restriction\\{x_{\eta}\in
I:{\ell\kern-1.07639ptg}(\eta)\leq n^{*}\\}$, and
$\mathbf{f}\restriction\\{x_{(\eta,\varrho)}:\eta\in Z\\}$ are defined, and

$J^{1}=\bigcup\limits_{n\in[n^{*},\omega)}\\{x_{\eta}:\eta\in Z_{n}\\}\cup
X\cup Z$

where $Z_{n}=\\{\eta\in J^{1}:{\ell\kern-1.07639ptg}(\eta)=n+1\\}$, it will
suffice to choose $\mathbf{f}\restriction\\{x_{\eta}:\eta\in Z_{n}\\}$ for
each $n\in[n^{*},\omega)$ to finish the definition of $\mathbf{f}$.

Let $Y_{n}=\\{\nu\restriction n:\nu\in Z_{n}\\}$, and for $\eta\in Y_{n}$ let
$X_{n,\eta}=\\{\nu\in Z_{n}:\nu\restriction n=\eta\\}$. Clearly $\langle
X_{n,\eta}:\eta\in Y_{n}\rangle$ is a partition of $Z_{n}$.

For $\eta\in Y_{n}$ let $\mathscr{P}_{n,\eta}=\\{\nu\restriction(n+1):\nu\in
Z,\ \nu\restriction n=\eta\\}$ and

$\mathcal{S}_{n,\eta}=\big{\\{}(\rho\restriction n){\char
94\relax}\langle\rho(n)(\ell)\rangle:\ell<k(n)\big{\\}}.$

By the assumption on $Z$, for every $\eta\in Y_{n}$ the set
$\mathscr{P}_{n,\eta}$ has $<h(n)/2$ elements. Now:

  1. $(*)$

For $\eta\in Y_{n}$ there is a function
$f_{\eta}:\mathcal{S}_{n,\eta}\to\\{0_{\mathbf{B}^{*}},1_{\mathbf{B}^{*}}\\}$
such that if $\nu\in\mathscr{P}_{n,\eta}$ is equal to $\eta{\char
94\relax}\langle\alpha_{0},\ldots,\alpha_{h(n)-1}\rangle$ then for some
$\ell<(h(n)-1)/4$ we have $f_{\eta}(x_{\eta{\char
94\relax}\langle\alpha_{2\ell}\rangle})=1_{\mathbf{B}^{*}}$ and
$f_{\eta}(x_{\eta{\char
94\relax}\langle\alpha_{2\ell+1}\rangle})=0_{\mathbf{B}^{*}}$.

[Why is this possible? By finite cardinality considerations.]

Now define $\mathbf{f}\restriction Z_{n}$ as follows: if $\nu\in Z_{n}$ then
$\nu\in\mathscr{P}_{n,\eta}$ for some $\eta\in Y_{n}$, and so we let
$\mathbf{f}(x_{\nu})=f_{\eta}(x_{\nu})$.

Now check. ∎

######  Discussion 3.9.

1) In the proof of clause $(B)^{+}$, the successor case we use the fact that
$h(n)$ converges to $\infty$, as when the level increases we need more
$\eta\in P^{I_{\beta}}_{\omega}$ to see non-freeness.

2) The proof there for limit $\alpha$ uses just “$\langle\mathbf{B}_{i}:i\leq
2^{\aleph_{0}}\rangle$ is $\lessdot$-increasing continuous with projections”
(i.e. 3.4(5)), and the induction hypothesis.

3) We can vary the construction in some ways. We can demand that each
$\bar{a}^{\alpha}$ is a maximal antichain — no difference so far. We may like
to use $\langle I_{\beta}:\beta<2^{\aleph_{0}}\rangle$ such that $I_{\beta}$
is not super unembeddable into $\sum\limits_{\gamma\neq\beta}I_{\gamma}$. We
can construct our Boolean algebra to be monorigid (i.e., with no one-to-one
endomorphism), and even get $2^{2^{\aleph_{0}}}$ such Boolean algebras, no one
embeddable to another: even restricting to non-zero elements, even not
embeddable into the completion of another. To carry this out we need the
following for $\lambda=2^{\aleph_{0}}$: there is $\bar{I}=\langle
I_{\alpha}:\alpha<\lambda\rangle$ exemplifying that
$K^{\omega}_{\mathrm{tr}(h)}$ has the full
$(\lambda,\lambda,\aleph_{1},\aleph_{1})$-super bigness property, such that
for at least one $\beta$, $I_{\beta}$ satisfies $(*)_{I_{\beta}}$ from the
beginning of the proof of 3.6. Now such a $\bar{I}$ does exist (with
$(*)_{I_{\beta}}$ for every $\beta$); this may be elaborated elsewhere.

4) Of course the proof works for $\mu=2^{\aleph_{0}}$.

5) We can separate some parts of the proof to independent claims. We can ask
for “$\mathbf{B}$ has length $\mu$, but no chain of cardinality $\mu$” (i.e.
the supremum is not obtained) for $\mu$ limit. It is natural to demand
$\mathrm{cf}(\mu)>\aleph_{0}$. Next, we address this.

######  Claim 3.10.

1) Assume $2^{\aleph_{0}}\geq\mu$ and
$\aleph_{0}<\kappa=\mathrm{cf}(\mu)<\mu$. Then there is a Boolean algebra
$\mathbf{B}$ such that $|\mathbf{B}|=2^{\aleph_{0}}$, $\mathbf{B}$ has no
homomorphic image of cardinality $\in[\aleph_{0},2^{\aleph_{0}})$, and
$\mathrm{length(\mathbf{B})}=\mu$, but the supremum is not obtained (i.e.
$\mathrm{length}^{+}(\mathbf{B})=\mu$ and every infinite homomorphic image
$\mathbf{B}^{\prime}$ of $\mathbf{B}$ has length $\geq\mu$).

2) Similarly, but slightly modifying the assumption to
$\aleph_{0}<\kappa=\mathrm{cf}(\mu)=\mu$.

###### Proof..

Like 3.6.

1) Let $\mu=\sum\limits_{i<\kappa}\mu_{i}$ with
$\langle\mu_{i}:i<\kappa\rangle$ be increasing continuous and
$\kappa<\mu_{i}<\mu$. For $\varepsilon<\kappa$, let $\mathbf{B}^{\varepsilon}$
be a subalgebra of $\mathcal{P}(\omega)$ of cardinality $\mu_{\varepsilon}$
and length $\mu_{\varepsilon}$. Let $\langle
I_{\alpha}:\alpha<2^{\aleph_{0}}\times\kappa\rangle$ be as in the proof of
3.6. We define $\mathbf{B}_{\alpha}$ (for $\alpha\leq
2^{\aleph_{0}}\times\kappa$) similarly to the proof of 3.6. Specifically:

$\begin{array}[]{l}\mathbf{B}_{0}=\mathbf{B}^{0},\text{ the trivial Boolean
algebra,}\\\ \mathbf{B}_{2^{\aleph_{0}}\times\varepsilon+1}\mbox{ is the free
product
}\mathbf{B}_{2^{\aleph_{0}}\times\varepsilon}*\mathbf{B}^{\varepsilon},\\\
\mathbf{B}_{\alpha}\mbox{ is increasing continuous in }\alpha,\\\
\mathbf{B}_{\alpha+1}=\mathrm{ba}[\mathbf{B}_{\alpha},\bar{a}^{\alpha},I_{\alpha}]\mbox{
for }\alpha<2^{\aleph_{0}}\times\kappa,\
\alpha\notin\bigl{\\{}2^{\aleph_{0}}\times\varepsilon:\varepsilon<\kappa\bigr{\\}},\end{array}$

where $\bar{a}_{\alpha}$ is $\langle a_{\alpha,n}:n<\omega\rangle$,
$a_{\alpha,n}\in\mathbf{B}_{\alpha},a_{\alpha,n}>0$, $n_{1}\neq
n_{2}\Rightarrow a_{\alpha,n_{1}}\cap a_{\alpha,n_{2}}=0$. The choice of the
$\bar{a}_{\alpha}$-s (i.e. the bookkeeping) is as in the proof of 3.6 above.

So, by the proof of 3.6:

  1. $(*)$

If $0<\alpha<2^{\aleph_{0}}\times\kappa$ then

    1. $(\alpha)$

$\mathbf{B}_{\alpha}$ satisfies the strong $\lambda$-cc if
$\lambda=\mathrm{cf}(\lambda)>\aleph_{0}$.

    2. $(\beta)$

$\mathbf{B}_{1+\alpha}$ has length
$\mu_{0}+\sum\bigl{\\{}\mu_{\varepsilon}:2^{\aleph_{0}}\times\varepsilon<\alpha\bigr{\\}}<\mu$.

    3. $(\gamma)$

If $\alpha=2^{\aleph_{0}}\times\varepsilon$ with $\varepsilon$ a successor
ordinal, then $\mathbf{B}_{\alpha}$ has no homomorphic image of cardinality
$\in[\aleph_{0},2^{\aleph_{0}})$.

    4. $(\delta)$

If $\alpha<\beta\leq 2^{\aleph_{0}}\times\kappa$ and
$b\in\mathbf{B}_{\beta}\setminus\\{0_{\mathbf{B}_{\beta}}\\}$ then for some
$a\in\mathbf{B}_{\alpha}$ we have $\mathbf{B}_{\beta}\models b\leq a$ and if
$\mathbf{B}_{\alpha}\models 0<a^{\prime}\leq a$ then $a^{\prime}\cap
b>0_{\mathbf{B}_{\beta}}$.

[Note: for clause $(\beta)$ we use the proof of (B)+ of 3.6. For
$\alpha=2^{\aleph_{0}}\times\varepsilon+1$ for clause $(\delta)$ we have a new
clause, but easy one].

It follows that

  1. $(**)$

$\mathbf{B}=\mathbf{B}_{2^{\aleph_{0}}\times\kappa}$ has length $\mu$.

Now we just need to show

  1. $(***)$

For $J\subseteq\mathbf{B}$ (with $|J|=\mu$) a chain we get a contradiction.

Let
$\mathbf{B}^{*}_{\varepsilon}=\mathbf{B}_{2^{\aleph_{0}}\times\varepsilon}$.
Let $c_{\alpha}\in J$ (for $\alpha<\mu$) be pairwise distinct.

By clause $(*)(\delta)$, for each $\varepsilon<\kappa$ and
$\alpha<\mu^{+}_{\varepsilon}$ we can find
$b^{\varepsilon}_{\alpha}\in\mathbf{B}^{*}_{\varepsilon}$ such that:

  1. (a)

$c_{\alpha}\leq b^{\varepsilon}_{\alpha}$

  2. (b)

$0<x\leq b^{\varepsilon}_{\alpha}\wedge x\in\mathbf{B}^{*}_{\varepsilon}\
\Rightarrow\ x\cap c_{\alpha}\neq 0$

Note:

  1. (c)

$b^{\varepsilon}_{\alpha}$ is unique, and

  2. (d)

$c_{\alpha}\leq c_{\beta}\ \Rightarrow\ b^{\varepsilon}_{\alpha}\leq
b^{\varepsilon}_{\beta}$.

As $\mathbf{B}^{*}_{\varepsilon}$ has length $\leq\mu_{\varepsilon}$ and $J$
is a chain, necessarily for some
$Y_{\varepsilon}\subseteq\mu^{+}_{\varepsilon}$ with
$|Y_{\varepsilon}|=\mu^{+}_{\varepsilon}$ we have

  1. (e)

$b^{\varepsilon}_{\alpha}=b^{\varepsilon}$ for $\alpha\in Y_{\varepsilon}$.

We can apply clause $(*)(\delta)$ to $-c_{\alpha}$ (for $\alpha\in
Y_{\varepsilon}$ and $\mathbf{B}^{*}_{\varepsilon}$, and possibly shrinking
$Y_{\varepsilon}$) to get
$a^{\varepsilon}_{\alpha}\in\mathbf{B}^{*}_{\varepsilon}$ such that:

  1. (f)

$(-c_{\alpha})\leq a^{\varepsilon}_{\alpha}$ and $0<x\leq
a^{\varepsilon}_{\alpha}\wedge x\in\mathbf{B}^{*}_{\varepsilon}\Rightarrow
x\cap(-c_{\alpha})\neq 0$.

As above, without loss of generality, shrinking $Y_{\varepsilon}$ further we
get

  1. (g)

$a^{\varepsilon}_{\alpha}=a^{\varepsilon}$ for $\alpha\in Y_{\varepsilon}$.

As the length of $\mathbf{B}^{*}_{\varepsilon}$ is
$\leq\mu_{\varepsilon}<\mu_{\varepsilon}^{+}=|Y_{\varepsilon}|$, for some
$\alpha\in Y_{\varepsilon}$ we have
$c_{\alpha}\notin\mathbf{B}^{*}_{\varepsilon}$; as

$a^{\varepsilon}_{\alpha}\geq-c_{\alpha},\quad b^{\varepsilon}_{\alpha}\geq
c_{\alpha},\quad
b^{\varepsilon}_{\alpha}\in\mathbf{B}^{*}_{\varepsilon}\quad\mbox{and}\quad
a^{\varepsilon}_{\alpha}\in\mathbf{B}^{*}_{\varepsilon},$

necessarily $a^{\varepsilon}_{\alpha}\cap b^{\varepsilon}_{\alpha}\neq 0$.

Hence:

  1. (h)

$b^{\varepsilon}\cap a^{\varepsilon}\neq 0$.

Let
$g(\varepsilon)=\min\\{\zeta<\varepsilon:a^{\varepsilon},b^{\varepsilon}\in\mathbf{B}^{*}_{\zeta}\\}$,
so for limit $\varepsilon$, $g(\varepsilon)<\varepsilon$. Hence on some
stationary $S\subseteq\kappa$ and $\zeta(*)<\kappa$ the function
$g\restriction S$ is constantly $\zeta(*)$, and without loss of generality

$\zeta<\varepsilon\in S\Rightarrow\big{|}\\{\alpha\in
Y_{\zeta}:c_{\alpha}\in\mathbf{B}^{*}_{\varepsilon}\\}\big{|}=\mu^{+}_{\zeta}.$

As $\mathbf{B}$ satisfies the ccc we can find
$\varepsilon_{1}<\varepsilon_{2}$ in $S$ such that

$b^{\varepsilon_{1}}\cap a^{\varepsilon_{1}}\cap b^{\varepsilon_{2}}\cap
a^{\varepsilon_{2}}\neq 0.$

Choose $\alpha\in Y_{\varepsilon_{1}}$ such that
$c_{\alpha}\in\mathbf{B}_{\varepsilon_{1}}$ and $\beta\in
Y_{\varepsilon_{2}}$. Now $\\{c_{\alpha},c_{\beta}\\}$ is independent: a
contradiction.

2) Similarly. ∎

######  Remark 3.11.

We may further ask: is the restriction “$\mathrm{cf}(\mu)>\aleph_{0}$” in
(3.10) necessary?

######  Observation 3.12.

Assume that the infinite Boolean algebra $\mathbf{B}$ has the length $\mu$,
$\mathrm{cf}(\mu)=\aleph_{0}$. Then the length is obtained.

###### Proof..

Let
${\dot{\mathcal{I}}}=\\{b\in\mathbf{B}:\mathrm{length}(\mathbf{B}\restriction
b)<\mu\\}$.

Easily

$b_{1}\leq b_{2}\wedge b_{2}\in{\dot{\mathcal{I}}}\Rightarrow
b_{1}\in{\dot{\mathcal{I}}}.$

Also clearly ${\dot{\mathcal{I}}}$ is closed under unions. [Why? If
$b_{1},b_{2}\in{\dot{\mathcal{I}}}$, $b=b_{1}\cup
b_{2}\notin{\dot{\mathcal{I}}}$ then there is a chain $\langle c_{t}:t\in
J\rangle$, $J$ a linear order of cardinality $\mu$, $(s<_{J}t\ \Rightarrow\
c_{s}<_{\mathbf{B}}c_{t})$ and $c_{t}\leq b$.

Let

$\mathbf{E}_{l}=\\{(t,s)\in J\times J:c_{t}\cap b_{l}=c_{s}\cap b_{l}\\}.$

Then $\mathbf{E}_{l}$ is a convex equivalence relation on $J$; if
$|J/\mathbf{E}_{l}|=\mu$ then $\\{c_{t}\cap b_{l}:t\in J\\}$ exemplifies
$b_{l}\notin{\dot{\mathcal{I}}}$, a contradiction. So
$|J/\mathbf{E}_{l}|<\mu$. Hence $\mathbf{E}=\mathbf{E}_{1}\cap\mathbf{E}_{2}$
is a convex equivalence relation with
$\leq|J/\mathbf{E}_{1}|\times|J/\mathbf{E}_{2}|<\mu$ classes, but as
$b=b_{1}\cup b_{2}$ it is the equality.]

If $\mathbf{B}/{\dot{\mathcal{I}}}$ is infinite then we can find $\langle
a_{n}/{\dot{\mathcal{I}}}:n<\omega\rangle$ pairwise disjoint non-zero. Now
$b_{n}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=a_{n}-\bigcup\limits_{\ell<n}a_{\ell}$
are pairwise disjoint members of $\mathbf{B}$ not in ${\dot{\mathcal{I}}}$.
Let $\mu=\sum\limits_{n<\omega}\mu_{n}$, $\mu_{n}<\mu$. Let $\langle
c^{n}_{t}:t\in J_{n}\rangle$ be an increasing chain in $\mathbf{B}\restriction
b_{n}$, $|J_{n}|\geq\mu_{n}$ (note that we can invert $J_{n}$). Let
$J=\sum\limits_{n<\omega}J_{n}$ (without loss of generality, $n<m\
\Rightarrow\ J_{n}\cap J_{m}=\varnothing$) and for $t\in J_{n}$ let
$c^{*}_{t}=b_{0}\cup\ldots\cup b_{n-1}\cup c^{n}_{t}$. Now $\langle
c^{*}_{t}:t\in J\rangle$ exemplifies that the length is obtained. So
$\mathbf{B}/{\dot{\mathcal{I}}}$ is finite, so without loss of generality
${\dot{\mathcal{I}}}$ is a maximal ideal. Try to choose
$a_{n}\in{\dot{\mathcal{I}}}$ satisfying $\bigwedge\limits_{\ell<n}a_{n}\cap
a_{\ell}=0$ such that $\mathrm{length}(\mathbf{B}\restriction a_{n})>\mu_{n}$.
If we succeed, then we may repeat the proof for the case
“$\mathbf{B}/{\dot{\mathcal{I}}}$ is infinite,” hence we necessarily fail.
Hence for some $n$ (replacing $\mathbf{B}$ by
$\mathbf{B}\restriction-(a_{0}\cup\ldots\cup a_{n-1})$) we have

$b\in{\dot{\mathcal{I}}}\ \Rightarrow\ \mathrm{length}(\mathbf{B}\restriction
b)\leq\mu_{n}.$

Let $J\subseteq\mathbf{B}$ be linearly ordered, $|J|>\mu^{+}_{n}$. Possibly
shrinking $J$, without loss of generality $J\subseteq{\dot{\mathcal{I}}}\vee
J\subseteq\mathbf{B}\setminus{\dot{\mathcal{I}}}$. As we can replace $J$ by
$\\{1_{\mathbf{B}}-b:b\in J\\}$ without loss of generality
$J\subseteq{\dot{\mathcal{I}}}$, so for some $b\in J$ we have $|\\{c\in
J:c\leq b\\}|\geq\mu_{n}^{+}$, and hence
$\mathrm{length}(\mathbf{B}\restriction b)\geq\mu_{n}^{+}$, a contradiction. ∎

######  Remark 3.13.

We may wonder if we can replace $\aleph_{0}$ in 3.10 by another cardinals.
Most natural are $\kappa$ strong limit of cofinality $\omega$.

######  Claim 3.14.

Assume $\kappa\leq\mu<2^{\kappa}$, $\kappa$ strong limit and
$\mathrm{cf}(\kappa)=\aleph_{0}$. Then there is a Boolean algebra $\mathbf{B}$
such that:

  1. $(\alpha)$

$|\mathbf{B}|=2^{\kappa}$

  2. $(\beta)$

$\mathbf{B}\models$ ccc

  3. $(\gamma)$

$\mathbf{B}$ has length $\mu$ (and satisfies clause (B)+ of 3.6)

  4. $(\delta)$

$\mathbf{B}$ has no homomorphic image $\mathbf{B}^{\prime}$ with
$|\mathbf{B}^{\prime}|\in[\kappa,2^{\kappa})$.

###### Proof..

Let $h\in{}^{\omega}\omega$ be $h(n)=2(n+1)$. Let
$\mathbf{B}_{0}\subseteq\mathcal{P}(\kappa)$ have cardinality $\mu$ and length
$\mu$,

$\displaystyle I^{0}_{\alpha}=\big{\\{}\eta:$ $\displaystyle\ \eta\mbox{ is an
$\omega$-sequence, $\eta(n)$ is an increasing }\ $ $\displaystyle\mbox{
sequence of ordinals $<2^{\kappa}$ of length }h(n)\big{\\}},$

and

$I_{\alpha}=I^{0}_{\alpha}\cup\\{\mathrm{Res}^{\ell}_{n}(\eta):n<\omega,\
\ell<h(n),\ \eta\in I^{0}_{\eta}\\}$

so $|I_{\alpha}|=2^{\kappa}$. Let
$\mathbf{B}_{\alpha+1}=\mathrm{ba}[\mathbf{B}_{\alpha},\bar{a}_{\alpha},I_{\alpha}]$,
$\mathbf{B}_{\alpha}$ increasing continuous for $\alpha\leq 2^{\kappa}$.
(Again, $\bar{a}_{\alpha}$ is an $\omega$-sequence of pairwise disjoint non-
zero elements of $\mathbf{B}_{\alpha}$ such that each sequence appears
$2^{\kappa}$ times.)

Again, for $\alpha<\beta$, $\mathbf{B}_{\alpha}\lessdot\mathbf{B}_{\beta}$
(and even the conclusion of 3.4(5) holds). The proof that
$\mathbf{B}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=\mathbf{B}_{(2^{\kappa})}$
satisfies 3.14$(\alpha)$, $(\beta)$, $(\gamma)$ is as in the proof of 3.6.

For $(\delta)$ we need 3.15 below. ∎

######  Observation 3.15.

Assume that $\kappa$ is a strong limit cardinal of countable cofinality. 1) If
$\mathbf{B}^{\prime}$ is a Boolean Algebra of cardinality $\geq\kappa$ but
$<2^{\kappa}$ then:

  1. $(a)$

There are pairwise disjoint non-zero $b_{n}$ (for $n<\omega$) in
$\mathbf{B}^{\prime}$ such that

    1. $(*)$

for no $c\in\mathbf{B}^{\prime}$, $\bigwedge\limits_{n<\omega}(b_{2n}\leq
c)\wedge\bigwedge\limits_{n<\omega}(b_{2n+1}\cap c=0)$.

2) For a Boolean algebra $\mathbf{B}^{\prime}$, a sufficient condition for
$\mathbf{B}^{\prime}$ to satisfy $(a)$ (i.e., the existence of a sequence
$\langle b_{n}:n<\omega\rangle$ of pairwise disjoint elements of
$\mathbf{B}^{\prime}$ satisfying $(*)$ above) is:

  1. $(b)$

$\mathbf{B}^{\prime}$ has cardinality $<2^{\kappa}$ and there are
$b_{n}\in\mathbf{B}^{\prime}$ such that $\bigwedge\limits_{n<m}b_{n}\cap
b_{m}=0$ and $\kappa=\liminf\limits_{n}|\mathbf{B}\restriction b_{n}|$.

We first prove that 3.15 suffices (for finishing the proof of 3.14). Toward
contradiction assume that $\mathbf{B}^{\prime}$ is a Boolean algebra of
cardinality $<2^{\kappa}$ but $\geq\kappa$, and $\mathbf{B}^{\prime}$ is a
homomorphic image of $\mathbf{B}$. If clause (a) is satisfied by
$\mathbf{B}^{\prime}$, then the proof is very similar to the earlier proof of
3.6: for a homomorphism $\mathbf{f}:\mathbf{B}\to\mathbf{B}^{\prime}$ from
$\mathbf{B}$ onto $\mathbf{B}^{\prime}$ we can find pairwise disjoint
$a_{n}\in\mathbf{B}$ (for $n<\omega$) such that $f(a_{n})=b_{n}$. So, for some
$\alpha$ we have $\bar{a}_{\alpha}=\langle a_{n}:n<\omega\rangle$, and we
repeat the relevant part of 3.6. Using clauses (b),(c) of Definition 3.3 we
get a contradiction. We are left with proving 3.15. First, the second part.

###### Proof..

Proof of Observation 3.15(2):

We can find $\bar{c}^{\zeta}=\langle c^{\zeta}_{n}:n<\omega\rangle$,
$c^{\zeta}_{n}\in\mathbf{B}^{\prime}$, $c^{\zeta}_{n}\leq b_{n}$ for
$\zeta<2^{\kappa}$ such that the sequences $\langle
c^{\zeta}_{n}:n<\omega\rangle$ are pairwise distinct for $\zeta<2^{\kappa}$.
For each $\zeta$ let $b^{\zeta}_{2n}=c^{\zeta}_{n}$,
$b^{\zeta}_{2n+1}=b_{n}-c^{\zeta}_{n}$, so if clause (a) fails then for every
$\zeta<2^{\kappa}$ there is $y_{\zeta}\in\mathbf{B}^{\prime}$ such that for
every $n<\omega$ we have

$b^{\zeta}_{2n}\leq y_{\zeta},\quad b^{\zeta}_{2n+1}\cap y_{\zeta}=0.$

So $\bigwedge\limits_{n}y_{\zeta}\cap b_{n}=c^{\zeta}_{n}$ and hence
$\zeta<\xi<2^{\kappa}\Rightarrow\ y_{\zeta}\neq y_{\xi}$, which contradicts
$|\mathbf{B}^{\prime}|<2^{\kappa}$.

Proof of Observation 3.15(1):

Assume that the conclusion fails. For a cardinal $\mu$, let

${\dot{\mathcal{I}}}_{\mu}={\dot{\mathcal{I}}}_{\mu}[\mathbf{B}^{\prime}]\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=\\{b\in\mathbf{B}^{\prime}:\mathbf{B}^{\prime}\restriction
b\mbox{ has cardinality}<\mu\\}.$

Clearly it is an ideal of $\mathbf{B}^{\prime}$ increasing with $\mu$ and
$1_{\mathbf{B}^{\prime}}\in{\dot{\mathcal{I}}}_{\mu}\Leftrightarrow\mu>|\mathbf{B}|$.
If $\mathbf{B}^{\prime}/{\dot{\mathcal{I}}}_{\kappa}[\mathbf{B}^{\prime}]$ is
infinite then we can easily get condition (B) of part (2), and we are done. If
it is finite, but
${\dot{\mathcal{I}}}_{\mu}[\mathbf{B}]\neq{\dot{\mathcal{I}}}_{\kappa}[\mathbf{B}^{\prime}]$
for every $\mu<\kappa$, then let $\kappa=\sum\limits_{n<\omega}\mu_{n}$,
$\mu_{n}<\mu_{n+1}$, and choose
$b_{n}\in{\dot{\mathcal{I}}}_{\kappa}[\mathbf{B}^{\prime}]\setminus{\dot{\mathcal{I}}}_{\mu_{n}}[\mathbf{B}^{\prime}]$.
But
${\dot{\mathcal{I}}}_{\kappa}[\mathbf{B}^{\prime}]=\bigcup\limits_{\mu<\kappa}{\dot{\mathcal{I}}}_{\mu}[\mathbf{B}^{\prime}]$,
so
$\bigwedge\limits_{n}\bigvee\limits_{m}b_{n}\in{\dot{\mathcal{I}}}_{\mu_{m}}[\mathbf{B}^{\prime}]$.
So without loss of generality
$b_{n}\in{\dot{\mathcal{I}}}_{\mu_{n+1}}[\mathbf{B}^{\prime}]\setminus{\dot{\mathcal{I}}}_{\mu_{n}}[\mathbf{B}^{\prime}]$
and hence $\langle b_{n}-\bigcup\limits_{l<n}b_{l}:n<\omega\rangle$ are as
required. We are left with the case that for some $\mu(*)<\kappa$,

${\dot{\mathcal{I}}}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}={\dot{\mathcal{I}}}_{\mu(*)}[\mathbf{B}^{\prime}]={\dot{\mathcal{I}}}_{\kappa}[\mathbf{B}^{\prime}]$

and without loss of generality
${\dot{\mathcal{I}}}={\dot{\mathcal{I}}}_{\mu(*)}[\mathbf{B}^{\prime}]$ is a
maximal ideal.

Without loss of generality $2^{\mu(*)}<\mu_{n}<\mu_{n+1}$ for $n<\omega$. Let
$b_{i}\in{\dot{\mathcal{I}}}$ (for $i<\kappa$) be distinct (these exist as
$|\mathbf{B}^{\prime}|\geq\kappa$ and ${\dot{\mathcal{I}}}$ is a maximal ideal
of $\mathbf{B}^{\prime}$). By the proof of Erdős–Tarski theorem, without loss
of generality $\langle b_{i}:i<\kappa\rangle$ are non-zero pairwise disjoint.

[Why? For example, apply the $\Delta$-system lemma to

$\big{\\{}\\{x:x\leq b_{i}\\}:i<(2^{\mu_{n}})^{+}\big{\\}},$

and get $Y_{n}\subseteq(2^{\mu_{n}})^{+}$ of cardinality $(2^{\mu_{n}})^{+}$
and a set $A_{n}$ of cardinality $\leq 2^{\mu(*)}$ such that

$i,j\in Y_{n}\wedge i\neq j\ \Rightarrow\ \\{x:x\leq b_{i}\\}\cap\\{x:x\leq
b_{j}\\}=A_{n}.$

So $|A_{n}|\leq\mu(*)$. Pick $Y^{\prime}_{n}\subseteq Y_{n}$ of cardinality
$(2^{\mu_{n}})^{+}$ such that

$i,j\in Y_{n}^{\prime}\wedge i\neq j\ \Rightarrow\ \\{x:x\leq
b_{i}\\}\cap\bigcup_{m<n}A_{m}=\\{x:x\leq
b_{j}\\}\cap\bigcup_{m<n}A^{\prime}_{m},$

where $A^{\prime}_{n}=\big{\\{}x:(\exists i\in Y_{n}^{\prime})[x\leq
b_{i}]\big{\\}}$. Let $i(n)=\min(Y_{n}^{\prime})$. Then

$X_{n}=\\{x_{i}-x_{i(n)}:i\in Y_{n},\
i>i(n)\\}\subseteq\mathbf{B}\setminus\\{0\\}$

is an antichain, and $\bigcup\limits_{n}X_{n}$ is as required.]

Let

$\mathscr{P}_{0}=\big{\\{}Y\in[\kappa]^{\aleph_{0}}:\mbox{there is
}b\in{\dot{\mathcal{I}}}\mbox{ such that }(\forall i\in Y)[b_{i}\leq
b]\big{\\}}.$

This is a subset of $[\kappa]^{\aleph_{0}}$ of cardinality
$\leq|{\dot{\mathcal{I}}}|\cdot\mu(*)^{\aleph_{0}}\leq|\mathbf{B}^{\prime}|+\kappa=|\mathbf{B}^{\prime}|$,
but $[\kappa]^{\aleph_{0}}=2^{\kappa}>|\mathbf{B}^{\prime}|$, so there is
$Y_{0}\in[\kappa]^{\aleph_{0}}\setminus\mathscr{P}_{0}$.

Let

$\mathscr{P}_{1}=\big{\\{}Y\in[\kappa]^{\aleph_{0}}:Y\subseteq\kappa\setminus
Y_{0}\mbox{ and }(\exists b\in{\dot{\mathcal{I}}})(\forall i\in Y)[b_{i}\leq
b]\big{\\}}.$

By cardinality considerations as above there is
$Y_{1}\in[\kappa]^{\aleph_{0}}\setminus\mathscr{P}_{1}$ disjoint to $Y_{0}$.
By assumption above (i.e., clause (a) fails) there is
$b\in\mathbf{B}^{\prime}$ such that $\bigwedge\limits_{i\in Y_{0}}b_{i}\leq b$
and $\bigwedge\limits_{i\in Y_{1}}b_{i}\leq(1-b)$. If
$b\in{\dot{\mathcal{I}}}$ we get contradiction to the choice of $Y_{0}$, if
not then $1_{\mathbf{B}}-b\in{\dot{\mathcal{I}}}$ contradicts the choice of
$Y_{1}$. Hence the observation holds and hence the Observation 3.15 is proven.
Hence Claim 3.14 is proven. ∎

######  Remark 3.16.

In other words 3.15 says

  1. $(*)$

If $\kappa$ is strong limit, $\mathrm{cf}(\kappa)=\aleph_{0}$ and $\mathbf{B}$
is a Boolean algebra of cardinality $\geq\kappa$ with $\aleph_{1}$-separation
(i.e., (a) of the observation fails) then $|\mathbf{B}|\geq 2^{\kappa}$.

##  4\. Using subtrees of $({}^{\omega\geq}2,\lhd)$ and theories unstable in
$\aleph_{0}$

######  Theorem 4.1.

Suppose $T\subseteq T_{1}$ are first order theories, $T_{1}$ is countable, $T$
is complete, superstable but $\aleph_{0}$-unstable. Then for
$\lambda>\aleph_{0}$ we have

${\dot{I}\dot{E}}(\lambda,T_{1},T)\geq\min\\{2^{\lambda},\beth_{2}\\}.$

######  Remark 4.2.

The reader is not required to know anything on superstable theories, just to
believe a result quoted below. So we can just assume $(*)$ from the proof.

###### Proof..

The assumption that the theory is superstable and not totally transcendental
($=\aleph_{0}$-stable) is used to obtain $m_{a},m_{b}<\omega$ and a countable
set of definable (without parameters) equivalence relations
$\\{\varphi_{n}(\bar{x};\bar{y}):n<\omega\\}\subseteq\mathbb{L}(\tau_{T})$
such that:141414We may write $\bar{x}\;\varphi_{n}\;\bar{y}$ instead of
$\varphi_{n}(\bar{x},\bar{y}))$.

  1. $(*)$

    1. (a)

${\ell\kern-1.07639ptg}(\bar{x})={\ell\kern-1.07639ptg}(\bar{y})=m_{a}+m_{b}$

    2. (b)

If $M$ is a model of $T$ and $\bar{a}\in{}^{m_{a}}|M|$ then the set
$\\{\bar{a}{\char 94\relax}\bar{b}/\varphi_{n}:\bar{b}\in{}^{m_{b}}|M|\\}$ is
finite.

    3. (c)

If for $\ell=1,2$, ${\ell\kern-1.07639ptg}(\bar{a}_{\ell})=m_{a}$,
${\ell\kern-1.07639ptg}(\bar{b}_{\ell})=m_{b}$, and $(\bar{a}_{1}{\char
94\relax}\bar{b}_{1})\ \varphi_{n}\ (\bar{a}_{2}{\char 94\relax}\bar{b}_{2})$
then $\bar{a}_{1}=\bar{a}_{2}$.

    4. (d)

$\varphi_{n+1}$ refines $\varphi_{n}$: i.e. for every $n<\omega$, $\bar{x}\
\varphi_{n+1}\ \bar{y}$ implies $\bar{x}\ \varphi_{n}\ \bar{y}$.

    5. (e)

There are (in some model $M$ of $T$) $\bar{c}_{\eta}$ for
$\eta\in{}^{\omega>}2$ such that:

$\begin{array}[]{c}\big{[}{\ell\kern-1.07639ptg}(\eta)\geq
n\wedge{\ell\kern-1.07639ptg}(\nu)\geq n\ \mbox{ implies }\
\bar{c}_{\eta}\;\varphi_{n}\;\bar{c}_{\nu}\Leftrightarrow\eta\restriction
n=\nu\restriction n\big{]},\\\ \bar{c}_{\eta}\restriction
m_{a}=\bar{c}_{\nu}\restriction
m_{a},\quad{\ell\kern-1.07639ptg}(\bar{c}_{\eta})=m_{a}+m_{b}.\end{array}$

The existence of this set of equivalence relations was proved in Chapter III,
5.1-5.3 of both [She78] and [She90].

Clearly, without loss of generality we may expand the theory $T_{1}$. Let

$\\{c_{\ell}:\ell<m_{1}\\}\cup\\{c_{\eta,\ell}:\ell\in[m_{1},m_{1}+m_{b}]\text{
and }\eta\in{}^{\omega>}2\\}$

be new constants in $T_{1}$. We let $\bar{c}_{\eta}=\langle
c_{\ell}:\ell<m_{a}\rangle{\char 94\relax}\langle
c_{\eta,\ell}:\ell\in[m_{a},m_{a}+m_{b})\rangle$. and suppose

$\displaystyle T_{1}\supseteq$ $\displaystyle\ \big{\\{}(\bar{c}_{\eta}\
\varphi_{n}\
\bar{c}_{\nu}):{\ell\kern-1.07639ptg}(\eta),{\ell\kern-1.07639ptg}(\nu)\geq
n,\ \eta\restriction n=\nu\restriction n\big{\\}}\ \cup$ $\displaystyle\
\big{\\{}\neg(\bar{c}_{\eta}\ \varphi_{n}\
\bar{c}_{\nu}):{\ell\kern-1.07639ptg}(\eta),{\ell\kern-1.07639ptg}(\nu)\geq
n,\ \eta\restriction n\neq\nu\restriction n\big{\\}}.$

Also without loss of generality, suppose that $T_{1}$ has Skolem functions (so
the axioms saying it has Skolem functions belong to $T_{1}$).

We will use the following fact. [For a sequence $\bar{\eta}$ let
$\bar{\eta}=\big{\langle}\bar{\eta}[\ell]:\ell<{\ell\kern-1.07639ptg}(\bar{\eta})\big{\rangle}$
and $\bar{a}_{\bar{\eta}}=\bar{a}_{\bar{\eta}[0]}{\char
94\relax}\bar{a}_{\bar{\eta}[1]}{\char
94\relax}\bar{a}_{\bar{\eta}[2]}\ldots$.] ∎

######  Fact 4.3.

1) Suppose

  1. (A)

$T\subseteq T_{1}$ are first order theories, $T$ complete and superstable,
unstable in $|T_{1}|$, $\tau=\tau(T)$ and $\tau_{1}=\tau(T_{1})$, and $T_{1}$
has Skolem functions.

  2. (B)

$\tau_{1}$ is countable, or at least $\mathrm{MA}_{\mu}$ holds for
$\mu=|T_{1}|$.

  3. (C)

$\varphi_{n}$ (for $n<\omega$), $m_{a},m_{b}$ are as in $(*)$ above, and
$m_{*}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=m_{a}+m_{b}$.

  4. (D)

$\varphi_{n}\in\tau$ is a $(2m_{*})$-place predicate,

$\Delta=\\{\varphi_{n}:n<\omega\\},\quad\tau^{+}_{1}=\tau_{1}\cup\\{d_{n}:n<\omega\\},$

$\tau\subseteq\tau_{1}$, and $|\tau_{1}|\leq\mu<2^{\aleph_{0}}$.

Then there are $M_{1}$, $\bar{a}_{\eta}$ ($\eta\in{}^{\omega}2$) such that:

  1. $(\alpha)$

$M_{1}$ is a model of $T_{1}$ and $\varphi_{n}^{M_{1}}$ is an equivalence
relation such that $\varphi^{M_{1}}_{n+1}$ refines $\varphi_{n}^{M_{1}}$.

  2. $(\beta)$

$\tau(M_{1})=\tau_{1}^{+}$,
$\\{\bar{a}_{\eta}:\eta\in{}^{\omega}2\\}\subseteq{}^{m_{*}}|M_{1}|$, and

${\ell\kern-1.07639ptg}(\eta)\geq n\wedge{\ell\kern-1.07639ptg}(\nu)\geq n\
\Rightarrow\ \big{[}\eta\restriction n=\nu\restriction
n\Leftrightarrow(\bar{a}_{\eta}\ \varphi_{n}\ \bar{a}_{\nu})\big{]}.$

For $\bar{\eta}\in{}^{m}({{}^{\omega}2})$, let
$\bar{a}_{\bar{\eta}}=\bar{a}_{\eta_{0}}{\char 94\relax}\ldots{\char
94\relax}\bar{a}_{\eta_{m-1}}$.

  3. $(\beta)_{1}$

$\bar{a}_{\eta}\restriction m_{a}=\bar{a}_{\nu}\restriction m_{a}=\langle
c^{M_{1}}_{\ell}:\ell<m_{a}\rangle$,
${\ell\kern-1.07639ptg}(\bar{a}_{\eta})=m_{*}$, and if $n<\omega$,
${\ell\kern-1.07639ptg}(\bar{a})=m_{a}<m_{*}$ then
$\big{|}\big{\\{}{\bar{a}}{\char
94\relax}{\bar{b}}/\varphi_{m}:\bar{b}\in{}^{m_{b}}(M_{1})\big{\\}}\big{|}<k_{m}$.

  4. $(\gamma)$

For every formula $\varphi(\bar{x})$ from $\mathbb{L}(\tau_{1})$ such that
$m_{*}$ divides ${\ell\kern-1.07639ptg}(\bar{x})$, there is $\eta_{\varphi}$
such that for $n\in[\eta_{\varphi},\omega)$:

    1. $(*)^{1}_{\varphi,n}$

If $\bar{\eta}$, $\bar{\nu}\in{}^{m}({}^{\omega}2)$,
${\ell\kern-1.07639ptg}(\bar{\eta})={\ell\kern-1.07639ptg}(\bar{\nu})=m=\dfrac{{\ell\kern-1.07639ptg}(\bar{x})}{m_{a}+m_{b}}$
(so
${\ell\kern-1.07639ptg}(\bar{a}_{\bar{\eta}})={\ell\kern-1.07639ptg}(\bar{x})$),
and

$\big{\langle}\eta_{\ell}\restriction\eta:\ell<{\ell\kern-1.07639ptg}(\bar{\eta})\big{\rangle}=\big{\langle}\nu_{\ell}\restriction
n:\ell<{\ell\kern-1.07639ptg}(\bar{\nu})\big{\rangle}$

is without repetitions, then
$M_{1}\models\varphi[\bar{a}_{\bar{\eta}}]=\varphi[\bar{a}_{\bar{\nu}}]$.

  5. $(\delta)$

$\langle d_{n}:n<\omega\rangle$ is an indiscernible sequence over
$\\{\bar{a}_{\eta}:\eta\in{}^{\omega}2\\}$ in $M_{1}\restriction\tau_{1}$.

  6. $(\delta)^{+}$

$d_{n}\neq d_{m}$ for $n\neq m$.

2) If
$M_{1},\tau,\tau_{1},\tau_{1}^{+},m_{a},m_{b},\langle\varphi_{n}:n<\omega\rangle$
are as in $(\alpha)$, $(\beta)$, $(\beta)_{1}$,
$(\gamma),(\delta),(\delta)^{+}$ above and $\mu=\aleph_{0}$ (or at least
$\mathrm{MA}_{\mu}$) then, replacing ${}^{\omega}2$ by a subtree, replacing
$\langle\varphi_{n}:n<\omega\rangle$ by a sub-sequence and renaming,
decreasing $M_{1}$, we can add to part (1):

  1. $(\gamma)^{+}$

For every sequence of terms $\bar{\sigma}(\bar{x})$ from $\tau^{+}_{1}$, if
$m\times(m_{a}+m_{b})={\ell\kern-1.07639ptg}(\bar{x})$,
$m_{a}+m_{b}={\ell\kern-1.07639ptg}(\bar{\sigma})$,
$\bar{\sigma}(\bar{x})\restriction m_{a}=(\bar{\sigma}\restriction
m_{a})(\bar{x}\restriction m_{d})$, $m_{e}<m_{a}$,
$m_{d}=m_{e}\times(m_{a}+m_{b})$, [i.e.
$\bar{\sigma}(\bar{a}_{\bar{\eta}})\restriction
m_{a}=(\bar{\sigma}\restriction m_{a})(\bar{a}_{\bar{\eta}\restriction
m_{e}})$ for $\bar{\eta}\in{}^{m}(^{\omega}2)$], then there exists
$n_{\bar{\sigma}}<\omega$ such that:

    1. (a)

For $n\geq n_{\bar{\sigma}}$ and $\bar{\eta},\bar{\nu}\in{}^{m}({}^{\omega}2)$
with no repetitions, $\bar{\eta}\restriction m_{e}=\bar{\nu}\restriction
m_{e}$, we have:

       * •

If $\ell\neq k\Rightarrow\bar{\eta}[\ell]\restriction
n\neq\bar{\eta}[k]\restriction n$ and
$(\forall\ell<m)\big{[}\bar{\eta}[\ell]\restriction
n=\bar{\nu}[\ell]\restriction n\big{]}$ then for every
$\bar{\rho}\in{}^{m}({}^{\omega}2)$, $\bar{\rho}\restriction
m_{e}=\bar{\eta}\restriction m_{e}$ implies

$\big{(}\bar{\sigma}(\bar{a}_{\bar{\eta}})\ \varphi_{n}\
\bar{\sigma}(\bar{a}_{\bar{\rho}})\big{)}\Leftrightarrow\big{(}\bar{\sigma}(\bar{a}_{\nu})\
\varphi_{n}\ \bar{\sigma}(\bar{a}_{\bar{\rho}})\big{)}.$

    2. (b)

For $n\geq\eta_{\bar{\sigma}}$ and $\bar{\eta},\bar{\nu}\in{}^{m}({}^{n}2)$
each with no repetition and

$\bar{\eta}\restriction m_{e}=\bar{\nu}\restriction m_{e},$

we have:

       * •

If there are $k\geq n$ and
$\bar{\eta}_{1},\bar{\nu}_{1}\in{}^{m}({}^{\omega}2)$ such that
$\neg\varphi_{k}(\bar{\sigma}(\bar{a}_{\bar{\eta}_{1}}),\bar{\sigma}(\bar{a}_{\bar{\nu}_{1}}))$,
for $\ell<m$, $\bar{\eta}_{1}[\ell]\restriction n=\bar{\eta}[\ell]$,
$\bar{\nu}_{1}[\ell]\restriction n=\bar{\nu}[\ell]$, and

$(\forall\ell,i<m)\big{[}\bar{\eta}_{1}[\ell]=\bar{\nu}_{1}[i]\Leftrightarrow\bar{\eta}[\ell]=\bar{\nu}[i]\big{]},$

then for every $\bar{\eta}^{*},\bar{\nu}^{*}\in{}^{m}({}^{\omega}2)$
satisfying $\bar{\eta}^{*}[\ell]\restriction n=\bar{\eta}[\ell]$,
$\bar{\nu}^{*}[\ell]\restriction n=\bar{\nu}[\ell]$ (for each $\ell<m$) and

$(\forall\ell,i<m)\big{[}\bar{\eta}^{*}[\ell]=\bar{\nu}^{*}[i]\Leftrightarrow\bar{\eta}[\ell]=\bar{\nu}[i]\big{]}$

we have $\neg\big{[}\bar{\sigma}(\bar{a}_{\bar{\eta}^{*}})\ \varphi_{n}\
\bar{\sigma}(\bar{a}_{\bar{\nu}^{*}})\big{]}$.

######  Remark 4.4.

1) This is the only place where countability (or $\mathrm{MA}_{|\tau_{1}|}$)
is used.

2) For alternative proof see 4.13.

###### Proof..

1) If we ignore $(\delta)^{+}$ (so can have $d_{n}=d_{0}$) use Theorem [She78,
Ch.VII,3.7]. In general, use [She78, Ch.VII,Ex.3.1]. What if $T_{1}$ is
uncountable but $\mathrm{MA}_{\mu}$? (The reader may ignore this proof or see
the proof of 4.13.)

Let $\mathbb{P}$ be the forcing notion of adding
$\lambda=\beth_{(2^{\mu})^{+}}$ Cohen reals,
$\langle\mathchoice{\oalign{$\displaystyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{i}:i<\lambda\rangle$,
$\mathchoice{\oalign{$\displaystyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\eta$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{i}\in{}^{\omega}2$. Let $\chi=(2^{\lambda})^{+}$ and let

$\Vdash_{\mathbb{P}}``\mathchoice{\oalign{$\displaystyle M$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\widetilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle M$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\widetilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle M$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\widetilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle M$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\widetilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\text{ is a model of $T_{1}$, the Skolem hull of
}\\{x_{i}:i<\lambda\\},\ \bar{x}_{i}\ \varphi_{m}\
\bar{c}_{\mathchoice{\oalign{$\displaystyle\eta$\crcr\vbox
to0.60275pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\eta$\crcr\vbox
to0.60275pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\eta$\crcr\vbox
to0.60275pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\eta$\crcr\vbox
to0.60275pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{i}\restriction m}".$

By the Omitting Type Theorem151515See, e.g., [She90, Ch.VII,§5]. there are
$\mathfrak{B}_{1}\prec\mathfrak{B}_{2}$ with
$\mathfrak{B}_{1}\prec(\mathcal{H}(\chi),\in,<^{*}_{\chi})$ and
$\|\mathfrak{B}_{1}\|=\mu$ such that $T_{1}$, $P$,  $\textstyle M$
$\textstyle{\widetilde{\mkern-3.0mu}\mkern 3.0mu}{}$ , $\langle
x_{i}:i<\lambda\rangle$ belong to $\mathfrak{B}_{1}$. Also in
$\mathfrak{B}_{2}$, $\langle a_{\rho}:\rho\in{}^{\omega}2\rangle$ is an
indiscernible sequence over $\mathfrak{B}_{1}$, and
$\mathfrak{B}_{2}\models``\mathchoice{\oalign{$\displaystyle a$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle a$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle a$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle a$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{i}$ is an ordinal $=\lambda$”.

Note that any set which $\mathfrak{B}_{2}$ considers a maximal antichain of
$\mathbb{P}^{\mathfrak{B}_{2}}$ really is so. Now we can naturally apply
$\mathrm{MA}_{\mu}$.

2) Satisfy requirement (a) by letting $\varphi^{\ell}_{n}(\bar{x}{\char
94\relax}\bar{z})\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=E_{n}\big{(}\bar{x}{\char
94\relax}\bar{z},F_{\ell}(\bar{z}){\char 94\relax}\bar{z}\big{)}$ for
$\ell<\ell^{*}_{n}<\omega$, where $F_{\ell}\in\tau^{+}$ are such that
$\\{F_{\ell}(\bar{x}):\ell<\ell^{*}_{n}\\}$ is a complete set of
representatives for $\\{\bar{x}{\char
94\relax}\bar{z}/\varphi_{n}:\bar{x}\\}$, possibly with repetitions. (Remember
$T_{1}$ has Skolem functions and there is $\ell^{*}_{n}$ which does not depend
on $\bar{z}$ by compactness). Requirement (b) is fulfilled by trimming the
perfect tree and renaming. ∎

######  Claim 4.5.

For $M_{1}$, $\bar{a}_{\eta}$ ($\eta\in{}^{\omega}2$), $\varphi_{n}$ as in the
conclusion of 4.3 we can conclude:

  1. $\otimes$

If $\nu\neq\rho$ are from ${}^{\omega}2$,
$\bar{\eta}_{\nu}=\langle\eta_{\nu,\ell}:\ell<\ell(*)\rangle$,
$\bar{\eta}_{\rho}=\langle\eta_{\rho,\ell}:\ell<\ell(*)\rangle$,
$\bar{x}=\langle x_{\ell}:\ell<\ell(*)\rangle$,
$\bar{\sigma}(\bar{x})=\langle\sigma_{m}(\bar{x}):m<m(*)\rangle$,
$\nu\restriction k=\rho\restriction k$,
$\eta_{\nu,\ell}\restriction k=\eta_{\rho,\ell}\restriction k$,
$\langle\eta_{\nu,\ell}:\ell<\ell(*)\rangle$ with no repetitions,
$k>n_{\bar{\sigma}}$, and

$\bigwedge\limits_{n<\omega}\big{[}\bar{a}_{\nu}\ \varphi_{n}\
\bar{a}_{\rho}\Leftrightarrow\bar{\sigma}(\bar{a}_{\bar{\eta}_{\nu}})\
\varphi_{n}\ \bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})\big{]}$

(moreover, the $\Delta$-type of $\bar{a}_{\nu}{\char 94\relax}\bar{a}_{\rho}$
and $\bar{\sigma}(\bar{a}_{\bar{\eta}_{\nu}}){\char
94\relax}\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})$ (in $M$) are equal for
every $n$) then
${\ell\kern-1.07639ptg}(\nu\cap\rho)\in\big{\\{}{\ell\kern-1.07639ptg}(\eta_{\nu,\ell}\cap\eta_{\rho,\ell}):\ell<\ell(*)\big{\\}}$.

###### Proof..

Assume not.

Let $n={\ell\kern-1.07639ptg}(\rho\cap\nu)$. Then
$\varphi_{n}(\bar{a}_{\rho},\bar{a}_{\nu})\wedge\neg\varphi_{n+1}(\bar{a}_{\rho},\bar{a}_{\nu})$.
We suppose first (for didactic reasons) for the sake of contradiction that for
every $\ell<n_{0}$ we have

$\bar{\eta}_{\nu}[\ell]\neq\bar{\eta}_{\rho}[\ell]\ \Rightarrow\
{\ell\kern-1.07639ptg}(\bar{\eta}_{\nu}[\ell]\cap\bar{\eta}_{\rho}[\ell])<n.$

By the equality of types
$\neg\varphi_{n+1}\bigl{(}\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}}),\bar{\sigma}(\bar{a}_{\bar{\eta}_{\nu}})\bigr{)}$,
now we can deduce by Fact 4.3(2) and the assumption that the conclusion of
$(\otimes)$ fails, that
$\neg\varphi_{n+1}\bigl{(}\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}}),\bar{\sigma}(\bar{a}_{\bar{\eta}_{\nu}})\bigr{)}$.
Again, by the equality of types
$\neg\varphi_{n}(\bar{a}_{\rho},\bar{a}_{\nu})$, a contradiction to
$\varphi_{n}(\bar{a}_{\rho},\bar{a}_{\nu})$.

Now we deal with the general case, i.e., we assume

  1. $(*)$

$(\forall\ell<n_{0})\big{[}{\ell\kern-1.07639ptg}(\bar{\eta}_{\nu}[\ell]\cap\bar{\eta}_{\rho}[\ell])\neq
n\big{]}$.

We shall derive a contradiction.

Define $\bar{\eta}\in{}^{n_{0}}({}^{\omega}2)$:

$\bar{\eta}[\ell]=\begin{cases}{\bar{\eta}}_{\rho}[\ell]&\mbox{ if
}\bar{\eta}_{\nu}[\ell]\restriction n\neq\bar{\eta}_{\rho}[\ell]\restriction
n,\\\ {\bar{\eta}}_{\nu}[\ell]&\mbox{ otherwise.}\end{cases}$

Clearly $\bar{\sigma}(\bar{a}_{\eta})\restriction
m_{a}=\bar{\sigma}(\bar{a}_{\eta_{\rho}})\restriction
m_{a}=\bar{\sigma}(\bar{a}_{\eta_{\nu}})\restriction m_{a}$ and
$\bar{\eta}\restriction m_{e}=\bar{\eta}_{\nu}\restriction
m_{e}=\bar{\eta}_{\rho}\restriction m_{e}$, and also $\bar{\eta}$ is with no
repetition and $\langle\bar{\eta}[\ell]\restriction n:\ell<n_{0}\rangle$ are
pairwise distinct.

Since, by the definition of $\bar{\eta}$, for which
$\bar{\eta}[\ell]\restriction n=\bar{\eta}_{\rho}[\ell]\restriction n$, using
$(*)$ we obtain

$\eta[\ell]\restriction(n+1)=\eta_{\rho}[\ell]\restriction(n+1).$

Let $\bar{b}=\bar{\sigma}(\bar{a}_{\bar{\eta}})$. By reflexivity of the
equivalence relation we have

$\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})\ \varphi_{n+1}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}}).$

By Fact 4.3(1), $\bar{\sigma}(\bar{a}_{\bar{\eta}})\ \varphi_{n+1}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})$; i.e. $\bar{b}\ \varphi_{n+1}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})$. Finally,161616As
$\neg\bigl{(}\bar{\sigma}(\bar{a}_{\bar{\eta}_{\nu}})\ \varphi_{n+1}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})\bigr{)}$. using transitivity of the
equivalence relation we have
$\neg\varphi_{n+1}\bigl{(}\bar{b},\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})\bigr{)}$.

By the definition of $\bar{\eta}$, for every $\ell<n_{0}$ we have

$\bar{\eta}[\ell]=\bar{\eta}_{\nu}[\ell]\ \mbox{ or }\
{\ell\kern-1.07639ptg}(\bar{\eta}[\ell]\cap\eta_{\nu}[\ell])<n.$

But since $n>k_{0}$, clearly

$\big{|}\big{\\{}\bar{\eta}[\ell]\restriction
k_{0}:\ell<n_{0}\big{\\}}\big{|}=\big{|}\big{\\{}\bar{\eta}_{\nu}[\ell]\restriction
k_{0}:\ell<n_{0}\big{\\}}\big{|}=n_{0}.$

So by Fact 4.3(2), as $\neg\bigl{(}\bar{b}\ \varphi_{n+1}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\nu}})\bigr{)}$ (see above), we have
$\neg\bigl{(}\bar{b}\ \varphi_{n}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\nu}})\bigr{)}$. But $\bar{b}\ \varphi_{n}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})$ (see above) and
$\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})\ \varphi_{n}\
\bar{\sigma}(\bar{a}_{\bar{\eta}_{\rho}})$, a contradiction. ∎

So for proving theorem 4.1 we can assume

######  Hypothesis 4.6.

$M_{1},\langle\varphi_{n}:n<\omega\rangle$, and $\bar{a}_{\eta}$ (for
$\eta\in{}^{\omega}2$) are as in $(\beta)+(\gamma)$ of 4.3(1) and $(\otimes)$
of 4.5.

######  Lemma 4.7.

Assume $\mu<\lambda\leq 2^{\aleph_{0}}$. We can find
$S_{\xi}\subseteq{}^{\omega}2$ for $\xi<2^{\aleph_{0}}$, pairwise disjoint,
each of cardinality $\lambda$, such that

  1. $\otimes$

If $\xi<2^{\aleph_{0}}$,
$f:S_{\xi}\to{}^{{\omega>}}\bigl{(}\mathscr{M}_{\mu,\omega}(\bigcup\limits_{\zeta\neq\xi}S_{\zeta})\bigr{)}$
and $\mathbf{n}$ is a function,

$\mathbf{n}:\big{\\{}{\bar{\sigma}}:(\exists\bar{x})\big{[}{\bar{\sigma}}=\langle\sigma_{\ell}(\bar{x}):\ell<\ell^{*}\rangle\big{]},\
\sigma_{\ell}\mbox{ a term of
}\mathbb{L}_{\mu,\aleph_{0}}(\tau)\big{\\}}\to\omega$

and $\tau$ is the vocabulary of
$\mathscr{M}_{\mu,\omega}(\bigcup\limits_{\zeta\neq\xi}S_{\xi})$, then we can
find $m^{*}$ (see below) $S^{*}\subseteq S_{\zeta}$, $k_{0}<\omega$,
$n_{0}=m_{a}+m_{b}<\omega$, a sequence
$\bar{\sigma}(\bar{x})=\langle\sigma_{\ell}(\bar{x}):\ell<{\ell\kern-1.07639ptg}(\bar{\sigma})\rangle$,
with ${\ell\kern-1.07639ptg}(\bar{x})=n_{0}$, $\langle\bar{\eta}_{\nu}:\nu\in
S^{*}\rangle$ and $\bar{\eta}_{0}\in{}^{n_{0}}({}^{\omega}2)$ with the
following properties. Letting $\eta_{\nu,\ell}=\bar{\eta}_{\nu}[\ell]$:

    1. $(A)$

$\eta\neq\nu\in S^{*}\ \Rightarrow\ {\ell\kern-1.07639ptg}(\eta\cap\nu)>k_{0}$

    2. $(B)$

For $\nu\in S^{*}$ we have ${\ell\kern-1.07639ptg}(\bar{\eta}_{\nu})=n_{0}$,
$(\forall\ell<n_{0})\left[\bar{\eta}_{\nu,\ell}\restriction
k_{0}=\bar{\eta}_{0,\ell}\restriction k_{0}\right]$, and
$\\{\bar{\eta}_{\nu,\ell}\restriction
k_{0}:\ell<n_{0}\\}\cup\\{\nu\restriction k_{0}\\}$ are pairwise distinct.

    3. $(C)$

$k_{0}>\mathbf{n}(\bar{\sigma})$

    4. $(D)$

For each $\ell<n_{0}$, either $\\{\bar{\eta}_{\nu,\ell}:\nu\in
S^{*}\\}=\\{\bar{\eta}_{0,\ell}\\}$ or $\\{\bar{\eta}_{\nu,\ell}:\nu\in
S^{*}\\}$ are pairwise distinct.

    5. $(E)$

The sets
$\\{{\ell\kern-1.07639ptg}(\nu_{1}\cap\nu_{2}):\nu_{1}\neq\nu_{2}\mbox{ from
}S^{*}\\}$ and

$\\{{\ell\kern-1.07639ptg}(\eta_{\nu_{1},\ell_{1}}\cap\eta_{\nu_{2},\ell_{2}}):\nu_{1},\nu_{2}\in
S^{*}\mbox{ and }\ell_{1},\ell_{2}<n_{0}\\}$

are disjoint.

    6. $(F)$

For every $\nu\in S^{*}$, $f(\nu)=\bar{\sigma}(\bar{\eta}_{\nu})$ (i.e. equal
to
$\big{\langle}\sigma_{\ell}(\langle\eta_{\nu,n}:n<n_{0}\rangle):l<m^{*}\big{\rangle}$).

    7. $(G)$

For $\nu_{1}\neq\nu_{2}\in S^{*}$, we have

$\eta_{\nu_{1},\ell}=\eta_{\nu_{2},\ell}\Leftrightarrow\ell<m_{a}\Leftrightarrow\eta_{\nu_{1},\ell}=\eta_{0,\ell}.$

    8. $(H)$

$S^{*}$ is $\mu^{+}$-large. (We say that $S\subseteq{}^{\omega}2$ is
$\chi$-_large_ iff for every $n<\omega$ and $\nu\in S$ we have
$\big{|}\\{\rho\in S:\rho\restriction n=\nu\restriction n\\}\big{|}\geq\chi$.)
We can replace $\mu^{+}$-large by $\lambda$-large if
$\mathrm{cf}(\lambda)>\aleph_{0}$.

    9. $(I)$

$\nu_{1},\nu_{2}\in
S^{*}\wedge\eta_{\nu_{1},\ell_{1}}=\eta_{\nu_{2},\ell_{2}}$ implies
$\ell_{1}=\ell_{2}$.

    10. $(J)$

For $\eta\in\bigcup\limits_{\xi}S_{\xi}$, let $\xi(\eta)$ be the unique $\xi$
such that $\eta\in S_{\xi}$. Now, if
$\xi(\eta_{\nu_{1},\ell_{1}})=\xi(\eta_{\nu_{2},\ell_{2}})$ with
$\ell_{1},\ell_{2}<n_{0}$ and $\nu_{1}\neq\nu_{2}\in S^{*}$, then

$\nu\in
S^{*}\Rightarrow\xi(\eta_{\nu_{1},\ell_{1}})=\xi(\eta_{\nu,\ell_{1}})=\xi(\eta_{\nu,\ell_{2}}).$

######  Remark 4.8.

1) This claim is a version of the “unembeddability” results;171717See
Definitions in [Shea, §2], results (for example) in VI, and here in §1 for the
tree ${}^{\omega\geq}2$. well, they are necessarily somewhat weaker than in §1
here.

2) Of course, we can replace $\bigcup\limits_{\zeta\neq\xi}S_{\zeta}$ by
$\sum\limits_{\zeta\neq\xi}S_{\zeta}$.

For proving 4.7 we will use the following combinatorial fact, which is
slightly stronger than Sierpiński’s lemma on almost disjoint sets of integers:

######  Fact 4.9.

There are $W(*)$, $\\{W_{\eta}:\eta\in{}^{\omega}2\\}$, and
$\\{U_{\eta}:\eta\in{}^{\omega}2\\}$ such that for all $\eta\in{}^{\omega}2$:

  1. (A)

$W(*),W_{\eta}$ are infinite subsets of $\omega$.

  2. (B)

$U_{\eta}$ is a perfect tree; i.e. $U_{\eta}\subseteq{}^{\omega>}2$ is
downward closed, $\langle\ \rangle\in U_{\eta}$, and

$(\forall\rho\in U_{\eta})(\exists\nu\in
U_{\eta})\big{[}\rho\unlhd\nu\wedge\nu{\char 94\relax}\langle 0\rangle\in
U_{\eta}\wedge\nu{\char 94\relax}\langle 1\rangle\in U_{\eta}\big{]}.$

  3. (C)

If $\rho,\nu\in U_{\eta}$, $\rho\neq\nu$, and
${\ell\kern-1.07639ptg}(\rho)=\ell(\nu)$ then
${\ell\kern-1.07639ptg}(\rho\cap\nu)\in W_{\eta}$, where $\rho\cap\nu$ is the
largest common initial segment of $\rho$ and $\nu$; i.e.

${\ell\kern-1.07639ptg}(\rho\cap\nu)\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=\max\\{n<\omega:\rho\restriction
n=\nu\restriction n\\}.$

  4. (D)

For all $\eta_{1}\neq\eta_{2}\in{}^{\omega}2$ and every $\rho\in
U_{\eta_{1}}$, $\nu\in U_{\eta_{2}}$, there are three possibilities:

    1. (a)

${\ell\kern-1.07639ptg}(\rho\cap\nu)\in W_{\eta_{1}}\cap W_{\eta_{2}}$

    2. (b)

${\ell\kern-1.07639ptg}(\rho\cap\nu)\in W(*)$ and
$\big{(}\forall\ell<{\ell\kern-1.07639ptg}(\rho\cap\nu)\big{)}\big{[}\ell\in
W_{\eta_{1}}\equiv\ell\in W_{\eta_{2}}\big{]}$.

    3. (c)

$\rho\unlhd\nu$ or $\nu\lhd\rho$.

  5. (E)

$W(*)\cap W_{\eta}=\varnothing$

  6. (F)

For distinct $\eta,\nu$ from ${}^{\omega}2$, we have:

    1. (a)

$W_{\eta}\cap W_{\nu}$ is finite (in fact, an initial segment of both).

    2. (b)

If $\ell\in W\\!(*)$ is above $W_{\eta}\cap W_{\nu}$ then $U_{\eta}\cap
U_{\nu}$ is finite, contained in ${}^{\ell^{\prime}>}2$ if
$\ell<\ell^{\prime}\in W_{\eta}\cup W_{\nu}$, and has no splitting of level
$\geq\ell$; i.e.

$\neg(\exists\rho\in{}^{\omega>}2)\big{[}{\ell\kern-1.07639ptg}(\rho)\geq\ell\wedge\\{\rho{\char
94\relax}\langle 0\rangle,\rho{\char 94\relax}\langle 1\rangle\\}\subseteq
U_{\eta}\cap U_{\nu}\big{]}.$

    3. (c)

If $\ell\in W(*)$ and $\ell<\sup(W_{\eta}\cap W_{\nu})$ then
$U_{\eta}\cap{}^{\ell\geq}2=U_{\nu}\cap{}^{\ell\geq}2$.

###### Proof..

By induction on $n$, define $k(n)=k_{n}<\omega$, the set $W_{n}(*)\subseteq
k(n)$ and the sets $U_{\eta}\subseteq{}^{k(n)\geq}2$, $W_{\eta}\subseteq
k(n)$, such that in the end (this imposes natural restrictions on them):

$\eta\in{}^{\omega}2\ \Rightarrow\ W_{\eta}\cap k_{n}=W_{\eta\restriction
n},\quad U_{\eta}\cap{}^{k(n)\geq}2=U_{\eta\restriction n},\quad W(*)\cap
k(n)=W_{n}(*).$

For $n=0$, let $k_{0}=0$, $W_{n}(*)=\varnothing$ and $W_{\eta}=\varnothing$,
$U_{\eta}=\varnothing$ for $\eta\in{}^{n}2$. For the induction step, choose
$k^{1}(n)=k(n)+n+1$ and for $\eta\in{}^{n}2$ let

$U^{1}_{\eta}=U_{\eta}\cup\\{\nu{\char 94\relax}(\eta\restriction\ell):\nu\in
U_{\eta}\cap{}^{k(n)}2,\ \ell\leq n\\}.$

Thus

$\big{(}\forall\nu\in{}^{k(n)}2\cap
U_{\eta}\big{)}\big{(}\exists!\rho\in{}^{k^{1}(n)}2\cap
U^{1}_{\eta}\big{)}\big{[}\nu\unlhd\rho\big{]}.$

Define $W_{n+1}(*)=W_{n}(*)\cup\big{[}k(n),k^{1}(n)\big{)}$. Fix an
enumeration $\\{\eta_{k}:k<2^{n+1}\\}$ of ${}^{n+1}2$. Let
$k(n+1)\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=k^{1}(n)+2^{n+1}$.
For $\eta\in{}^{n+1}2$, there is a unique $k<2^{n}$ such that $\eta=\eta_{k}$.
Let

$\displaystyle
U_{\eta_{k}}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=U^{1}_{\eta_{k}\restriction
n}\cup\big{\\{}\nu\in{}^{k(n+1)\geq}2:$ $\displaystyle\ \nu\restriction
k^{1}(n)\in U^{1}_{\eta_{k}\restriction n},\mbox{ and for }\ell<2^{n}\mbox{ we
have }$ $\displaystyle\
k^{1}(n)+\ell<{\ell\kern-1.07639ptg}(\nu)\wedge(\ell\neq
2k+1)\Rightarrow\nu(k^{1}(n)+\ell)=0\big{\\}}$

and $W_{\eta_{k}}=W_{\eta_{k}}\cup\\{k^{1}(n)+2k+1\\}.$ It is easy to verify
that the construction provides a family of sets as required. ∎

Proof of Lemma 4.7: Let $W(*)$, $U_{\eta}$, $W_{\eta}$ be as in 4.9. Fix an
enumeration $\\{\eta_{\xi}:\xi<2^{\aleph_{0}}\\}={}^{\omega}2$ and let
$W^{\xi}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=W_{\eta_{\xi}}$.
Let

$S_{\xi}\subseteq\lim(U_{\eta_{\xi}})\
\big{(}=\big{\\{}\rho\in{}^{\omega}2:(\forall n<\omega)[\rho\restriction n\in
U_{\eta_{\xi}}]\big{\\}}\big{)}$

be of cardinality $\lambda$. Fix $\\{\rho^{\xi}_{i}:i<\lambda\\}=S_{\xi}$, and
without loss of generality $S_{\xi}$ is $\chi$-large.181818Recall that we say
$S\subseteq{}^{\omega}2$ is $\chi$-large if for every $n<\omega$ and $\nu\in
S$,
$\big{|}\\{\rho\in S:\rho\restriction n=\nu\restriction n\\}\big{|}\geq\chi$.
If $\chi\geq(|\tau_{1}|+\aleph_{0})^{+}$ we may omit it.

Note that for every $S\subseteq{}^{\omega}2$ of cardinality $>\mu$, for some
$S_{1}\subseteq S$, $|S_{1}|\leq\mu$ and $S\setminus S_{1}$ is
$\mu^{+}$-large. Let $U^{\zeta}=U_{\eta_{\zeta}}$; note that by 4.9(B)+(D),
the sets $S_{\xi}\setminus S$ are pairwise disjoint.

So let $\xi,f,\mathbf{n}$ be as in the assumption of 4.7$\otimes$.

For $\nu\in S_{\xi}$ let $f(\nu)={\bar{\sigma}}_{\nu}(\bar{\eta}_{\nu})$,
where $\bar{\sigma}_{\nu}$ is a finite sequence of terms and
$\bar{\eta}_{\nu}$ is a finite sequence of members of
$\bigcup\limits_{\zeta\neq\xi}S_{\zeta}$ with no repetitions. So there are
$S^{*}\subseteq S_{\xi}$ which is $\mu^{+}$-large, and $\bar{\sigma}$, and an
integer $n_{0}$ such that

$\nu\in S^{*}\ \Rightarrow\
\bar{\sigma}_{\nu}=\bar{\sigma}\wedge{\ell\kern-1.07639ptg}(\bar{\eta}_{\nu})=n_{0},$

and without loss of generality, for some $m_{a}\leq m_{b}<\omega$, we have
$\bar{\sigma}(\bar{\eta}_{\nu})\restriction m_{a}=\bar{\eta}^{*}$ and

$\\{\eta^{*}_{\ell}:\ell<m_{a}\\}\cup\\{\eta_{\nu,\ell}:\nu\in S^{*}\mbox{ and
}\ell\in[m_{a},m_{b})\\}$

is without repetition (this is possible by the $\Delta$-system argument).

As $S_{\xi}\cap\bigcup\limits_{\zeta\neq\xi}S_{\zeta}=\varnothing$, clearly
the sequence $\langle\nu\rangle{\char 94\relax}\bar{\eta}_{\bar{\nu}}$ is
without repetitions for any $\nu\in S^{*}$. So for some $k=k_{\nu}<\omega$
large enough, we have:

  1. $(i)$

$\langle\nu\restriction k\rangle{\char
94\relax}\langle\eta_{\nu,\ell}\restriction k:\ell<\ell(*)\rangle$ is without
repetitions.

  2. $(ii)$

Letting $\eta_{\nu,\ell}\in S_{\zeta(\nu,\ell)}$, we have $W^{\xi}\cap
W^{\zeta(\eta_{\nu,\ell})}\subseteq\\{0,\ldots,k_{\nu}-1\\}$. Moreover,
$k_{\nu}>\min(W^{\xi}\setminus W^{\zeta(\eta_{\nu},\ell)})$ (remember clause
(F) of 4.7).

As we can shrink $S^{*}$ as long as it is $\mu^{+}$-large, without loss of
generality for some $k$:

  1. $(iii)$

$\nu_{1}\neq\nu_{2}\in S^{*}\ \Rightarrow\
{\ell\kern-1.07639ptg}(\nu_{1}\cap\nu_{2})>k$

  2. $(iv)$

$\nu\in S^{*}\ \Rightarrow\ k_{\nu}<k<\omega$.

So for $\nu_{1}\neq\nu_{2}\in S^{*}$, on the one hand
${\ell\kern-1.07639ptg}(\nu_{1}\cap\nu_{2})\in W^{\xi}\setminus k$ (as
$\nu_{1},\nu_{2}\in S_{\xi}\subseteq\lim(U_{\eta_{\xi}})$; see clause (iii)
above and 4.9(C)) and on the other hand

${\ell\kern-1.07639ptg}(\eta_{\nu_{1}\cap\ell},\eta_{\nu_{2},\ell})\in
W(*)\cup U^{\zeta(\nu_{1},\ell)}\cup U^{\zeta(\nu_{2},\ell)}$

which is disjoint to $W^{\xi}\setminus k$. So we have proved clause (E) of
4.7; the other clauses can be checked. ∎

######  Claim 4.10.

If clauses $(\beta)$, $(\gamma)$, $(\delta)$ of 4.3(1) hold, and
4.5$(\otimes)$ does as well, then for $\lambda\leq 2^{\aleph_{0}}$:

  1. $(*)_{\lambda}$

There is a family $\mathscr{P}$ of subsets of ${}^{\omega}2$ each of
cardinality $\lambda$ (even their union has cardinality $\lambda$) with
$|\mathscr{P}|=2^{\lambda}$, such that (letting $N^{1}_{S}$ be the Skolem Hull
of $\\{\bar{a}_{\eta}:\eta\in S\\}$ for $S\in\mathscr{P}$) we have:

     * •

For $Y_{1}\neq Y_{2}$ from $\mathscr{P}$, $N^{1}_{Y_{1}}$ has no
$\Delta$-embedding into $N^{1}_{Y_{2}}$.

     * •

$\|N^{1}_{Y}\|=\lambda$ for $Y\in\mathscr{P}$.

###### Proof..

For $X\subseteq\lambda$, let $M^{1}_{X}$ be the Skolem Hull of
$\\{\bar{a}_{\eta}:\eta\in\bigcup\limits_{\xi\in X}S_{\xi}\\}$ and
$M_{X}\mathrel{\vbox{\hbox{\scriptsize.}\hbox{\scriptsize.}}}=M^{1}_{X}\restriction\tau_{T}$.

In order to prove the theorem it is enough to assume $X,Y\subseteq\lambda$ and
$X\not\subseteq Y$, and show there does not exist an elementary embedding $f$
from $M_{X}$ into $M_{Y}$. Let $\xi\in X\setminus Y$. For the sake of
contradiction suppose $f:M_{X}\to M_{Y}$ is an elementary embedding, or just
one preserving the satisfaction of $\varphi_{n},\neg\varphi_{n}$.

We can represent $M_{Y}$ in
$\mathscr{M}_{\mu,\omega}\bigl{(}\bigcup\limits_{\zeta\neq\xi}S_{\zeta}\bigr{)}$,
and let us define
$f^{\prime}:S_{\xi}\to\mathscr{M}_{\mu,\omega}\bigl{(}\bigcup\limits_{\zeta\neq\xi}S_{\zeta}\bigr{)}$
by $f^{\prime}(\nu)=f(\bar{a}_{\nu})$, let $\mathbf{n}$ be essentially as in
4.3, but translated. Apply lemma 4.7 to $f^{\prime}$ and $\mathbf{n}$, and get
$S^{*}$, $k_{0}$, $n_{0}$, $m_{a}$, $m_{b}$, $\bar{\sigma}$,
$\langle\bar{\eta}_{\nu}:\nu\in S^{*}\rangle$ as there. Of course $n_{0}$,
$m_{a}$, $m_{b}$ are predetermined as in 4.3.

So we are done proving 4.10. ∎

###### Proof..

Proof of Theorem 4.1:

When $\lambda\leq 2^{\aleph_{0}}$, the result follows from 4.10 by 4.5.

So the proof of Theorem 4.1 for the case $\lambda\leq 2^{\aleph_{0}}$ is
completed. How to deal with the case $\lambda>2^{\aleph_{0}}$? We just need to
use $(\delta)^{+}$; i.e. use 4.12 (and Definition 4.11) below. ∎

######  Definition 4.11.

For any cardinal $\kappa$ and $M_{1}$ as in 4.3(1)$(\beta)$-$(\delta)^{+}$, we
define a model $M_{1,\kappa}$ as follows: it is a $\tau_{1}$-model generated
by $\\{\bar{a}_{\eta}:\eta\in{}^{\omega}2\\}\cup\\{d_{i}:i<\kappa\\}$ such
that for every $n<\omega$, $i_{1}<\ldots i_{n}<\kappa$, and
$\eta_{1},\ldots,\eta_{m}\in{}^{\omega}2$, the quantifier-free type of
${\bar{a}}_{\eta_{1}}{\char 94\relax}\ldots{\char
94\relax}{\bar{a}}_{\eta_{m}}{\char 94\relax}\langle
d_{i_{1}},\ldots,d_{i_{n}}\rangle$ in $M_{1,\kappa}$ is equal to the
quantifier-free type of ${\bar{a}}_{\eta_{1}}{\char 94\relax}\ldots{\char
94\relax}{\bar{a}}_{\eta_{m}}{\char 94\relax}\langle
d_{1},\ldots,d_{n}\rangle$ in $M_{1}$. (So if $M_{1}$ has Skolem functions
then $M_{1}=M_{1,\mu}$ and they realize the same types.)

######  Claim 4.12.

If clauses $(\beta),(\gamma),(\delta),(\delta)^{+}$ of 4.3(1) hold, and
4.5$(\otimes)$ does as well, then for $\lambda\geq 2^{\aleph_{0}}$:

  1. $(*)_{\lambda}$

There is a family $\mathscr{P}$ of subsets of ${}^{\omega}2$ each of
cardinality $2^{\aleph_{0}}$ with $|\mathscr{P}|=\beth_{2}$ such that, letting
$N^{\lambda}_{S}$ be the Skolem Hull of $\\{\bar{a}_{\eta}:\eta\in
S\\}\cup\\{d_{i}:i<\kappa\\}$ in $M_{1,\lambda}$ with $S\in\mathscr{P}$ (so
$\|N^{\lambda}_{S}\|=\lambda$), we have:

    1. $(*)$

For $Y_{1}\neq Y_{2}$ from $\mathscr{P}$, $N^{1}_{Y_{1}}$ has no
$\Delta$-embedding into $N^{1}_{Y_{2}}$ (i.e. no function from $N^{1}_{Y_{1}}$
into $N^{1}_{Y_{2}}$ preserves all the relations $\pm\varphi_{n})$.

We may consider using relations $\varphi_{n}$ which are not equivalence
relations, and we may like to give another proof when $\mu>\aleph_{0}$ but
still $\mathrm{MA}_{\mu}$ holds.

######  Claim 4.13.

$[$Assume $\mathrm{MA}_{\mu}.]$

Suppose $M_{1}$, $\tau_{1}$,
$\langle\bar{a}_{\eta}:\eta\in{}^{\omega}2\rangle$, $\varphi_{n}$ (for
$n<\omega$), $\langle d_{n}:n<\omega\rangle$ satisfy clauses $(a)$, $(b)$,
$(\beta)$, $(\gamma)$, $(\delta)$ of 4.3, and $M_{1}$ is a $\tau_{1}$-model of
the complete first order theory $T_{1}$. Also suppose
$\bar{a}_{\eta}\in{}^{k}(M_{1})$ for $\eta\in{}^{\omega>}2$ are such that if
$n<m<\omega$ and $\eta,\nu\in{}^{m}2$ then $\eta\restriction n=\nu\restriction
n\Leftrightarrow M_{1}\models\bar{a}_{\eta}\ \varphi_{n}\ \bar{a}_{\nu}$. (So
$\varphi_{n}$ is not necessarily an equivalence relation and $|\tau_{1}|=\mu$
is not necessary countable).

$1)$ If we replaced ${}^{\omega\geq}2$ by a perfect subtree (splitting
determined by level only) and replaced $\langle\varphi_{n}:n<\omega\rangle$ by
a subsequence, then we could add the statement of 4.5$(\otimes)$ to the
assumptions.

$2)$ So the conclusion of 4.10 holds, and if we further assume $(\delta)^{+}$
of 4.3, the conclusion of 4.12 also holds.

###### Proof..

We use Carlson and Simpson [CS84].

Let $W^{*}$ be the set of $\omega$-sequences $\eta$ from
$\\{0,1\\}\cup\\{x_{i}:i<\omega\\}$ such that each $x_{i}$ appears infinitely
often. For $\eta\in W^{*}$, let

$W_{\eta}=\\{\nu\in
W^{*}:\eta(\ell)\in\\{0,1\\}\Rightarrow\nu(\ell)=\eta(\ell),\
\eta(\ell_{1})=\eta(\ell_{2})\Rightarrow\nu(\ell_{1})=\nu(\ell_{2})\\}.$

As a set, $W\subseteq W^{*}$ is _large_ if it contains some $W_{\eta}$. Let

$\displaystyle I_{W}=\big{\\{}\nu\in{}^{\omega>}2:$ $\displaystyle\text{ for
some $\eta\in W$, for every
}\ell,\ell_{1},\ell_{2}<{\ell\kern-1.07639ptg}(\nu),$ $\displaystyle\
\eta(\ell)\in\\{0,1\\}\Rightarrow\nu(\ell)=\eta(\ell)\wedge\eta(\ell_{1})=\eta(\ell_{2})\Rightarrow\nu(\ell_{1})=\nu(\ell_{2})\big{\\}}.$

Let

$\mathrm{lev}(W)=\big{\\{}\ell:\text{for some }\eta\in W,\
\eta(\ell)\notin\\{0,1\\}\text{ but
}\eta(0),\ldots,\eta(\ell-1)\in\\{0,1\\}\big{\\}}.$

We say $W_{1}\subseteq^{*}W_{2}$ if for some $n$,
$\\{\nu\restriction[n,\omega):\nu\in
W_{1}\\}\supseteq\\{\nu\restriction(n,\omega):\nu\in W_{2}\\}$. By
$\mathrm{MA}_{\mu}$, if $\langle W_{i}:i<\delta\leq\mu\rangle$ is
$\subseteq^{*}$-decreasing sequence then there is $W$ such that
$\bigwedge\limits_{i}W_{i}\subseteq^{*}W$.

By the partition theorem there, if $n<\omega$,
$\eta_{1},\ldots,\eta_{k}\in{}^{n}2$ are pairwise distinct and
${\bar{\sigma}}^{1}$, ${\bar{\sigma}}^{2}$ are $\tau^{+}_{1}$-terms then we
can find large $W_{1}\subseteq W$ such that $W_{1}\restriction n=W\restriction
n$ and:

  1. $\circledast^{n}_{W_{1},\bar{\sigma}}$

If $n<m\in\mathrm{lev}(W_{1})$,
${}\rho^{\nu}_{\ell}\in\mathcal{T}_{W_{1}}\cap{}^{m}2$ for $\ell=1,\ldots,k$,
and $\nu_{\ell}=\eta_{\ell}{\char 94\relax}\rho_{2}\restriction[n,\omega)$,
then the truth value of
${\bar{\sigma}}^{1}(\bar{a}_{\nu_{1}},\ldots,\bar{a}_{\nu_{k}})\ \varphi_{n}\
{\bar{\sigma}}^{2}(\bar{a}_{\nu_{1}},\ldots,\bar{a}_{\nu_{k}})$ is constant.

Repeating it, we can get $W_{1}$ such that
$\circledast^{n}_{W_{1},\bar{\sigma}}$ for every $n$.

  1. $(i)$

Either $g$ is constant $<\min(\mathrm{lev}(W_{1})\setminus n)$ or

$n\in\mathrm{lev}(W_{1})\ \Rightarrow\
\big{[}g(n),n\big{)}\cap\mathrm{lev}(W_{1})=\varnothing.$

  2. $(ii)$

If $n<m\in\mathrm{lev}(W_{1})$ and
$\eta_{\ell}\lhd\nu_{\ell}\in\mathcal{T}_{W_{1}}\cap{}^{m}2$ then

$\min\\!\big{\\{}i:\neg\big{[}\sigma^{1}(\bar{a}_{\nu_{1}},\ldots,\bar{a}_{\nu_{k}})\
\varphi_{i}\
\sigma^{2}(\bar{a}_{\nu_{1}},\ldots,\bar{a}_{\nu_{k}})\big{]}\big{\\}}=g(m).$

We apply such reasoning to the following statement: “Given
$\eta_{1},\ldots,\eta_{k}\in\mathcal{T}_{W_{1}}\cap{}^{n}2$ pairwise distinct
and $n<m\in\mathrm{lev}(W_{1})$, and assuming
$\eta_{\ell}\lhd\nu_{\ell}^{i}\in\mathcal{T}_{W_{1}}\cap{}^{m}2$ for
$\ell\in\\{0,1,\ldots,k\\}$ and $i\in\\{0,1\\}$, we have

$\bar{\sigma}(\bar{a}_{\nu^{0}_{1}},\ldots,\bar{a}_{\nu^{0}_{k}})\
\varphi_{\ell}\
\bar{\sigma}(\bar{a}_{\nu^{1}_{1}},\ldots,\bar{a}_{\nu^{1}_{k}})."$

We get that this depends only on
${\ell\kern-1.07639ptg}(\nu^{0}_{\ell}\cap\nu^{1}_{\ell})$ and
$\nu^{i}_{\ell}\big{(}{\ell\kern-1.07639ptg}(\nu^{0}_{\ell}\cap\nu^{1}_{\ell})\big{)}$.
∎

######  Discussion 4.14.

The parallel (for a module $\dot{\mathbb{M}}$) concerning “a surgery at” is
extending the ring $\dot{\mathbf{R}}$ to $\dot{\mathbf{R}}^{+}$; e.g. by
$\\{x_{t}:t\in I\\}$ freely except some equation involving $x$ and the
$x_{i}$-s and “below $x$” is replaced by the ideal generated by those
equations.

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