The gap that Hausdorff constructed satisfied the following property.

######  Definition 3.1.6 (Hausdorff condition).

Let $(\mathcal{L},\mathcal{R})$ be an $(\omega_{1},\omega_{1})$-pregap on
$\omega$. We say that $(\mathcal{L},\mathcal{R})$ is Hausdorff if
$\\{R\in\mathcal{R}\,:\,rank(R)<rank(L)\textit{ and }L\cap R\subseteq k\\}$ is
finite for each $L\in\mathcal{L}$ and $k\in\omega.$

######  Definition 3.1.7 (Luzin condition).

Let $(\mathcal{L},\mathcal{R})$ be an $(\omega_{1},\omega_{1})$-pregap on an
infinite set $X$. We say that $(\mathcal{L},\mathcal{R})$ is Luzin if
$\\{R\in\mathcal{R}\,:\,rank(R)<rank(L)\textit{ and }|L\cap R|\leqslant k\\}$
is finite for each $L\in\mathcal{L}$ and $k\in\omega.$

######  Remark 3.1.8.

Note that the Luzin condition is stronger than the Hausdroff condition. It is
a standard excercise to prove that any $(\omega_{1},\omega_{1})$-pregap
satisfying the Hausdorff condition is in fact a gap. Under the $P$-ideal
Dichotomy ($PID$), if $(\mathcal{L},\mathcal{R})$ is an
$(\omega_{1},\omega_{1})$-gap then there are cofinal
$\mathcal{L}^{\prime}\subseteq\mathcal{L}$ and
$\mathcal{R}^{\prime}\subseteq\mathcal{R}$ so that
$(\mathcal{L}^{\prime},\mathcal{R}^{\prime})$ is a Hausdorff gap. This result
was proved by Uri Abraham and Stevo Todorčević in [4]. The same result follows
from $\mathfrak{m}>\omega_{1}.$

In the following theorem we construct a Hausdorff gap using a $2$-construction
scheme (that is, a construction scheme of type $\langle
m_{k},2,r_{k+1}\rangle_{k\in\omega}$). The first construction of a Hausdorff
gap using morasses was performed by Daniel Velleman in [127]. In [69] and
[110] the reader may find a construction of such a gap using construction
schemes and ordinal metrics respectively. First, a quick remark.

######  Remark 3.1.9.

If $\langle L_{\alpha}\rangle_{\alpha\in\omega_{1}}$ and $\langle
R_{\alpha}\rangle_{\alpha\in\omega_{1}}$ are $\omega_{1}$-towers and
$L_{\alpha}\cap R_{\alpha}=^{*}\emptyset$ for each $\alpha\in\omega_{1}$ then
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is a pregap. This is because
for each $\alpha<\beta\in\omega_{1}$, $L_{\alpha}\cap
R_{\beta}\subseteq^{*}L_{\beta}\cap R_{\beta}.$

######  Theorem 3.1.10.

Let $\mathcal{F}$ be a $2$-construction scheme. For each
$\alpha\in\omega_{1}$, define

$L_{\alpha}=\\{2k+\Xi_{\alpha}(k)\,:\,k\in\omega\backslash 1,\textit{ and
}\Xi_{\alpha}(k)\geqslant 0\\},$
$R_{\alpha}=\\{2k+(1-\Xi_{\alpha}(k))\,:\,k\in\omega\backslash 1,\textit{ and
}\Xi_{\alpha}(k)\geqslant 0\\}.$

Then, $(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is a Hausdorff gap.

###### Proof.

Since $r_{k}=0$ for infinitely many $k^{\prime}s$, it should be clear that
each $L_{\alpha}$ and $R_{\alpha}$ are infinite. By virtue of the part (c) in
Lemma 2.3.9, we have that if $\alpha<\beta$ then

$L_{\alpha}\backslash
L_{\beta}\subseteq\\{2k+\Xi_{\alpha}(k)\,:\,k\leqslant\rho(\alpha,\beta)\textit{
and }\Xi_{\alpha}(k)\not=\Xi_{\beta}(k)\,\\},$ $R_{\alpha}\backslash
R_{\beta}\subseteq\\{2k+(1-\Xi_{\alpha}(k))\,:\,k\leqslant\rho(\alpha,\beta)\textit{
and }\Xi_{\alpha}(k)\not=\Xi_{\beta}(k)\,\\}.$

As the sets on the right are finite, this means $\langle
L_{\alpha}\rangle_{\alpha\in\omega_{1}}$ and $\langle
R_{\alpha}\rangle_{\alpha\in\omega_{1}}$ are both $\omega_{1}$-towers.
Furthermore by definition we have that $L_{\alpha}\cap R_{\alpha}=\emptyset$
for each $\alpha\in\omega_{1}$. This implies that
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is a pregap due to the Remark
3.1.9.

The only thing left to prove is that the Hausdorff condition is satisfied. For
this purpose take $\beta\in\omega_{1}$ and $k\in\omega$. We claim
$\\{\alpha<\beta\,:\,L_{\beta}\cap R_{\alpha}\subseteq
k\\}\subseteq(\beta)_{k}$. For this, take an arbitrary $\alpha<\beta$
satisfying $\rho(\alpha,\beta)\geqslant k.$ By the part (b) of Lemma 2.3.9,
$\Xi_{\alpha}(\rho(\alpha,\beta))=0$ and $\Xi_{\beta}(\rho(\alpha,\beta))=1$.
This means $2\rho(\alpha,\beta)+1\in L_{\beta}\cap R_{\alpha}$, so we are
done. ∎

An interesting feature of the proof given above is that not only we prove that
there is a gap using a $2$-construction scheme, but such gap can be explicitly
defined from it.

For a countably infinite set $X$, we say that a family
$\mathcal{A}\subseteq[X]^{\omega}$ is an almost disjoint family if $A\cap
B=^{*}\emptyset$ whenever $A,B\in\mathcal{A}$ are different. Almost disjoint
families are one of the central objects of study in modern combinatorial set
theory. Constructing almost disjoint families with special properties is
usually difficult and in most cases had lead to the development of powerful
tools (see [6], [15], [21], [22], [31], [38],[84] and [96]). Almost disjoint
families have also played a central roll in the solution of many problems of
Topology and Analysis. An example of this is the solution of the selection
problem posed by Jan van Mill and Evert Wattel in [126] and solved by Michael
Hrušák and Iván Martínez-Ruíz in [50]. The reader interested in learning more
about almost disjoint families is refered to [46], [48] and [49].

Our next goal is to use almost disjoint families with the aim of proving that,
basically, there are gaps of any possible type for which the cofinality of the
two partial ordered sets involved is $\omega_{1}$.

######  Definition 3.1.11.

Let $\mathcal{A}$ be an almost disjoint family of size $\omega_{1}$ over a set
$X$. We say that:

  * $\blacksquare$

$\mathcal{A}$ is inseparable if for any two disjoint
$\mathcal{L},\mathcal{R}\in[\mathcal{A}]^{\omega_{1}}$, the pair
$(\mathcal{L},\mathcal{R})$ forms a gap.

  * $\blacksquare$

$\mathcal{A}$ is Luzin if we can enumerate it as $\langle
A_{\alpha}\rangle_{\alpha\in\omega_{1}}$ in such way that
$\\{\alpha\in\beta\,:\,|A_{\alpha}\cap A_{\beta}|<n\\}$ is finite for each
$\beta\in\omega_{1}$ and $n\in\omega$.

  * $\blacksquare$

$\mathcal{A}$ is Jones if for any two disjoint
$\mathcal{L}\in[\mathcal{A}]^{\omega}$ and
$\mathcal{R}\in[\mathcal{A}]^{\omega_{1}}$, the pregap
$(\mathcal{L},\mathcal{R})$ can be separated. If $\mathcal{A}$ is indexed as
$\langle A_{\alpha}\rangle_{\alpha\in\omega_{1}}$ then $\mathcal{A}$ is Jones
if and only if for any $\beta\in\omega_{1}$ the pregap $(\langle
A_{\alpha}\rangle_{\alpha\leqslant\beta},\langle
A_{\delta}\rangle_{\delta>\beta})$ can be separated.

  * $\blacksquare$

$\mathcal{A}$ is Luzin-Jones if it is both Luzin and Jones.

######  Remark 3.1.12.

It is easy to check that any Luzin family is in fact inseparable. In [71],
Judith Roitman and Lajos Soukup showed that under $\mathfrak{m}>\omega_{1}$
any uncountable $AD$ family either contains a Luzin family or contains no
inseparable family.

The first construction of a Luzin family was done in [71] by Nikolái
Nikoláyevich Luzin. A Jones family was implicitly constructed by F. B. Jones
in [57]. Although, at first glance, Luzin and Jones properties seem to be
incompatible, a construction of a Luzin-Jones family was obtained in [42] by
Osvaldo Guzmán , Michael Hrušák and Piotr Koszmider (building from work by
Koszmider in [63]). We would like to point out that the highly complex
construction of the mentioned family is carried out through the use of
simplified morasses. Here we give an elementary construction of such object.

######  Theorem 3.1.13.

There is a Luzin-Jones family.

###### Proof.

Let $\mathcal{F}$ be a $2$-construction scheme. For every
$k\in\omega\backslash 1$, let

$N_{k}=\\{k\\}\times(m_{k-1}\backslash r_{k})\times[k\cdot m_{k-1},\,k\cdot
m_{k}).$

We aim to construct a Luzin-Jones family $\langle
A_{\alpha}\rangle_{\alpha\in\omega_{1}}$ over the union of the $N_{k}$’s,
namely $N$. For this, take an arbitrary $\alpha\in\omega_{1}$. Given
$k\in\omega\backslash 1$, we define $A_{\alpha}^{k}\subseteq N_{k}$ as
follows:

  1. $(a)$

If $\Xi_{\alpha}(k)=-1$, let $A^{k}_{\alpha}=\emptyset$.

  2. $(b)$

If $\Xi_{\alpha}(k)=0$, let
$A^{k}_{\alpha}=\\{k\\}\times\\{\lVert\alpha\rVert_{k}\\}\times[k\cdot
m_{k-1},\,k\cdot m_{k})$.

  3. $(c)$

If $\Xi_{\alpha}(k)=1$, let
$A^{k}_{\alpha}=\\{k\\}\times[r_{k},m_{k-1})\times\big{[}k\cdot\lVert\alpha\rVert_{k},\,k(\lVert\alpha\rVert_{k}+1)\big{)}.$

Now, we define $A_{\alpha}$ as $\bigcup\limits_{k\in\omega\backslash
1}A^{k}_{\alpha}$ and $\mathcal{A}$ as $\langle
A_{\alpha}\rangle_{\alpha\in\omega_{1}}$.

Our first task is to show that $\mathcal{A}$ is an almost disjoint family.
First observe that since $r_{k}=0$ infinitely often it follows that each
$A_{\alpha}$ is in fact infinite. Now let $\alpha<\beta\in\omega_{1}$ and take
an arbitrary $k>\rho(\alpha,\beta)$. In virtue of the part (c) by Lemma 2.3.9
we have that $\Xi_{\alpha}(k)=-1$ or $\Xi_{\alpha}(k)=\Xi_{\beta}(k)$. In
either case it follows that $A^{k}_{\alpha}\cap A^{k}_{\beta}=\emptyset.$
Therefore we conclude that

$A_{\alpha}\cap A_{\beta}=\bigcup_{k\in\omega\backslash
1}\big{(}A^{k}_{\alpha}\cap
A^{k}_{\beta}\big{)}\subseteq\bigcup_{k\leqslant\rho(\alpha,\beta)}N_{k}.$

Now we prove that $\mathcal{A}$ is Luzin. We claim that whenever
$\alpha<\beta$ and $k=\rho(\alpha,\beta)$ then $|A_{\alpha}\cap
A_{\beta}|\geqslant k$. If this happens then
$\\{\alpha<\beta\,:\,|A_{\alpha}\cap A_{\beta}|<k\\}\subseteq(\beta)_{k}$ for
any $\beta\in\omega_{1}$. Let $\alpha,\beta$ and $k$ be as previously stated.
The part (b) of Lemma 2.3.9 assures $\Xi_{\alpha}(k)=0<\Xi_{\beta}(k)=1$. In
this way

$A^{k}_{\alpha}\cap
A^{k}_{\beta}=\\{k\\}\times\\{|(\alpha)^{-}_{k}|\\}\times\big{[}k\cdot\lVert\alpha\rVert_{k},\,k(\lVert\alpha\rVert_{k}+1)\big{)}.$

The cardinality of this set is $k$ and it is contained in $A_{\alpha}\cap
A_{\beta}$. Thus, we are done.

In here, $\Xi_{\alpha}(k)=0$ and $\Xi_{\beta}(k)=1$.

We will end the proof by showing that $\mathcal{A}$ is a Jones family. For
this purpose define $C_{\beta}$ as

$\bigcup\limits_{k\in\omega}\big{(}\bigcup\limits_{\alpha\in(\beta)_{k}}A^{k+1}_{\alpha}\big{)}$

for each $\beta\in\omega_{1}$. We will show that $C_{\beta}$ separates
$(\langle A_{\alpha}\rangle_{\alpha\leqslant\beta},\langle
A_{\alpha}\rangle_{\delta>\beta})$. Indeed, if $\alpha\leqslant\beta$ then
$A^{k+1}_{\alpha}\subseteq C_{\beta}$ for each $k\geqslant\rho(\alpha,\beta)$,
so in particular $A_{\alpha}\subseteq^{*}C_{\beta}$. On the other hand, if
$\delta>\beta$ then $A^{k+1}_{\delta}\cap A^{k+1}_{\alpha}=\emptyset$ for any
$k\geqslant\rho(\delta,\beta)$ and each $\alpha\in(\beta)_{k}$. This is
because for any such $\alpha$ we have that $\rho(\alpha,\delta)\leqslant k$.
Consequently $A_{\delta}\cap C_{\beta}=^{*}\emptyset$. This finishes the
proof. ∎

The previous construction yields an almost disjoint family which, in
principle, satisfies a lot more properties than just being Luzin-Jones. In the
rest this section, we will analyse those properties. The following definitions
are essential for that task.

######  Definition 3.1.14 (Succesor-like elements).

Let $(X,\leqslant)$ be a partial order. We say that $x\in X$ is succesor-like
if:

  * $\blacksquare$

$pred(x)$ is finite,

  * $\blacksquare$

for any $y<x$ there is $z\in pred(x)$ for which $y\leqslant z$.

######  Definition 3.1.15 (Luzin representation).

Let $(X,\leqslant)$ be a partial order of size $\omega_{1}$. A Luzin
representation of $X$ is an ordered pair $(\mathcal{T},\mathcal{A})$
consisting of two families of infinite subsets of $\omega$ indexed as $\langle
T_{x}\rangle_{x\in X}$ and $\langle A_{x}\rangle_{x\in X}$ respectively.
Furthermore, $\mathcal{A}$ is a Luzin family and for all $x,y\in X$ the
following properties hold:

  1. $(a)$

$A_{x}\subseteq T_{x}.$

  2. $(b)$

If $y\not\leqslant x$, then $A_{y}\subseteq^{*}T_{y}\backslash T_{x}$.

  3. $(c)$

If $\inf(x,y)$ exists, then $T_{x}\cap T_{y}=^{*}T_{\inf(x,y)}.$

  4. $(d)$

If $x$ is succesor-like, then

$T_{x}\backslash\big{(}\bigcup\limits_{z\in pred(x)}T_{z}\big{)}=^{*}A_{x}.$

  5. $(e)$

If $x$ and $y$ are incompatible, then $T_{x}\cap T_{y}=^{*}\emptyset.$

In particular $(\mathcal{T},\subseteq^{*})$ is order isomorphic to $X$ by
virtue of the points $(b)$ and $(c)$. In the case there is
$(\mathcal{T},\mathcal{A})$ a Luzin representation of $X$ we say that
$\mathcal{A}$ codes $X$.

######  Definition 3.1.16 ($\omega_{1}$-like order).

We say that a partial order $(X,\leqslant)$ is $\omega_{1}$-like if it is
well-founded, $|X|=\omega_{1}$ and $|(-\infty,x)|\leqslant\omega$ for all
$x\in X$.

######  Proposition 3.1.17.

Let $(Y,\leqslant)$ be a partial order of cardinality $\omega_{1}$. There is a
well-founded cofinal $X\subseteq Y$ with $|(-\infty,x)_{X}|\leqslant\omega$
for each $x\in X$.

###### Proof.

Enumerate $Y$ as $\langle y_{\alpha}\rangle_{\alpha\in\omega_{1}}$. Define $X$
as the set of all $y_{\beta}$’s such that $y_{\beta}\not\leqslant y_{\alpha}$
for each $\alpha<\beta$. Of course $X$ is well-founded as for each $x,z\in X$
with $x=y_{\alpha}$ and $z=y_{\beta}$ it happens that if $x<z$ then
$\alpha<\beta$. Because of this we also have that
$(-\infty,\,y_{\beta})_{X}\subseteq\\{y_{\alpha}\,:\,\alpha<\beta\\}$ for each
$y_{\beta}\in X$. Lastly $X$ is cofinal in $Y$ because for each
$\beta\in\omega_{1}$ the element $y_{\xi_{\beta}}\in X$ where
$\xi_{\beta}=\min(\,\alpha\leqslant\beta\,:\,y_{\alpha}\geqslant
y_{\beta}\,)$. ∎

The following lemma is easy.

######  Lemma 3.1.18.

Let $(X,\leqslant)$ be $\omega_{1}$-like. Then there is a bijection
$\phi:X\longrightarrow\omega_{1}$ with $\phi(x)<\phi(y)$ for all $x<y\in X.$

Let $(X,\leqslant)$ $\omega_{1}$-like and $\phi:X\longrightarrow\omega_{1}$ as
above. In the next lemma, in some sense, we will pull back a construction
scheme from $\omega_{1}$ to some sort of construction scheme on $X$. This
suggests that the theory of construction schemes could be generalized to other
partial orders of size $\omega_{1}$. The extend and utility of this idea
remains mainly unexplored.

######  Lemma 3.1.19.

Let $\mathcal{F}$ be a $2$-construction scheme and let $X$ and $\phi$ be as in
Lemma 3.1.18. For each $x\in X$ and $k\in\omega$ define
$M^{k}_{x}=\\{\,z\leqslant x\,:\,\phi(z)\in(\phi(x))_{k}\,\\}$. The following
properties hold for each $x,y\in X$ and $k\in\omega$:

  1. $(1)$

If $\inf(x,y)$ exists and $k>\rho^{\phi[\\{x,y,\inf(x,y)\\}]}$, then
$M^{k}_{x}\cap M^{k}_{y}=M^{k}_{\inf(x,y)}.$

  2. $(\frac{3}{2})$

If $y\leqslant x$ and $k>\rho^{\phi[\\{x,y\\}]}$, then $M^{k}_{y}\subseteq
M^{k}_{x}$.

  3. $(2)$

If $y\not\leqslant x$, then $y\in M^{k}_{y}\backslash M^{k}_{x}.$

  4. $(3)$

If $x$ is succesor-like and $k>\rho^{\phi[pred(x)\cup\\{x\\}]}$, then
$M^{k}_{x}\backslash\big{(}\bigcup\limits_{z\in
pred(x)}M^{k}_{z}\big{)}=\\{x\\}$.

  5. $(4)$

If $x$ and $y$ are incompatible, then $M^{k}_{x}\cap M^{k}_{y}=\emptyset.$

###### Proof.

The points (2) and (4) are trivial and the point ($\frac{3}{2}$) follows
directly from (1). Therefore, we will only prove (1) and (3).

###### Proof of $(1)$.

Since $\phi$ preserves the order then
$\phi(\inf(x,y))\leqslant\min(\phi(x),\phi(y))$ which implies that
$(\phi(\inf(x,y))_{k}\sqsubseteq(\phi(x))_{k}\cap(\phi(y))_{k}$. In this way,
if $z\in M^{k}_{\inf(x,y)}$ then $z\leqslant x,y$ by definition of the
infimum, and $z$ belongs to both $(\phi(x))_{k}$ and $(\phi(y))_{k}$ because
$z\in(\phi(\inf(x,y)))_{k}$. Therefore $z\in M^{k}_{x}\cap M^{k}_{y}$. On the
other hand, if $z\in M^{k}_{x}\cap M^{k}_{y}$ then $z\leqslant x$ and
$z\leqslant y$, so $z\leqslant\inf(x,y)$. Furthermore,
$\phi(z)\in(\phi(x))_{k}\cap(\phi(y))_{k}$ and
$\phi(z)\leqslant\phi(\inf(x,y))$. As $(\phi(\inf(x,y)))_{k}$ is an initial
segment of such intersection then $\phi(z)$ belongs to it. Consequently $z\in
M^{k}_{\inf(x,y)}$. ∎

###### Proof of $(3)$.

Note that since $x\not\leqslant z$ for any $z\in pred(x)$ then $x\in
M^{k}_{x}\backslash M^{k}_{z}$ for any such $z$. This proves the inclusion
from right to left. To show that the one from left to right also holds, let

$w\in M^{k}_{x}\backslash\big{(}\bigcup\limits_{z\in
pred(x)}M^{k}_{z}\big{)}.$

Suppose towards a contradiction that $w<x$ and consider $z\in pred(x)$ with
$w\leqslant z$. Since $k>\rho(\phi(x),\phi(z))$ we conclude that
$(\phi(z))_{k}\sqsubseteq(\phi(x))_{k}$. But then $\phi(w)\in(\phi(z))_{k}$
because $\phi(w)\leqslant\phi(z)$ and $\phi(w)\in(\phi(x))_{k}$. Consequently
$w\in M^{k}_{z}$ which is a contradiction. This finishes the proof. ∎

∎

In the next theorem we show that not only a great variety of partial orders
posses a Luzin representation, but that the particular Luzin family that we
constructed in the Theorem 3.1.13 codes all of them. For that reason, it will
be convenient to recall some basic facts regarding such family.

We have a countable set $N$ and $\langle N_{k}\rangle_{k\in\omega}$ a
partition of $N$ into finite sets. For each $\alpha\in\omega_{1}$ we have
$A_{\alpha}^{k}\subseteq N_{k}$ and we defined
$A_{\alpha}=\bigcup\limits_{k\in\omega}A^{k+1}_{\alpha}$ and
$\mathcal{A}=\langle A_{\alpha}\rangle_{\alpha\in\omega_{1}}$. One key feature
of $\mathcal{A}$ is that:

If $\alpha,\beta\in\omega_{1}$ are distinct and $k>\rho(\alpha,\beta)$ then
$A^{k}_{\alpha}\cap A^{k}_{\beta}=\emptyset.$

######  Theorem 3.1.20.

Let $\mathcal{A}$ be the Luzin family constructed in Theorem 3.1.13 and let
$(X,\leqslant)$ an $\omega_{1}$-like order. Then there is a family
$\mathcal{T}=\langle T_{x}\rangle_{x\in X}\subseteq[\omega]^{\omega}$ and a
re-indexing of $\mathcal{A}$ as $\langle\hat{A}_{x}\rangle_{x\in X}$ so that
$(\mathcal{T},\mathcal{A})$ is a Luzin representation of $X$.

###### Proof.

Let $\phi:X\longrightarrow\omega_{1}$ be as in Lemma 3.1.18 and $M^{k}_{x}$ be
as in Lemma 3.1.19 for any $x\in X$ and $k\in\omega$. Fix $x\in X$ and let
$\hat{A}_{x}=A_{\phi(x)}.$ This defines the re-indexing of $\mathcal{A}$. Now,
for each $k\in\omega$ and $x\in X$, let

$T^{k}_{x}=\bigcup\\{\,A_{\phi(z)}^{k+1}\,:\,z\in
M^{k}_{x}\,\\}=\bigcup\\{\,A^{k+1}_{\xi}\,:\,\xi\in\phi[M^{k}_{x}]\,\\}$
$T_{x}=\bigcup\limits_{k\in\omega}T^{k}_{x}.$

We claim that $(\mathcal{T},\mathcal{A})$ is a Luzin representation of $X$. In
the following paragraphs we will show that the points $(a)$, $(b)$, $(c)$,
$(d)$ and $(e)$ of Definition 3.1.15 are satisfied for such pair.

###### Proof of $(a)$.

Given $k>0$ we have that $A^{k}_{\phi(x)}\subseteq T^{k-1}_{x}$ because $x$ is
always an element of $M^{k}_{x}$. Therefore $\hat{A}_{x}=A_{\phi(x)}\subseteq
T_{x}$. ∎

Before proving the remaining points let us fix some notation. Given
$P\in[\omega_{1}]^{<\omega}$ and $k\in\omega$, let
$A^{k}_{P}=\bigcup\\{A^{k}_{\alpha}\,:\,\alpha\in P\\}$. Note that if
$P,Q\in[\omega_{1}]^{<\omega}$ and $k>\rho^{P\cup Q}$ then, by virtue of the
key feature of $\mathcal{A}$ highlighted just before this theorem, we have the
following:

$A^{k}_{P}\cap A^{k}_{Q}=A^{k}_{P\cap Q},$ $A^{k}_{P}\cup
A^{k}_{Q}=A^{k}_{P\cup Q},$ $A^{k}_{P}\backslash A^{k}_{Q}=A^{k}_{P\backslash
Q}.$

We will apply these three equalities for finite sets of the form
$\phi[M^{k}_{x}]$. More precisely, observe that if $x,y\in X$ and
$k>\rho(\phi(x),\phi(y))$ then
$\rho^{\phi[M^{k}_{x}]\cup\phi[M^{k}_{y}]}<k+1$. This is because if we take
$F\in\mathcal{F}_{k}$ with $\phi(x),\phi(y)\in F$, then
$\phi[M^{k}_{x}]\cup\phi[M^{k}_{y}]=\phi[M^{k}_{x}\cup
M^{k}_{y}]\subseteq(\phi(x))_{k}\cup(\phi(y))_{k}\subseteq F.$

###### Proof of $(c)$.

Let $x,y\in X$ and suppose that $\inf(x,y)$ exists. Take an arbitrary
$k>\rho^{\phi[\\{x,y,\inf(x,y)\\}]}$. By means of the point $(1)$ of Lemma
3.1.19, $M^{k}_{x}\cap M^{k}_{y}=M^{k}_{\inf(x,y)}$. In this way:

$\displaystyle T^{k}_{x}\cap T^{k}_{y}=A^{k+1}_{M^{k}_{x}}\cap
A^{k+1}_{M^{k}_{y}}=A^{k+1}_{M^{k}_{x}\cap
M^{k}_{y}}=A^{k+1}_{M^{k}_{\inf(x,y)}}=T^{k}_{\inf(x,y)}.$

Therefore, $T_{x}\cap T_{y}=\bigcup\limits_{k\in\omega}\big{(}T^{k}_{x}\cap
T^{k}_{y}\big{)}=^{*}\bigcup\\{T^{k}_{\inf(x,y)}\,:\,k>\rho^{\phi[x,y,\inf(x,y)]}\\}=^{*}T_{\inf(x,y)}.$
∎

The remaining parts of the theorem are proved in a completely similar way.
Because of this, we leave the calculations to the reader. ∎

The following result greatly extends Hausdorff’s theorem about the existence
of an $(\omega_{1},\omega_{1})$-gap.

######  Theorem 3.1.21.

Let $(X,<_{X})$ and $(Y,<_{Y})$ be two partial orders with
$cof(X)=cof(Y)=\omega_{1}$. Then there are cofinal $X^{\prime}\subseteq X$ and
$Y^{\prime}\subseteq Y$ for which there is an $(X^{\prime},Y^{\prime})$-gap.

###### Proof.

Without loss of generality we may assume that $X\cap Y=\emptyset$. By applying
Proposition 3.1.17 we get $X^{\prime}\subseteq X$ and $Y^{\prime}\subseteq Y$
two $\omega_{1}$-like cofinal subsets of $X$ and $Y^{\prime}$ respectively.
Let $Z=X^{\prime}\cup Y^{\prime}$ and $<_{Z}$ be the partial order over $Z$
induced by $X^{\prime}$ and $Y^{\prime}$. That is, $x<_{Z}y$ if and only if
$x,y\in X^{\prime}$ and $x<_{X}y$ or $x,y\in Y^{\prime}$ and $x<_{Y}y$. It is
straightforward that $(Z,<_{Z})$ is an $\omega_{1}$-like order. Furthermore,
if $x\in X^{\prime}$ and $y\in Y^{\prime}$ then there is no $z\in Z$ with
$z\leqslant x,y.$ In virtue of Theorem 3.1.20, there is a Luzin representation
of $Z$, namely $(\mathcal{T},\mathcal{A})$. Due to the point (e) of Definition
3.1.15 it follows that $(\langle T_{x}\rangle_{x\in X^{\prime}},\langle
T_{y}\rangle_{y\in Y^{\prime}})$ forms an $(X^{\prime},Y^{\prime})$-pregap. We
claim that this is in fact a gap. For this, just note that any set separating
$(\langle T_{x}\rangle_{x\in X^{\prime}},\langle T_{y}\rangle_{y\in
Y^{\prime}})$ would also separate $(\langle A_{x}\rangle_{x\in
X^{\prime}},\langle A_{y}\rangle_{y\in Y^{\prime}})$ by the point (a) of
Definition 3.1.15. But this is impossible since $\mathcal{A}$ is Luzin (hence
inseparable) and both $X^{\prime}$ and $Y^{\prime}$ are uncountable. ∎

####  3.1.1 Donut-separability

In this subsection we will explore in more depth the relation between almost
disjoint families and gaps. When applying the argument of Theorem 3.1.21 to
the orders $X=Y=\omega_{1}$ we get the following result.

######  Theorem 3.1.22.

There is an $(\omega_{1},\omega_{1})$-pregap
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ such that the set

$\\{L_{\alpha+1}\backslash
L_{\alpha}\,:\,\alpha\in\omega_{1}\\}\cup\\{R_{\alpha+1}\backslash
R_{\alpha}\,:\,\alpha\in\omega_{1}\\}$

forms a Luzin family. In particular
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is a gap.

###### Proof.

Let $Z=\omega_{1}\times 2$ and define the order over $Z$ given by
$(\alpha,i)<_{Z}(\beta,j)$ if and only if $\alpha<\beta$ and $i=j$.
$(Z,<_{z})$ is an $\omega_{1}$-like order so that for any
$\alpha,\beta\in\omega_{1}$, there is no $z\in Z$ with
$z\leqslant(\alpha,0),(\beta,1)$. In particular, by Theorem 3.1.13 there is a
Luzin representation $(\mathcal{T},\mathcal{A})$ of $Z$. For any
$\alpha\in\omega_{1}$, let $L_{\alpha}=T_{(\alpha,0)}$ and
$R_{\alpha}=T_{(\alpha,1)}$. Then
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is a pregap due to the same
arguments given in Theorem 3.1.21. In this particular case, note that for any
$\alpha\in\omega_{1}$ and $i\in 2$, the set $pred((\alpha+1,i))$ satisfies the
conditions of point (d) in Definition 3.1.15. Even more,
$pred((\alpha+1,i))=\\{(\alpha,i)\\}$. In this way, $L_{\alpha+1}\backslash
L_{\alpha}=^{*}A_{(\alpha,0)}$ and $R_{\alpha+1}\backslash
R_{\alpha}=^{*}A_{(\alpha,1)}$. We conclude that

$\\{L_{\alpha+1}\backslash
L_{\alpha}\,:\,\alpha\in\omega_{1}\\}\cup\\{R_{\alpha+1}\backslash
R_{\alpha}\,:\,\alpha\in\omega_{1}\\}$

is an uncountable subset of $\mathcal{A}$. Therefore, it must also be a Luzin
family. ∎

The core idea of Theorems 3.1.21 and 3.1.22 is the following: If
$(\mathcal{L},\mathcal{R})$ is a pregap and for each $L\in\mathcal{L}$ and
$R\in\mathcal{R}$ we can find infinite sets $A_{L}\subseteq L$ and
$A_{R}\subseteq R$ such that the family
$\\{A_{L}\\}_{L\in\mathcal{L}}\cup\\{A_{R}\\}_{R\in\mathcal{R}}$ forms an
inseparable family, then the original pregap is in fact a gap. Such phenomenom
was studied in [59] by Piotr Kalemba and Szymon Plewik for the case of
$(\omega_{1},\omega_{1})$-gaps. The main difference between the gaps they
studied and the one we constructed in the previous theorem, is that the
associated Luzin family can be explicitly defined from such gap. This leads to
the following definition.

######  Definition 3.1.23 (donut-separable gaps).

We call an $(\omega_{1},\omega_{1})$-gap
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ donut-separable if there is a
set separating $(L_{\alpha+1}\backslash L_{\alpha},R_{\alpha+1}\backslash
R_{\alpha})_{\alpha\in\omega_{1}}$. If no such set exists we call the gap
donut-inseparable.

######  Corollary 3.1.24.

There is a donut-inseparable Hausdorff gap.

In contrast, there is also (in ZFC) a Hausdorff gap which is donut-separable.

######  Theorem 3.1.25.

The Hausdorff gap $(\mathcal{A},\mathcal{B})$ constructed in Theorem 3.1.10 is
donut-separable.

###### Proof.

Let $\alpha\in\omega_{1}$. Now consider an arbitrary
$k>\rho(\alpha,\alpha+1)$. Note that
$\lVert\alpha\rVert_{k}+1=\lVert\alpha+1\rVert_{k}$. By the point (c) of Lemma
2.3.9 we know that either $\Xi_{\alpha+1}(k)=\Xi_{\alpha}(k)$ or these two
numbers are different and $\Xi_{\alpha}(k)=-1$. It follows that in the latter
case $r_{k}-1=\lVert\alpha\rVert_{k}<\lVert\alpha+1\rVert_{k}=r_{k}<m_{k}$,
which means that $\Xi_{\alpha+1}(k)=0$. From this fact we conclude that

$\displaystyle L_{\alpha+1}\backslash L_{\alpha}$
$\displaystyle=^{*}\\{2k+\Xi_{\alpha+1}(k)\,:k>\rho(\alpha,\alpha+1)\textit{
and }\Xi_{\alpha}(k)\not=\Xi_{\alpha+1}(k)\\}$
$\displaystyle\subseteq\\{2k\,:k\in\omega\textit{ and
}\Xi_{\alpha+1}(k)=0\\}\subseteq\\{2k\,:k\in\omega\\}.$

In the same way, we have that $R_{\alpha+1}\backslash
R_{\alpha}\subseteq^{*}\\{2k+1\,:k\in\omega\\}.$ Hence, the set of even
numbers separates $(L_{\alpha+1}\backslash L_{\alpha},R_{\alpha+1}\backslash
R_{\alpha})_{\alpha\in\omega_{1}}$. ∎

######  Lemma 3.1.26.

Let $(\mathcal{L},\mathcal{R})$ be an $(\omega_{1},\omega_{1})$-gap and
$\mathcal{L}^{\prime}$, $\mathcal{R}^{\prime}$ be cofinal subsets of
$\mathcal{L}$ and $\mathcal{R}$ respectively. If $(\mathcal{L},\mathcal{R})$
is donut-inseparable then so is $(\mathcal{L}^{\prime},\mathcal{R}^{\prime}).$

The previous lemma motivates the following definition.

######  Definition 3.1.27 (Strongly donut-separable gaps).

Let $(\mathcal{L},\mathcal{R})$ be an $(\omega_{1},\omega_{1})$-gap. We say
that $(\mathcal{L},\mathcal{R})$ is strongly donut-separable if for any two
cofinal subsets of $\mathcal{L}$ and $\mathcal{R}$, say $\mathcal{L}^{\prime}$
and $\mathcal{R}^{\prime}$, the gap
$(\mathcal{L}^{\prime},\mathcal{R}^{\prime})$ is donut-separable.

It is natural to wonder if there is a strongly donut-separable gap. We will
prove that this statement is independent from $ZFC.$

######  Lemma 3.1.28.

Let $(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ be an
$(\omega_{1},\omega_{1})$-gap. The following statements are equivalent:

  1. $(1)$

$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is strongly donut-separable.

  2. $(2)$

For any $S\in[\omega_{1}]^{\omega_{1}}$, the gap
$(L_{\alpha},R_{\alpha})_{\alpha\in S}$ is donut-separable.

  3. $(3)$

For any club $S\subseteq\omega_{1}$, the gap
$(L_{\alpha},R_{\alpha})_{\alpha\in S}$ is donut-separable.

###### Proof.

The only nontrivial part of the proof is to show that $(3)$ implies $(1)$. For
this, let $X$ and $Y$ be uncountable subsets of $\omega_{1}$. Our goal is to
prove that there is a $C\in[\omega]^{\omega}$ which separates
$(L_{X(\alpha+1)}\backslash L_{X(\alpha)},R_{Y(\alpha+1)}\backslash
R_{Y(\alpha)})_{\alpha\in\omega_{1}}$. Let $\langle
M_{\alpha}\rangle_{\alpha\in\omega_{1}}$ be a continuous chain of elementary
submodels of a largely enough $H(\lambda)$ so that $X,Y\in M_{0}$. Now define
$S$ as $\\{0\\}\cup\\{M_{\alpha}\cap\omega_{1}\,:\,\alpha\in\omega_{1}\\}$.
Then $S$ is a club. Furthermore, by elementarity it is straightforward that
for any $\alpha\in\omega_{1}$ there is $\beta\in S$ with:

  * $\blacksquare$

$S(\beta)\leqslant X(\alpha)<X(\alpha+1)\leqslant S(\beta+1)$.

  * $\blacksquare$

$S(\beta)\leqslant Y(\alpha)<Y(\alpha+1)\leqslant S(\beta+1).$

By the hypotheses, there is $C\in[\omega]^{\omega}$ separating
$(L_{S(\alpha+1)}\backslash L_{S(\alpha)},R_{S(\alpha+1)}\backslash
R_{S(\alpha)})_{\alpha\in\omega_{1}}$. Note that if $\alpha\in\omega_{1}$ and
$\beta\in S$ are as previously stated then $L_{X(\alpha+1)}\backslash
L_{X(\alpha)}\subseteq^{*}L_{S(\beta+1)}\backslash L_{S(\beta)}\subseteq^{*}C$
and $R_{Y(\alpha+1)}\backslash
R_{Y(\alpha)}\subseteq^{*}R_{Y(\beta+1)}\backslash
R_{Y(\beta)}\subseteq^{*}\omega\backslash C$. Therefore $C$ separates the
pregap $(L_{X(\alpha+1)}\backslash L_{X(\alpha)},R_{Y(\alpha+1)}\backslash
R_{Y(\alpha)})_{\alpha\in\omega_{1}}$. This finishes the proof. ∎

######  Corollary 3.1.29.

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap which is strongly donut-separable. If
$\mathbb{P}$ is a $ccc$ forcing, then

$\mathbb{P}\Vdash\text{\say{ $(\mathcal{L},\mathcal{R})$ is strongly donut-
separable}}.$

###### Proof.

Suppose towards a contradiction that there is $G$ a $\mathbb{P}$-generic
filter over $V$ such that, in $V[G]$, $(\mathcal{L},\mathcal{R})$ is not
strongly donut-separable. According to the point (3) of Lemma 3.1.28, there is
a club $S\in V[G]$ such that the gap $(L_{\alpha},R_{\alpha})_{\alpha\in S}$
is donut-inseparable. Since $\mathbb{P}$ is a $ccc$ forcing, we can find a
club $C\in V$ for which $C\subseteq S$. is donut-inseparable and belongs to
$V$. This contradiction ends the proof. ∎

Suppose that $(\mathcal{L},\mathcal{R})$ is a pregap over $\omega$ and $C$ is
an infinite subset separating it. Define
$s:\mathcal{L}\cup\mathcal{R}\longrightarrow\omega$ as:

$s_{C}(X)=\begin{cases}\min(n\,:X\backslash n\subseteq C)&\textit{ if
}X\in\mathcal{L}\\\ \min(n\,:X\backslash n\cap C=\emptyset)&\textit{ if
}X\in\mathcal{R}\end{cases}$

Then $L\cap R\subseteq\max(s_{C}(L),s_{C}(R))$ for any $L\in\mathcal{L}$ and
$R\in\mathcal{R}.$ This motivates the following definition.

######  Definition 3.1.30 (Separating functions).

Let $(\mathcal{L},\mathcal{R})$ be a pregap. We say that
$s:\mathcal{L}\cup\mathcal{R}\longrightarrow\omega$ is separating if $L\cap
R\subseteq\max(s(L),s(R))$ for any $L\in\mathcal{L}$ and $R\in\mathcal{R}$.

######  Lemma 3.1.31.

Let $(\mathcal{L},\mathcal{R})$ be a pregap. If there is a separating
$s:\mathcal{L}\cup\mathcal{R}\longrightarrow\omega$ then there is
$C\in[\omega]^{\omega}$ which separates $(\mathcal{L},\mathcal{R})$.

###### Proof.

Let $s$ be as in the hypotheses. We define $C$ as

$\\{n\in\omega\,:\,\exists L\in\mathcal{L}\,(n\in L\textit{ and
}s(L)<n)\\}=\bigcup\limits_{L\in\mathcal{L}}L\backslash(s(L)+1).$

Note that $L\subseteq^{*}C$ for any $L\in\mathcal{L}$. For $R\in\mathcal{R}$
we claim that $R\backslash(s(R)+1)\cap C=\emptyset$ which in particular
implies that $R\cap C=^{*}\emptyset$. Suppose towards a contradiction that
this is not the case and let $n$ be an element in the intersection of both
sets. On one hand, since $n\in C$ there is $L\in\mathcal{L}$ with $n\in L$ and
$s(L)<n$. In particular, $n\in L\cap R$. On the other hand, since $s$ is
separating then $L\cap R\subseteq\max(s(R),s(L))\subseteq n$. This is a
contradiction. Therefore the claim is true, which means that $C$ separates
$(\mathcal{L},\mathcal{R})$. ∎

######  Definition 3.1.32.

Let $(\mathcal{D},\mathcal{E})=(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$
be a pregap (not necessarily of type $(\omega_{1},\omega_{1})$) so that
$D_{\alpha}\cap E_{\alpha}=\emptyset$ for each $\alpha$. We define the forcing
$\mathbb{P}(\mathcal{D},\mathcal{E})=\mathbb{P}(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$
as the set of all functions $p;\omega_{1}\longrightarrow\omega$ with finite
domain and such that $D_{\alpha}\cap
E_{\beta}\subseteq\max(p(\alpha),p(\beta))$ for all $\alpha,\beta\in dom(p)$.
The order is given by

$p\leqslant q\text{ if and only if }q\subseteq p.$

######  Remark 3.1.33.

If $(\mathcal{D},\mathcal{E})=(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$
is as in the previous definition and $\beta\in\omega_{1}$ then
$\mathcal{M}_{\beta}=\\{p\in\mathbb{P}(\mathcal{D},\mathcal{E})\,:\,\beta\in
dom(p)\,\\}$ is dense in $\mathbb{P}(\mathcal{E},\mathcal{D})$. If
$\mathcal{G}$ is a filter intersecting each $\mathcal{M}_{\beta}$ then the
function $s:\mathcal{D}\cup\mathcal{E}\longrightarrow\omega_{1}$ given by:

$s(D_{\alpha})=s(E_{\alpha})=\bigcup\mathcal{G}(\alpha)$

is well defined and separating.

The following proposition generalizes a well-known result of Kenneth Kunen (we
reiterate that our pregaps do not need to be linearly ordered).

######  Proposition 3.1.34.

Let $(E_{\alpha},D_{\alpha})_{\alpha\in\omega_{1}}$ be a pregap with
$E_{\alpha}\cap D_{\alpha}=\emptyset$ for each $\alpha$. The following
statements are equivalent:

  1. $(a)$

$\mathbb{P}=\mathbb{P}(E_{\alpha},D_{\alpha})_{\alpha\in\omega_{1}}$ is $ccc$.

  2. $(b)$

There is $\mathbb{Q}$ $ccc$ with $\mathbb{Q}\Vdash\text{\say{
$(E_{\alpha},D_{\alpha})_{\alpha\in\omega_{1}}\textit{ can be separated }$}}$.

  3. $(c)$

There is $W$ a transitive model of $ZFC$ extending $V$ with
$\omega_{1}^{W}=\omega_{1}^{V}$ where

$W\models(E_{\alpha},D_{\alpha})_{\alpha\in\omega_{1}}\textit{ can be
separated.}$

###### Proof.

Trivially $(a)$ implies $(b)$ and $(b)$ implies $(c)$. In order to prove that
$(c)$ implies $(a)$ take an arbitrary uncountable subset $\mathbb{P}$ in $V$,
say $\mathcal{A}$. For any $p\in\mathcal{A}$ let $d_{p}=dom(p)$. Without loss
of generality we can suppose that $\mathcal{A}$ is a $\Delta$-system with root
$R$, there is $n\in\omega$ so that $|p|=n$ for every $p\in\mathcal{A}$ and
$p(d_{p}(i))=q(d_{q}(i))$ for all $p,q\in\mathcal{A}$ and each $i<n$.

Let $W$ be as in the hypotheses of $(c)$ and let $C\in W$ which separates
$(E_{\alpha},D_{\alpha})_{\alpha\in\omega_{1}}$. Since
$\omega_{1}^{V}=\omega_{1}^{W}$ and $V$ models that $\mathcal{A}$ is
uncountable, we can find an uncountable
$\mathcal{A}^{\prime}\subseteq\mathcal{A}$ in $W$, so that
$E_{d_{p}(i)}\backslash C=E_{d_{q}(i)}\backslash C$ and $D_{d_{p}(i)}\cap
C=D_{d_{q}(i)}\cap C$ for all $p,q\in\mathcal{A}^{\prime}$ and $i<n$. Fix two
distinct $p,q\in\mathcal{A}^{\prime}$. We claim that $p\cup q\in\mathbb{P}$.
Indeed, let $i,j<n$. Then

$\displaystyle E_{d_{p}(i)}\cap D_{d_{q}(j)}$
$\displaystyle=((E_{d_{p}(i)}\backslash C)\cap
D_{d_{q}(j)})\cup(E_{d_{p}(i)}\cap(D_{d_{q}(j)}\cap C))$
$\displaystyle=((E_{d_{q}(i)}\backslash C)\cap
D_{d_{q}(j)})\cup(E_{d_{p}(i)}\cap(D_{d_{p}(j)}\cap C))$
$\displaystyle\subseteq\max(q(d_{q}(i)),q(d_{q}(j))\cup\max(p(d_{p}(i)),p(d_{p}(j)))$
$\displaystyle=\max(p(d_{p}(i)),p(d_{q}(j))).$

This finishes the proof. ∎

Consider the gap $(L_{\alpha},R_{\alpha})_{\alpha\in\omega}$ constructed in
Theorem 3.1.10 and for any $k\in\omega$ let $N_{k}=2(k+1)=\\{0,\dots,2k+1\\}$.
The following properties follow directly from the proof of that theorem and
the definition of the $\Delta$-function.

######  Proposition 3.1.35.

Let $\alpha<\beta\in\omega_{1}$. Then:

  1. $(0)$

$L_{\alpha}\cap R_{\alpha}=\emptyset.$

  2. $(1)$

For each $k\in\omega$, both $\\{2k,2k+1\\}\cap L_{\alpha}$ and
$\\{2k,2k+1\\}\cap R_{\alpha}$ have at most one point.

  3. $(2)$

All of the sets $L_{\alpha}\backslash L_{\beta}$, $R_{\alpha}\backslash
R_{\beta}$, $L_{\alpha}\cap R_{\beta}$ and $L_{\beta}\cap R_{\alpha}$ are
subsets of $N_{\rho(\alpha,\beta)}$.

  4. $(3)$

$L_{\alpha}\cap N_{\Delta(\alpha,\beta)-1}=L_{\beta}\cap
N_{\Delta(\alpha,\beta)-1}$ and $R_{\alpha}\cap
N_{\Delta(\alpha,\beta)-1}=R_{\beta}\cap N_{\Delta(\alpha,\beta)-1}$.

Let us briefly recall Definition 2.4.17: If $\mathcal{F}$ is a $2$-capturing
construction scheme over $\omega_{1}$, the parametrized Martin’s number
$\mathfrak{m}_{\mathcal{F}}$ associated to $\mathcal{F}$, is the minimun of
the Martin’s numbers of $ccc$ forcing notions which force $\mathcal{F}$ to be
$2$-capturing in the generic extension.
In the following theorem, we will show that the existence of
$(\omega_{1},\omega_{1})$-gaps which are strongly donut-separable is
consistent with $ZFC$. The proof we provide can be simplified by the use of
Lemma 4.0.3. We decided to keep it this way in order to motivate such lemma.

######  Theorem 3.1.36.

Let $\mathcal{F}$ be a $2$-capturing $2$-construction scheme for which
$\mathfrak{m}_{\mathcal{F}}>\omega_{1}$. The Hausdorff gap constructed from
$\mathcal{F}$ in Theorem 3.1.10 is strongly donut-separable.

###### Proof.

Let $S\in[\omega_{1}]^{\omega_{1}}$. We will show that the gap
$(L_{S(\alpha)},R_{S(\alpha)})_{\alpha\in\omega_{1}}$ is donut-separable. In
other words, we will show that pregap
$(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$ can be separated where
$D_{\alpha}=L_{S(\alpha+1)}\backslash L_{S(\alpha)}$ and
$E_{\alpha}=R_{S(\alpha+1)}\backslash R_{S(\alpha)}$ for each
$\alpha\in\omega_{1}$. This is enough due to the Lemma 3.1.28. In view of
Remark 3.1.33 and since we are assuming that
$\mathfrak{m}_{\mathcal{F}}>\omega_{1}$, it is sufficient to prove that for
$\mathbb{P}=\mathbb{P}(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$ we have:

  1. $(1)$

$\mathbb{P}$ is a $ccc$ forcing,

  2. $(2)$

$\mathbb{P}\Vdash\text{\say{ $\mathcal{F}\textit{ is }2\textit{-capturing}$
}}.$

###### Proof of $(1)$.

For this consider an arbitrary $\mathcal{A}\in[\mathbb{P}]^{\omega_{1}}$.
Given $p\in\mathcal{A}$ define $Z_{p}=\\{S(\alpha),S(\alpha+1)\,:\,\alpha\in
dom(p)\\}$. By refining $\mathcal{A}$ we can assume without loss of generality
that for any $p,q\in\mathcal{A}$ the following conditions hold:

  1. $(a)$

$|p|=|q|$,

  2. $(b)$

If $f:dom(p)\longrightarrow dom(q)$ is the increasing bijection then
$p(\alpha)=q(f(\alpha))$ for each $\alpha\in dom(p).$ In particular, there is
$k\in\omega$ with $k>\max(im(p))$ for each $p\in\mathcal{A}$.

Furthermore, we can suppose that the set $\\{dom(p)\,:\,p\in\mathcal{A}\\}$
forms a root-tail-tail $\Delta$-system with root $R$ satisfying the following
properties for any two distinct $p,q\in\mathcal{A}$:

  1. (c)

If $R\not=\emptyset$, then $\max(R)+1<\min(dom(p)\backslash R)$.

  2. (d)

$\max(dom(p))+1\not\in dom(q)$.

As the set $\\{Z_{p}\,:\,p\in\mathcal{A}\\}$ is uncountable and $\mathcal{F}$
is assumed to be $2$-capturing then there are distinct $p,q\in\mathcal{A}$ for
the set $\\{Z_{p},Z_{q}\\}$ is captured at some level $l>k$. We affirm that
$p$ and $q$ are compatible. This will follow from the next claim.

Claim 1: $p\cup q$ is a condition of $\mathbb{P}$.

###### Proof of claim.

First note that if $F\in\mathcal{F}_{l}$ is such that $Z_{p}\cup
Z_{q}\subseteq F$ then $Z_{p}\cap Z_{q}=Z_{p}\cap R(F)=Z_{q}\cap R(F)$ by
means of Lemma 2.4.5.

Now, let us consider $f:dom(p)\longrightarrow dom(q)$ and
$h:F_{0}\longrightarrow F_{1}$ the increasing bijections. In order to prove
that $p\cup q$ is a condition it is enough to take $\alpha\in dom(p)$ and
$\beta\in dom(q)$ and show that both $E_{\alpha}\cap D_{\beta}$ and
$D_{\alpha}\cap E_{\beta}$ are contained in $\max(p(\alpha),q(\beta))$. If
either $\alpha$ or $\beta$ belong to the intersection of $dom(p)\cap dom(q)$
there is nothing to do. So we can assume that $\alpha\in dom(p)\backslash
dom(q)$ and $\beta\in dom(q)\backslash dom(p)$. Then
$\\{S(\alpha),S(\alpha+1)\\}\subseteq Z_{p}\backslash Z_{q}\subseteq
F_{0}\backslash R(F)$ and $\\{S(\beta),S(\beta+1)\\}\subseteq Z_{q}\backslash
Z_{p}\subseteq F_{1}\backslash R(F)$ due to the points (c) and (d). Thus
$\rho(S(\alpha+1),S(\beta+1))=l$ and consequently $D_{\alpha}\cap
E_{\beta}\subseteq L_{S(\alpha+1)}\cap R_{S(\beta+1)}\subseteq N_{l}$ due to
the point (2) of Proposition 3.1.35. But
$\Xi_{S(\alpha)}(l)=\Xi_{S(\alpha+1)}(l)=0$. Therefore $2l\in
L_{S(\alpha)}\cap L_{S(\alpha+1)}$. In particular $2l\in L_{S(\alpha+1)}$, so
$2l+1\not\in L_{S(\alpha+1)}$ by the point (1) of Proposition 3.1.35. In this
way $\\{2l,2l+1\\}\cap D_{\alpha}=\emptyset$. Thus,

$D_{\alpha}\cap E_{\beta}\subseteq N_{l-1}.$

The next thing to note is that $h(S(\alpha))=S(f(\alpha))$ and
$h(S(\alpha+1))=S(f(\alpha)+1).$ This means that
$\Delta(S(\alpha),S(f(\alpha)))=l=\Delta(S(\alpha+1),S(f(\alpha)+1))$. In
virtue of the point (3) of proposition 3.1.35 we have

$L_{S(\alpha)}\cap N_{l-1}=L_{S(f(\alpha))}\cap N_{l-1},$ $L_{S(\alpha+1)}\cap
N_{l-1}=L_{S(f(\alpha)+1)}\cap N_{l-1}.$

Hence, $D_{\alpha}\cap N_{l-1}=D_{f(\alpha)}\cap N_{l-1}$. From all the
equations we have so far we deduce that $D_{\alpha}\cap
E_{\beta}=D_{f(\alpha)}\cap
E_{\beta}\subseteq\max(q(f(\alpha)),q(\beta))=\max(p(\alpha),q(\beta)).$ In a
completely similar way we can show that $E_{\alpha}\cap
D_{\beta}\subseteq\max(p(\alpha),q(\beta))$. We conclude that $p\cup
q\in\mathbb{P}$ so we are done. ∎

∎

###### Proof of $(2)$.

For this we will show that the equivalence of $2$-capturing stated in Lemma
2.4.15 is forced by $\mathbb{P}$. Let $\dot{X}$ be a name so that
$\mathbb{1}_{\mathbb{P}}\Vdash\text{\say{
$\dot{X}\in[\omega_{1}]^{\omega_{1}}$}}$ and let $q\in\mathbb{P}$. We shall
find $p\leqslant q$ so that

$p\Vdash\text{\say{ $\exists M\in[\dot{X}]^{2}\,(\,M\textit{ is captured })$
}}.$

As $\dot{X}$ is forced to be uncountable we can find for each
$\alpha\in\omega_{1}$ a condition $p_{\alpha}\leqslant q$ and
$\xi_{\alpha}>\alpha$ with $p_{\alpha}\Vdash\text{\say{
$\xi_{\alpha}\in\dot{X}$ }}.$ We can suppose without loss of generality that
the family $\mathcal{A}=\\{p_{\alpha}\,:\,\alpha\in\omega_{1}\\}$ satisfies
the conditions (a), (b), (c) and (d) of the previous paragraphs. We may also
assume that for any distinct $\alpha,\beta\in\omega_{1}:$

  1. $(d)$

$\xi_{\alpha}\not=\xi_{\beta}$ and
$|Z_{p_{\alpha}}\cup\\{\xi_{\alpha}\\}|=|Z_{p_{\beta}}\cup\\{\xi_{\beta}\\}|$,

  2. $(e)$

If $g:Z_{p_{\alpha}}\cup\\{\xi_{\alpha}\\}\longrightarrow
Z_{p_{\beta}}\cup\\{\xi_{\beta}\\}$ is the increasing bijection then
$g[Z_{p_{\alpha}}]=Z_{p_{\beta}}$ and $g(\xi_{\alpha})=\xi_{\beta}.$

Since $\mathcal{F}$ is $2$-capturing there are $\alpha<\beta\in\omega_{1}$ for
which
$\\{Z_{p_{\alpha}}\cup\\{\xi_{\alpha}\\},Z_{p_{\beta}}\cup\\{\xi_{\beta}\\}\\}$
is captured at some level $l$ greater than $k$. In virtue of the conditions
(d) and (e) it is easy to see that $\\{Z_{p_{\alpha}},Z_{p_{\beta}}\\}$ and
$\\{\xi_{\alpha},\xi_{\beta}\\}$ are also captured at level $l$. As in the
first part of the proof, this implies that $p=p_{\alpha}\cup p_{\beta}$ is a
condition of $\mathbb{P}$. To finish, just note that $p\leqslant q$ and
$p\Vdash\text{\say{ $\\{\xi_{\alpha},\xi_{\beta}\\}\in[\dot{X}]^{2}\textit{
and }\\{\xi_{\alpha},\xi_{\beta}\\}\textit{ is captured }$}}.$ This finishes
the proof. ∎

∎

######  Definition 3.1.37 (Adequate sets).

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap. We say that
$X\in[\omega_{1}]^{\omega_{1}}$ is adequate for $(\mathcal{L},\mathcal{R})$ if
for any $Y\in[\omega_{1}]^{\omega_{1}}$, the pregap

$(L_{X(\alpha+1)}\backslash L_{X(\alpha)},R_{X(\alpha+1)}\backslash
R_{X(\alpha)})_{\alpha\in Y}$

is a gap.

######  Remark 3.1.38.

If an $(\omega_{1},\omega_{1})$-pregap admits an adequate set, then such
pregap is not strongly donut-separable.

######  Theorem 3.1.39 (Under $CH$).

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap. Then there is a club
$X\in[\omega_{1}]^{\omega_{1}}$ adequate for $(\mathcal{L},\mathcal{R}).$ In
particular, there are no strongly-donut separable gaps.

###### Proof.

Since we are assuming $CH$ we can enumerate $[\omega]^{\omega}$ as $\langle
C_{\alpha}\rangle_{\alpha\in\omega_{1}}$. We will build $X$ be recursion in
such way that for any $\beta\in\omega_{1}$ and each $\alpha\leqslant\beta$ one
of the following conditions occur:

(A)

$L_{X(\beta+1)}\backslash L_{X(\beta)}\not\subseteq^{*}C_{\alpha}.$

(B)

$(R_{X(\beta+1)}\backslash R_{X(\beta)})\cap C_{\alpha}\not=^{*}\emptyset.$

If $\gamma$ is limit and we have constructed $X(\alpha)$ for any
$\alpha<\gamma$ we just define $X(\gamma)$ as
$\sup(X(\alpha)\,:\,\alpha<\gamma)$. The interesting case happens when we have
constucted $X(\beta)$ for some $\beta\in\omega_{1}$ and we want to define
$X(\beta+1).$ For this case, we have the following claim:

Claim: The pregap $(L_{\delta}\backslash L_{X(\beta)},R_{\delta}\backslash
R_{X(\beta)})_{\delta>X(\beta)}$ is a gap.

###### Proof of claim.

Suppose towards a contradiction that this is not the case and let $C$ be a set
separating $(L_{\delta}\backslash L_{X(\beta)},R_{\delta}\backslash
R_{X(\beta)})_{\delta>X(\beta)}$. Then $C\cup L_{X(\beta)}$ separates
$(\mathcal{L},\mathcal{R})$ which is a contradiction. ∎

Fix $\alpha\leqslant\beta$. By the previous claim, $C_{\alpha}$ does not
separate $(L_{\delta}\backslash L_{X(\beta)},R_{\delta}\backslash
R_{X(\beta)})_{\delta>X(\beta)}$. In this way, there is
$\delta_{\alpha}>X(\beta)$ so that either $L_{\delta_{\alpha}}\backslash
L_{X(\beta)}\not\subseteq^{*}C_{\alpha}$ or $R_{\delta_{\alpha}}\cap
C_{\alpha}\not=^{*}\emptyset.$ Let us define $X(\beta+1)$ as
$\sup(\delta_{\alpha}\,:\alpha\leqslant\beta)$. Then
$L_{\delta_{\alpha}}\backslash L_{X(\beta)}\subseteq L_{X(\beta+1)}\backslash
L_{X(\beta)}$ and $R_{\delta_{\alpha}}\backslash R_{X(\beta)}\subseteq
R_{X(\beta+1)}\backslash R_{X(\beta)}$ for each $\alpha\leqslant\beta$. In
this way we guarantee that either contition (A) or condition (B) will ocurr
for any such $\alpha.$ This finishes the recursion.

We will now prove that $X$ is as desired. Let $Y\in[\omega_{1}]^{\omega_{1}}$.
We need to show that $(L_{X(\alpha+1)}\backslash
L_{X(\alpha)},R_{X(\alpha+1)}\backslash R_{X(\alpha)})_{\alpha\in Y}$ is a
gap. For this let $C$ be an infinite subset of $\omega$. By the assumptions we
know there is $\alpha\in\omega_{1}$ so that $C=C_{\alpha}$. Let $\beta\in Y$
be such that $\beta>\alpha$. Therefore, either $L_{X(\beta+1)}\backslash
L_{X(\beta)}\not\subseteq^{*}C_{\alpha}$ or $(R_{X(\beta+1)}\backslash
R_{X(\beta)})\cap C_{\alpha}\not=^{*}\emptyset.$ In any case, $X(\beta)$
testifies that $C_{\alpha}$ does not separate the gap that we are considering.
This finishes the proof. ∎

By combining the previous result and Theorem 3.1.36, we conclude the
following:

######  Theorem 3.1.40.

The statement There is a strongly donut-separable gap is independent from
$ZFC.$

Now we will analyze the relationship between donut-separable gaps and the
Cohen forcing.

######  Lemma 3.1.41.

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap and let $X\in[\omega_{1}]^{\omega_{1}}$.
If $X$ is adequate for $(\mathcal{L},\mathcal{R})$ then
$\mathbb{C}\Vdash\text{\say{ $X\textit{ is adequate for
}(\mathcal{L},\mathcal{R})$}}.$

###### Proof.

Let $G$ be a $\mathbb{C}$-generic filter over $V$ and let
$Y\in[\omega_{1}]^{\omega_{1}}\cap V[G]$. Suppose towards a contradiction that
there is $C\in V[G]$ separating the pregap

$(L_{X(\alpha+1)}\backslash L_{X(\alpha)},R_{X(\alpha+1)}\backslash
R_{X(\alpha)})_{\alpha\in Y}.$

Since $Y$ is uncountable we can find $a,b\in[\omega]^{<\omega}$ and
$Y^{\prime}\in[Y]^{\omega_{1}}\cap V[G]$ so that for any $\alpha\in
Y^{\prime}$, $(L_{X(\alpha+1)}\backslash L_{X(\alpha)})\backslash C=a.$ Due to
well-known facts concerning the Cohen forcing $\mathbb{C}$, we know there is
$Z\in[\omega_{1}]^{\omega_{1}}\cap V$ with $Z\subseteq Y$. Let us define

$C^{\prime}=\bigcup\limits_{\alpha\in Z}L_{X(\alpha+1)}\backslash
L_{X(\alpha)}.$

Then $C^{\prime}\in V$ and trivially $L_{X(\alpha+1)}\backslash
L_{X(\alpha)}\subseteq C^{\prime}$ for any $\alpha\in Z$. Furthermore,
$C^{\prime}\subseteq C\cup a$. Therefore, $(R_{X(\alpha+1)}\backslash
R_{X(\alpha)})\cap C^{\prime}$ is finite for any $\alpha\in Z.$ We conclude
that, in $V$, there is a set separating the gap $(L_{X(\alpha+1)}\backslash
L_{X(\alpha)},R_{X(\alpha+1)}\backslash R_{X(\alpha)})_{\alpha\in Z}.$ This is
a contradiction to the hypotheses. Hence, the proof is over. ∎

######  Theorem 3.1.42 (Under $CH$).

Let $\kappa$ be an uncountable cardinal. Then

$\mathbb{C}_{\kappa}\Vdash\text{\say{ There are no strongly donut-separable
$(\omega_{1},\omega_{1})$-gaps }}.$

In particular, the statement There are no strongly donut-separable
$(\omega_{1},\omega_{1})$-gaps is consistent with an arbitrarily large
continuum.

###### Proof.

If $\kappa=\omega_{1}$ the argument is clear, since in this case
$\mathbb{C}_{\kappa}$ forces $CH$. So suppose that $\kappa>\omega_{1}$ and let
$G$ be a $\mathbb{C}_{\kappa}$-generic filter over $V$. Finally, let
$(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}\in
V[G]$ be an arbitrary $(\omega_{1},\omega_{1})$-gap. Since
$|\mathcal{L}|=|\mathcal{R}|=\omega_{1}$, then there is a
$\mathbb{C}_{\omega_{1}}$-generic filter over $V$, namely $H$, and a
$\mathbb{C}_{\kappa}$-generic filter over $V[H]$, namely $K$, so that
$V[H][K]=V[G]$ and $(\mathcal{L},\mathcal{R})\in V[H].$ By the hypotheses, $V$
is a model of $CH$. Therefore $V[H]$ models $CH$ too. According to Theorem
3.1.39 there is $X\in V[H]$ which is adequate for $(\mathcal{L},\mathcal{R}).$
We finish by proving the following claim.

Claim: $X$ testifies that $(\mathcal{L},\mathcal{R})$ is not strongly donut-
separable.

###### Proof of claim..

Suppose towards a contradiction that there is $C\in V[G]$ separating the
pregap

$(L_{X(\alpha+1)}\backslash L_{X(\alpha)},R_{X(\alpha+1)}\backslash
R_{X(\alpha)})_{\alpha\in\omega_{1}}.$

Again, since $|C|=\omega$ we can find $\mathbb{C}$-generic filter over $V[H]$,
say $H^{\prime}$, for which $C\in V[H][H^{\prime}]$. By virtue of Lemma
3.1.41, $V[H][H^{\prime}]\models X\textit{ is adequate for
}(\mathcal{L},\mathcal{R}).$ In particular $C$ can not separate the pregap
that we are considering. This contradiction finishes the proof. ∎

∎

As $\mathfrak{m}_{\mathcal{F}}>\omega_{1}$ is consistent with $\mathfrak{c}$
arbitrarily large, we conclude that both the statement there is a strongly
donut-separable gap and its negation are consistent with an arbitrarily large
continuum.
Our next goal is to show that $PFA$ also implies that there are no strongly
donut-separable gaps. For this sake, we need to consider the following
definitions.

######  Definition 3.1.43 (Almost Luzin gaps).

Let $(\mathcal{D},\mathcal{E})=(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$
be pregap (not necessarily of type $(\omega_{1},\omega_{1}))$. We say that
$(\mathcal{D},\mathcal{E})$ is almost Luzin if
$(\mathcal{D}^{\prime},\mathcal{E}^{\prime})$ is a gap for any two uncountable
$\mathcal{D}^{\prime}\subseteq\mathcal{D}$ and
$\mathcal{E}^{\prime}\subseteq\mathcal{E}$.

######  Definition 3.1.44 (Highly adequate sets).

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap. We say that
$X\in[\omega_{1}]^{\omega_{1}}$ is highly adequate for
$(\mathcal{L},\mathcal{R})$ if the pregap

$(L_{X(\alpha+1)}\backslash L_{X(\alpha)},R_{X(\alpha+1)}\backslash
R_{(X(\alpha)})_{\alpha\in X}$

is almost Luzin.

######  Remark 1.

If $X$ is highly adequate for $(\mathcal{L},\mathcal{R})$ then it is also
adequate.

It is very easy check that the dichotomy satisfied by club $X$ that we
constructed in Theorem 3.1.39 already turns $X$ into a highly adequate set for
the gap $(\mathcal{L},\mathcal{R})$. Hence, we have the following proposition.

######  Proposition 3.1.45 (Under $CH$).

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap. Then there is a club
$X\in[\omega_{1}]^{\omega_{1}}$ highly adequate for
$(\mathcal{L},\mathcal{R})$.

######  Definition 3.1.46 (Normal pregaps).

Suppose that $(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$ is a pregap. We
say that it is normal if $D_{\alpha}\cap E_{\alpha}=\emptyset$ for each
$\alpha\in\omega_{1}.$

######  Remark 3.1.47.

Suppose that $(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$ is a pregap and
consider, for each $\alpha$, $D^{\prime}_{\alpha}=D_{\alpha}\backslash
E_{\alpha}$ and $E^{\prime}_{\alpha}=E_{\alpha}\backslash D_{\alpha}$. Then
$(D^{\prime}_{\alpha},E^{\prime}_{\alpha})_{\alpha\in\omega_{1}}$ is a normal
pregap. Furthermore, both gaps induce the same sets in
$\mathscr{P}(\omega)/\text{FIN}$. In particular,
$(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$ is strongly donut-separable if
and only if the same holds for
$(D^{\prime}_{\alpha},E^{\prime}_{\alpha})_{\alpha\in\omega_{1}}$.

######  Definition 3.1.48 (Biorthogonal gaps).

Let $(\mathcal{D},\mathcal{E})=(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$
be a normal pregap. We say that $(\mathcal{D},\mathcal{E})$ is biorthogonal if
$(D_{\alpha}\cap E_{\beta})\cup(D_{\beta}\cap E_{\alpha})\not=\emptyset$ for
all $\alpha\not=\beta\in\omega_{1}$.

######  Lemma 3.1.49.

If $(\mathcal{D},\mathcal{E})=(E_{\alpha},D_{\alpha})_{\alpha\in\omega_{1}}$
is a normal biorthogonal pregap, then it is a gap.

###### Proof.

Suppose towards a contradiction that there is $C\in[\omega]^{\omega}$ which
separates $(\mathcal{D},\mathcal{E})$. According to the pigheonhole principle,
there is $X\in[\omega_{1}]^{\omega_{1}}$ for which $D_{\alpha}\backslash
C=D_{\beta}\backslash C$ and $E_{\alpha}\cap C=E_{\beta}\cap C$ for all
$\alpha,\beta\in X.$ Fix two distinct $\alpha,\beta\in X$. Then

$\displaystyle D_{\alpha}\cap E_{\beta}$
$\displaystyle=(D_{\alpha}\cap(E_{\beta}\cap
C))\cup(E_{\beta}\cap(D_{\alpha}\backslash C))$
$\displaystyle=(D_{\alpha}\cap(E_{\alpha}\cap
C))\cup(E_{\beta}\cap(D_{\beta}\backslash C))=\emptyset$

By symmetry, we also have that $D_{\beta}\cap E_{\alpha}=\emptyset$, but this
is a contradiction to the normality of $(\mathcal{D},\mathcal{E})$. Thus, the
proof is over. ∎

######  Definition 3.1.50.

Let $(\mathcal{D},\mathcal{E})=(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$
be a normal gap. We define the forcing $\mathbb{L}(\mathcal{D},\mathcal{E})$
as the set of all $p\in[\omega_{1}]^{<\omega}$ so that $(D_{\alpha}\cap
E_{\beta})\cup(D_{\beta}\cap E_{\alpha})\not=\emptyset$ for all
$\alpha\not=\beta\in dom(p).$ The order is given by

$p\leqslant q\text{ if and only if }q\subseteq p.$

######  Proposition 3.1.51.

Let $(\mathcal{D},\mathcal{E})=(D_{\alpha},E_{\alpha})_{\alpha\in\omega_{1}}$
be a normal gap. If $(\mathcal{D},\mathcal{E})$ is almost Luzin, then
$\mathbb{L}(\mathcal{D},\mathcal{E})$ is $ccc$.

###### Proof.

Let $\mathcal{A}$ be an uncountable subset of
$\mathbb{L}(\mathcal{D},\mathcal{E})$. We will show that $\mathcal{A}$ is not
an antichain. By refining $\mathcal{A}$, we may assume without loss of
generality that there are $n,m\in\omega$ for which the following properties
hold for each $p,q\in\mathcal{A}$:

  1. $(1)$

$|p|=n$.

  2. $(2)$

For all $i,j<n$, $(D_{p(i)}\cap E_{p(j)})\cup(D_{p(j)}\cap E_{p(i)})\subseteq
m$.

  3. $(3)$

For all $i<n$, $D_{p(i)}\cap m=D_{q(i)}\cap m$ and $D_{p(i)}\cap
m=D_{q(i)}\cap m$.

Note that for any two given conditions in
$p,q\in\mathbb{L}(\mathcal{D},\mathcal{E})$, we have that $p$ and $q$ are
compatible if and only if $p\backslash q$ and $q\backslash p$ are compatible
conditions. Because of this and due to the $\Delta$-system Lemma, we may also
assume that the elements of $\mathcal{A}$ are pairwise disjoint. Now, we
proceed to find distinct $p,q\in\mathcal{A}$ for which $p\cup q$ is a
condition. This will be done after proving the following claims.

Claim 1: Let $p,q\in\mathcal{A}$ and $i\not=j<n$, then $(D_{p(i)}\cap
E_{q(j)})\cup(D_{q(j)}\cap E_{p(i)})\not=\emptyset.$

###### Proof of claim.

Just note that by the conditions (2) and (3), we have the following chain of
equalities:

$\displaystyle((D_{p(i)}\cap E_{q(j)})\cup(D_{q(j)}\cap E_{p(i)}))\cap m$
$\displaystyle=(D_{p(i)}\cap(E_{q(j)}\cap m))\cup((D_{q(j)}\cap m)\cap
E_{p(i)})$ $\displaystyle=(D_{p(i)}\cap(E_{p(j)}\cap m))\cup((D_{p(j)}\cap
m)\cap E_{p(i)})$ $\displaystyle=(D_{p(i)}\cap E_{p(j)})\cup(D_{p(j)}\cap
E_{p(i)})\not=\emptyset.$

This proves the claim. ∎

Claim 2: Let $\mathcal{A}^{\prime}\in[\mathcal{A}]^{\omega_{1}}$ and $i<n$.
Then there is $\mathcal{X}\in[\mathcal{A}^{\prime}]^{\omega_{1}}$ so that for
any $k\in\omega$, if $\\{p\in\mathcal{X}\,:\,k\in D_{p(i)}\\}$ is non-empty,
then it is uncountable. Analogously with the set $\\{p\in\mathcal{X}\,:\,k\in
E_{p(i)}\\}$.

###### Proof of claim.

Let $M$ be a countable elementary submodel of $H(\omega_{2})$ such that
$\mathcal{A}^{\prime}\in M$. We put
$\mathcal{X}=\mathcal{A}^{\prime}\backslash M$. Note that if $k\in\omega$ is
such that $k\in D_{q(i)}$ for some $q\in\mathcal{X}$, then the set
$\\{p\in\mathcal{A}^{\prime}\,:\,k\in D_{p(i)}\\}$ is an element of $M$ which
is not contained in it. By elementarity, it follows that this set is
uncountable. Therefore $\\{p\in\mathcal{X}\,:\,k\in
D_{p(i)}\\}=\\{p\in\mathcal{A}^{\prime}\,:\,k\in D_{p(i)}\\}\backslash M$ is
uncountable as well. The same argument holds for the set
$\\{p\in\mathcal{X}\,:\,k\in E_{p(i)}\\}$. Hence, the proof is over. ∎

Claim 3: Let $i<n$ and $\mathcal{X}$, $\mathcal{Y}$ be uncountable disjoint
subsets of $\mathcal{A}$ satisfying the conclusions of the Claim 2 when
applied to $i$. Then there are
$\mathcal{X}^{\prime}\in[\mathcal{X}]^{\omega_{1}}$ and
$\mathcal{Y}^{\prime}\in[\mathcal{X}]^{\omega_{1}}$ so that

$(D_{p(i)}\cap E_{q(i)})\cup(D_{q(i)}\cap E_{q(i)})\not=\emptyset$

for all $p\in\mathcal{X}^{\prime}$ and $q\in\mathcal{Y}^{\prime}$.

###### Proof of claim..

By the hypotheses of the proposition, we have that $(\langle
D_{p(i)}\rangle_{p\in\mathcal{X}},\langle E_{q(i)}\rangle_{q\in\mathcal{Y}})$
is a gap. In particular, there is $k\in\omega$, $p^{\prime}\in\mathcal{X}$ and
$q^{\prime}\in\mathcal{Y}$ for which $k\in D_{p^{\prime}(i)}\cap
E_{q^{\prime}(i)})\cup(D_{q^{\prime}(i)}\cap E_{q^{\prime}(i)})$. Without loss
of generality we may assume that $k\in D_{p^{\prime}(i)}\cap
E_{q^{\prime}(i)}$. We define

$\mathcal{X}^{\prime}=\\{p\in\mathcal{X}\,:,k\in D_{p(i)}\\},$
$\mathcal{Y}^{\prime}=\\{q\in\mathcal{Y}\,:\,k\in E_{q(i)}\\}.$

It is straightforward that $\mathcal{X}^{\prime}$ and $\mathcal{Y}^{\prime}$
are the sets that we are looking for. ∎

By applying multiple times the claims 2 and 3, we may build two sequences
$\mathcal{X}_{n-1}\subseteq\dots\subseteq\mathcal{X}_{0}\subseteq\mathcal{A}$
and
$\mathcal{Y}_{n-1}\subseteq\dots\subseteq\mathcal{Y}_{0}\subseteq\mathcal{A}$
of uncountable sets for which $\mathcal{X}\cap\mathcal{Y}=\emptyset$ and such
that $(D_{p(i)}\cap E_{q(i)})\cup(D_{q(i)}\cap E_{q(i)})\not=\emptyset$ for
all $i<n$, $p\in\mathcal{X}_{i}$ and $q\in\mathcal{Y}_{i}$. Using Claim 1, it
should be clear that if $p\in\mathcal{X}_{n-1}$ and $q\in\mathcal{Y}_{n-1}$,
then $p\cup q$ is a condition of $\mathbb{L}(\mathcal{D},\mathcal{E})$. Thus,
we are done.

∎

######  Corollary 3.1.52.

Let $(\mathcal{D},\mathcal{E})$ be an almost Luzin gap. Then there is a $ccc$
forcing $\mathbb{Q}$ so that $\mathbb{Q}\Vdash\text{\say{
$(\mathcal{D},\mathcal{E})$ has a biorthogonal subgap.}}$

######  Theorem 3.1.53 (Under $PFA$).

There are no strongly donut-separable $(\omega_{1},\omega_{1})$-gaps

###### Proof.

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap and consider the forcing
$\mathbb{P}=2^{<\omega_{1}}$. According to the Remark 3.1.47, we may assume
that $(\mathcal{L},\mathcal{R})$ is normal. Now, let $G\subseteq\mathbb{P}$ be
a $\mathbb{P}$-generic filter over $V$. Then $V[G]$ models $CH$. Furthermore,
$[\omega]^{\omega}\cap V=[\omega]^{\omega}\cap V[G]$. In this way,
$(\mathcal{L},\mathcal{R})$ is still a gap in $V[G]$. According to the
Proposition 3.1.45, we can find a club $X\subseteq\omega_{1}$ in $V[G]$ which
is highly adequate for $(\mathcal{L},\mathcal{R})$. That is, in $V[G]$, the
gap $(L_{X(\alpha+1)}\backslash L_{X(\alpha)},R_{X(\alpha+1)}\backslash
R_{X(\alpha)})_{\alpha\in\omega_{1}}$ is almost Luzin. Therefore, by virtue of
the Corollary 3.1.52 there is a $ccc$ forcing $\mathbb{Q}$ which forces this
gap to have a biorthogonal subgap. Let $H$ be a $\mathbb{Q}$-generic filter
over $V[G]$. Then, in $V[G][H]$, there is $Y\in[\omega_{1}]^{\omega_{1}}$ so
that the pregap

$(\langle L_{X(\alpha+1)}\backslash L_{X(\alpha)}\rangle,\langle
R_{X(\alpha+1)}\backslash R_{X(\alpha)}\rangle)_{\alpha\in Y}$

is biorthogonal.
Returning to $V$, let $\dot{\mathbb{Q}}$ be a $\mathbb{P}$ name for
$\mathbb{Q}$ in $V[G]$. Now, let $\dot{X}$ and $\dot{Y}$ be
$\mathbb{P}*\dot{Q}$-names for $X$ and $Y$ respectively. Finally, let
$p\in\mathbb{P}*\dot{\mathbb{Q}}$ which forces $\dot{X}$ and $\dot{Y}$ to have
the properties discussed in the previous paragraph. Given $\xi\in\omega_{1}$,
let

$D_{\xi}=\\{q\leqslant
p\>\,:\exists\xi<\alpha,\beta\in\omega_{1}\,(q\Vdash\text{\say{
$\alpha\in\dot{X}$, and $\beta\in\dot{Y}$)}}\,\\},$ $E_{\xi}=\\{q\leqslant
p\,:\,p\Vdash\text{\say{$\xi\not\in\dot{X}$}}\textit{ or there is
}\alpha\in\omega_{1}\textit{ such that
}p\Vdash\text{\say{$\dot{X}(\alpha)=\xi$}}\\}.$

It is straightforward that $D_{\xi}$ and $E_{\xi}$ are dense below $p$ for any
$\xi\in\omega_{1}$. Since $\mathbb{P}$ is $\sigma$-closed and
$\dot{\mathbb{Q}}$ is forced to be $ccc$, then $\mathbb{P}*\dot{\mathbb{Q}}$
is proper. In this way, there is a filter $F$ which intersects each $D_{\xi}$
and $E_{\xi}$. Let

$X^{\prime}=\\{\alpha\,:\,\exists q\in
F\,(q\Vdash\text{\say{$\alpha\in\dot{X}$}}\\},$
$Y^{\prime}=\\{\alpha\,:\,\exists q\in
F\,(q\Vdash\text{\say{$\alpha\in\dot{Y}$}}\\}.$

Note that the statement

$(L_{X^{\prime}(\alpha+1)}\backslash L_{X^{\prime}(\alpha)}\cap
R_{X^{\prime}(\beta+1)}\backslash
R_{X^{\prime}(\beta)})\cup(R_{X^{\prime}(\alpha+1)}\backslash
R_{X^{\prime}(\alpha)}\cap L_{X^{\prime}(\beta+1)}\backslash
L_{X^{\prime}(\beta)})=\emptyset$

is absolute. From this, it easily follows that that the pregap

$(L_{X^{\prime}(\alpha+1)}\backslash
L_{X^{\prime}(\alpha)},R_{X^{\prime}(\alpha+1)}\backslash
R_{X^{\prime}(\alpha)})_{\alpha\in Y}$

is biorthogonal. In particular, it is a gap by Lemma 3.1.49. Form this, it
follows directly that $(\mathcal{L},\mathcal{R})$ is not strongly donut-
separable.

∎

So far we have seen that both $CH$ and $PFA$ imply that the non-existence of
strongly donut-separable gaps. On the other hand, the existence of a
$2$-capturing construction scheme $\mathcal{F}$ for which
$\mathfrak{m}_{\mathcal{F}}>\omega_{1}$ implies the existence of a strongly
donut-separable gap. We will end this subsection by proving that $MA$ is not
sufficient to decide this problem. This result suggests that the cardinal
invariant $\mathfrak{m}_{\mathcal{F}}$ should not be treated just as a sort of
weakening of $MA$, but rather as an interesting principle on its own.

######  Theorem 3.1.54.

$MA$ is independent from the statement there are strongly donut-inseparable
gaps.

###### Proof.

Since $PFA$ implies that there are no strongly donut-inseparable gaps, then
$MA$ is consistent with that same statement. In order to show the consistency
of the negation, we start with a model $V$ in which there is a strongly donut-
inseparable gap. We now consider a $ccc$ forcing $\mathbb{P}$ which forces
$MA$ and $G$ a $\mathbb{P}$-generic filter over $V$. Then $V[G]$ is a model
$MA$. Furthermore, by means of the Corollary 3.1.29, $V[G]$ also models the
existence of a strongly donut-inseparable gap. Thus, the proof is over.

∎

####  3.1.2 The gap cohomology group

In the subsection we naturally extend the work done by Daniel Talayco in
[105]. In particular, we study the size of groups defined in a similar context
as Talayco and prove that, under certain assumptions, these groups are as big
as possible.

######  Definition 3.1.55 ($*$-lower semi-lattice).

Let $X$ be an infinite set. We say that a family
$\mathcal{T}\subseteq[X]^{\omega}$ is a $*$-lower semi-lattice if for any
$A,B\in\mathcal{T}$ there is $C\in\mathcal{T}$ so that $A\cap B=^{*}C.$ In
other words, $\\{[A]\,:\,A\in\mathcal{T}\\}$ is a lower semi-lattice in
$\mathscr{P}(X)/\text{FIN}$.

######  Remark 2.

Towers are particular cases of $*$-lower semi-lattices.

######  Definition 3.1.56 (Coherent Subsystems).

Let $\mathcal{T}$ be a $*$-lower semi-lattice over a countable set $X$. A
function $g:\mathcal{T}\longrightarrow\mathscr{P}(X)$ is said to be a coherent
subsystem of $\mathcal{T}$ if for any $A,B\in\mathcal{T}$, the following
happens:

  1. $(1)$

$g(A)\subseteq A$,

  2. $(2)$

if $B\subseteq^{*}A$ then $g(A)\cap B=^{*}g(B)$.

We say that $g$ is trivial if there is $C\in\mathscr{P}(X)$ so that $C\cap
A=^{*}g(A)$ for any $A\in\mathcal{T}$. In this case, we say that $C$
trivializes $g$. The set of all coherent subsystems of $\mathcal{T}$ is
denoted as $C(\mathcal{T})$ and the set of all trivial coherent subsystem is
denoted as $Tr(\mathcal{T}).$

######  Definition 3.1.57.

Let $\mathcal{T}$ be a $*$-lower semilattice over a countable set $X$ and $g$
be a coherent subsystem of $\mathcal{T}$. We denote the sequence
$(g(A),A\backslash g(A))_{A\in\mathcal{T}}$ as $\mathcal{C}(\mathcal{T},g).$

The following lemma is easy. We prove it just to emphasise the reason of
defining coherent subsystems just for $*$-lower semi-lattices.

######  Lemma 3.1.58.

Let $\mathcal{T}$ be a $*$-lower semi-lattice and let $g$ be a coherent
subsystem of $\mathcal{T}$. Then $\mathcal{C}(\mathcal{T},g)$ is a pregap.
Furthermore, $g$ is non-trivial if and only if $\mathcal{C}(\mathcal{T},g)$
forms a gap.

###### Proof.

Let $A,B\in\mathcal{T}$. We need to prove that $g(A)\cap(B\backslash
g(B))=^{*}\emptyset.$ For this let $C\in\mathcal{T}$ be so that $C=^{*}A\cap
B$. Then

$\displaystyle g(A)\cap B=^{*}(g(A)\cap A)\cap B=^{*}g(A)\cap
C=^{*}g(C)=^{*}g(B)\cap C\subseteq^{*}g(B).$

This finishes the argument. Now we will prove the proposed equivalence.

###### Proof of $\Rightarrow$.

We will prove this by contra-positive. Suppose that there is a set $C$
separating the pregap $\mathcal{C}(\mathcal{T},g)$. For any $A\in\mathcal{T}$
we have that $g(A)\subseteq^{*}C$ and $C\cap(A\backslash g(A))=^{*}\emptyset.$
In this way, $C\cap A=^{*}g(A)$. We conclude that $C$ trivializes $g$. ∎

###### Proof of $\Leftarrow$.

Again, by contra-positive. Suppose that there is $C$ which trivializes $g$.
Then $C\cap A=^{*}g(A)$ for any $A\in\mathcal{T}$. In particular
$g(A)\subseteq^{*}C$ and $C\cap(A\backslash g(A))=\emptyset$. We conclude that
$C$ separates $\mathcal{C}(\mathcal{T},g)$. ∎

∎

######  Definition 3.1.59.

Let $\mathcal{T}$ be a $*$-lower semi-lattice over a set $X$. For $f,g\in
C(\mathcal{T})$ we can define $f\cdot g\in C(\mathcal{T})$ given as:

$(f\cdot g)(A)=f(A)\Delta g(A).$

The following lemma is easy.

######  Lemma 3.1.60.

$C(\mathcal{T})$ is an abelian (Boolean)222A group $(G,\cdot,e)$ is Boolean if
$g\cdot g=e$ for any $g\in G$. Equivalently, $G$ is a vector space over
$\mathbb{Z}_{2}$. In particular, the isomorphism type of $G$ is completely
determined by its cardinality. group with respect to this operation and
$Tr(\mathcal{T})$ is a subgroup of it.

Since $C(\mathcal{T})$ is abelian, then $Tr(\mathcal{T})$ is a normal
subgroup. Therefore, we can consider the quotient induced by it.

######  Definition 3.1.61 (Gap cohomology group).

We define the gap cohomology group of $\mathcal{T}$ as the quotient of these
two groups. That is,

$G(\mathcal{T})=C(\mathcal{T})/Tr(\mathcal{T}).$

We say that $f,g\in C(\mathcal{T})$ are cohomologous if the class of $f$ is
equal to the class of $g$ inside $G(\mathcal{T})$, or equivalently, if $f\cdot
g\in Tr(\mathcal{T})$.

######  Remark 3.1.62.

If $\mathcal{T}$ has size $\omega_{1}$, then $|G(\mathcal{T})|\leqslant
2^{\omega_{1}}.$

In [105], Daniel E. Talayco defined the gap cohomology group for the
particular case of $\omega_{1}$-towers. In there he showed that the
$2^{\omega}\leqslant|G(\mathcal{T})|$ for any $\omega_{1}$-tower
$\mathcal{T}$. In order to do that, he proved the following Theorem.

######  Theorem 3.1.63 (The $\aleph_{0}$ gap theorem).

let $\mathcal{T}=\langle T_{\alpha}\rangle_{\alpha\in\omega_{1}}$ be an
$\omega_{1}$-tower. There is a sequence of functions $\langle
f_{\alpha}\rangle_{\alpha\in\omega_{1}}$ with the following properties:

  * $\blacksquare$

$\forall\alpha\in\omega_{1}\,(\,f_{\alpha}:T_{\alpha}\longrightarrow\omega\,)$,

  * $\blacksquare$

$\forall\alpha<\beta\in\omega_{1}\,(\,f_{\alpha}|_{T_{\alpha}\cap
T_{\beta}}=^{*}f_{\beta}|_{T_{\alpha}\cap T_{\beta}}\,),$

  * $\blacksquare$

$\forall
m<n\in\omega\,(\,(f_{\alpha}^{-1}[\\{m\\}],f_{\alpha}^{-1}[\\{n\\}]\,)_{\alpha\in\omega_{1}}\textit{
is a Hausdorff gap }).$

He also proved that the $\Diamond$-principle implies that the size of this
group is always $2^{\omega_{1}}$. Later, Stevo Todorčević noted that the
$\aleph_{0}$ gap theorem already gives you the previous conclusion without
assuming any extra axioms (see Theorem 32 in [106]). Namely, if
$2^{\omega}=2^{\omega_{1}}$ the result is clear due to Talayco’s calculations.
On the other hand, if $2^{\omega}<2^{\omega_{1}}$ then the result follows from
an easy counting argument involving the quotient structure of
$G(\mathcal{T}).$ In [80], Charles Morgan constructed an $\omega_{1}$-tower
whose gap cohomology group can be explicitly calculated. This construction was
carried through the use of morasses. Finally, in [36], Ilijas Farah improved
the $\aleph_{0}$ gap theorem. This was achieved by considering the following
concept.

######  Definition 3.1.64 (Coherent families of functions).

A coherent family of functions supported by an $\omega_{1}$-tower $\langle
T_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ is a family of
functions $\langle f_{\alpha}\rangle_{\alpha\in\omega_{1}}$ such that:

  1. $(1)$

$\forall\alpha\in\omega_{1}\backslash\omega\,(\,f_{\alpha}:T_{\alpha}\longrightarrow\alpha\,),$

  2. $(2)$

$\forall\alpha,\beta\in\omega_{1}\backslash\omega\,(\,f_{\alpha}|_{T_{\alpha}\cap
T_{\beta}}=^{*}f_{\beta}|_{T_{\alpha}\cap T_{\beta}}\,).$

Given such family, we define $L^{\xi}_{\alpha}$ as
$f^{-1}_{\alpha}[\\{\xi\\}]$ for all $\xi\in\omega_{1}$ and $\alpha>\xi$.
Additionally, we let $\mathcal{L}^{\xi}=\langle
L^{\xi}_{\alpha}\rangle_{\alpha>\xi}$

######  Lemma 3.1.65.

If $\langle f_{\alpha}\rangle_{\omega_{1}\backslash\omega}$ is as in the
previous definition and $\xi<\mu\in\omega_{1}$, the pair
$(L^{\xi}_{\alpha},L^{\mu}_{\alpha})_{\alpha\in\omega_{1}\backslash\mu}$ is a
pregap. We will denote it as $(\mathcal{L}^{\xi},\mathcal{L}^{\mu})$ although,
in principle, $\mathcal{L}^{\xi}=\langle L^{\xi}_{\alpha}\rangle_{\alpha>\xi}$
instead of $\langle L^{\xi}_{\alpha}\rangle_{\alpha>\mu}.$

######  Definition 3.1.66.

Let $\mathfrak{F}=\langle
f_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ be a coherent family
of functions supported by an $\omega_{1}$-tower. We say that $\mathfrak{F}$ is
Luzin (respectively Hausdorff) if $(\mathcal{L}^{\xi},\mathcal{L}^{\mu})$ is
Luzin (respectively Hausdorff) for each $\xi<\mu\in\omega_{1}$.

Ilijas Farah constructed a Haudorff coherent family of functions supported by
an $\omega_{1}$-tower by forcing it and then appealing to Keisler’s
completeness Theorem for $L^{\omega}(Q)$. Unlike Talayco’s result, the proof
idea behind the construction of Farah does not work for any
$\omega_{1}$-tower. This is due to the nature of Keisler’s Theorem.

Here we give a direct construction of a Luzin coherent family of functions
with the use of a $2$-construction scheme. No previous direct construction was
known.

######  Theorem 3.1.67.

There is a Luzin coherent family of functions supported by an
$\omega_{1}$-tower.

###### Proof.

Let $\mathcal{F}$ be a $2$-construction scheme. First we define
$\omega_{1}$-tower over the countable set

$N=\bigcup\limits_{k\in\omega}N_{k}$

where each $N_{k}$ is equal to $\\{k\\}\times k\times r_{k}\times r_{k}.$
Given $\alpha\in\omega_{1}\backslash\omega$ we define $T_{\alpha}$ as

$\bigcup\limits_{\Xi_{\alpha}(k)\geqslant
0}N_{k}=\\{(k,s,i,j)\in\omega^{4}\,:\,\Xi_{\alpha}(k)\geqslant 0\textit{ and
}(s,i,j)\in k\times r_{k}\times r_{k}\,\\}.$

Note that for each $\alpha<\beta\in\omega_{1}$ and every
$k>\rho(\alpha,\beta)$ we have that $T_{\alpha}\cap N_{k}\subseteq
T_{\beta}\cap N_{k}$ due to the part (c) of Lemma 2.3.9. Consequently,
$T_{\alpha}\subseteq^{*}T_{\beta}$. In this way $\mathcal{T}=\langle
T_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ is a tower.

Now we define a Luzin coherent family of functions supported by $\mathcal{T}$.
Let $\alpha\geqslant\omega$. Note that if $x\in T_{\alpha}$ then $x=(k,s,i,j)$
where $k>0$, $\lVert\alpha\rVert_{k}>r_{k}$ and $i,j<r_{k}$ (In particular,
$(\alpha)_{k}(i)$ and $(\alpha)_{k}(j)$ are defined). In this way, we can
define $f_{\alpha}:T_{\alpha}\longrightarrow\alpha$ as follows:

$f_{\alpha}(k,s,i,j)=\begin{cases}(\alpha)_{k}(i)&\textit{ if
}\Xi_{\alpha}(k)=0\\\ (\alpha)_{k}(j)&\textit{ if
}\Xi_{\alpha}(k)=1\end{cases}$

By definition, the family $\langle
f_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ satisfies the point
(1) of Definition 3.1.64. The two following claims will finish the proof.

Claim 1: $\langle f_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$
satisfies the point (2) of Definition 3.1.64.

###### Proof of claim.

Let $\alpha<\beta\in\omega_{1}\backslash\omega$ and $(k,s,i,j)\in
T_{\alpha}\cap T_{\beta}$ be such that $k>\rho(\alpha,\beta)$. By definition
of $T_{\alpha}$ and $T_{\beta}$ it follows that both $\Xi_{\alpha}(k)$ and
$\Xi_{\beta}(k)$ are non-negative numbers. In virtue of the part (c) of Lemma
2.3.9, it must happen that $\Xi_{\alpha}(k)=\Xi_{\beta}(k)$. Furthermore,
$(\alpha)_{k}\sqsubseteq(\beta)_{k}$ so $(\alpha)_{k}(i)=(\beta)_{k}(i)$ and
$(\alpha)_{k}(j)=(\beta)_{k}(j)$. Hence,
$f_{\alpha}(k,s,i,j,s)=f_{\beta}(k,s,i,j)$. Since all but finitely many
elements of $T_{\alpha}\cap T_{\beta}$ have their first coordinate bigger than
$\rho(\alpha,\beta)$, we have shown that $f_{\alpha}|_{T_{\alpha}\cap
T_{\beta}}=^{*}f_{\beta}|_{T_{\alpha}\cap T_{\beta}}$. ∎

Claim 2: $\langle f_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ is
Luzin.

###### Proof of claim.

Let $\xi<\mu\in\omega_{1}$, $\beta>\mu$ and $n\in\omega$. We need to prove
that the set

$\\{\alpha\in\beta\backslash\mu\,:\,|L^{\xi}_{\alpha}\cap
L^{\mu}_{\beta}|\leqslant n\\}$

is finite. We claim that this set is contained in $(\beta)_{l}$ where
$l=\max(n,\rho^{\\{\xi,\mu,\beta\\}})$. Indeed, take an
$\alpha\in\beta\backslash\mu$ such that $\alpha\notin(\beta)_{l}$ and let
$k=\rho(\alpha,\beta)$ (so obviously $l<k$). According to the part (b) of
Lemma 2.3.9, $\Xi_{\alpha}(k)=0$ and $\Xi_{\beta}(k)=1.$ Furthermore, since
$\alpha\notin(\beta)_{l}$ then $k\geqslant\rho^{\\{\xi,\mu,\beta\\}}$. This
means that both $\xi$ and $\mu$ belong to $(\alpha)_{k}$. Hence

$\Xi_{\xi}(k)\leqslant\Xi_{\mu}(k)\leqslant\Xi_{\alpha}(k)=0<\Xi_{\beta}(k).$

By virtue of the part (c) of Lemma 2.3.9, $\Xi_{\mu}(k)=\Xi_{\xi}(k)=-1$. In
other words, $\lVert\xi\rVert_{k}$ and $\lVert\mu\rVert_{k}$ are numbers
strictly smaller than $r_{k}$. Moreover,
$(\alpha)_{k}(\lVert\xi\rVert_{k})=(\beta)_{k}(\lVert\xi\rVert_{k})=\xi$ and
$(\alpha)_{k}(\lVert\mu\rVert_{k})=(\beta)_{k}(\lVert\mu\rVert_{k})=\xi$. From
this it follows that

$\\{k\\}\times
k\times\\{\lVert\xi\lVert_{k}\\}\times\\{\lVert\mu\lVert_{k}\\}\subseteq
f^{-1}_{\alpha}[\\{\xi\\}]\cap f^{-1}_{\beta}[\\{\mu\\}]=L^{\xi}_{\alpha}\cap
L^{\mu}_{\beta}.$

As this set has cardinality $k$ (which is bigger than $n$) we are done. ∎

∎

The gap cohomology group of $\omega_{1}$-towers with a Luzin coherent family
of functions can be explicitly calculated using Talayco’s ideas. We present
here such calculations for the sake of completeness.

######  Proposition 3.1.68.

Let $\mathcal{T}=\langle T_{\alpha}\rangle_{\alpha\in\omega_{1}}$ be an
$\omega_{1}$-tower and suppose there is a Luzin coherent family of functions
$\langle f_{\alpha}\rangle_{\alpha\in\omega_{1}}$ supported by $\mathcal{T}$.
Then $|G(\mathcal{T})|=2^{\omega_{1}}$.

###### Proof.

For each $S\in\mathscr{P}(\omega_{1})$, we define a coherent subsystem
$g_{S}:\mathcal{T}\longrightarrow\mathscr{P}(\omega)$ as follows:

$g_{S}(T_{\alpha})=f^{-1}_{\alpha}[S].$

Note that if $S,S^{\prime}\in\mathscr{P}(\omega_{1})$ then $g_{S}\cdot
g_{S^{\prime}}=g_{S\Delta S^{\prime}}$. Let
$\mathcal{S}\subseteq\mathscr{P}(\omega_{1})$ be a family of size
$2^{\omega_{1}}$ so that $S\Delta S^{\prime}$ and
$\omega_{1}\backslash(S\Delta S^{\prime})$ are infinite for any two distinct
$S,S^{\prime}\in\mathcal{S}$. Given such $S$ and $S^{\prime}$ we claim that
$g_{S}$ and $g_{S^{\prime}}$ are not cohomologous. Indeed, let $\xi\in S\Delta
S^{\prime}$ and $\mu\in\omega_{1}\backslash(S\Delta S^{\prime})$ with
$\xi<\mu$. Then $L^{\xi}_{\alpha}\subseteq g_{S\Delta S^{\prime}}(T_{\alpha})$
and $L^{\mu}_{\alpha}\subseteq g_{\omega_{1}\backslash(S\Delta
S^{\prime})}(T_{\alpha})=T_{\alpha}\backslash g_{S\Delta
S^{\prime}}(T_{\alpha})$ for any $\alpha>\mu$. By hypothesis we have that
$(L^{\xi}_{\alpha},L^{\mu}_{\alpha})_{\alpha\in\omega_{1}}$ is a Luzin-gap.
This implies that $G(\mathcal{T},g_{S\Delta S^{\prime}})$ is a gap as well.
That is, $g_{S\Delta S^{\prime}}=g_{S}\cdot g_{S^{\prime}}$ is a non-trivial
coherent subsystem of $\mathcal{T}$. We have proved that for any two distinct
$S,S^{\prime}\in\mathcal{S}$, $g_{S}$ and $g_{S^{\prime}}$ are not
cohomologous. Since $|S|=2^{\omega_{1}}$, then
$|G(\mathcal{T})|=2^{\omega_{1}}$. ∎

As a second application of the concept of Luzin representations, we use
Theorem 3.1.20 to generalize the results of Talayco, Todorčević, Morgan and
Farah regarding the existence of big gap cohomology groups. It is worth noting
that, as for now, it is unclear if there is an analogous to the $\aleph_{0}$
gap theorem for aribrary $*$-lower semi-lattices. Therefore, we can not argue
in the same way as Todorčević in order to conclude that the gap cohomology
group has always cardinality $2^{\omega_{1}}$.

######  Theorem 3.1.69.

Let $(X,<,\wedge)$ be an $\omega_{1}$-like lower semi-lattice. Then there is a
$*$-lower semi-lattice $\mathcal{T}$ isomorphic to $X$ with
$|G(\mathcal{T})|=2^{\omega_{1}}.$

###### Proof.

Consider $X\times 2$ with the order given by $(x,i)<(y,j)$ if and only if
$x<y$ and $i=j$. It is easy to see that $X\times 2$ is also $\omega_{1}$-like.
In virtue of Theorem 3.1.20 we can take $\mathcal{T^{\prime}}=\langle
T^{\prime}_{(x,i)}\rangle_{(x,i)\in X\times 2}$ and $\mathcal{A}=\langle
A_{(x,i)}\rangle_{(x,i)\in X\times 2}$ so that
$(\mathcal{T}^{\prime},\mathcal{A})$ is a Luzin representation of $X\times 2$.
For any $x\in X$ let $T_{x}=T^{\prime}_{(x,0)}\cup T^{\prime}_{(x,1)}.$ It is
straightforward that the following properties hold for any $x,y\in X$:

  1. $(a)$

$A_{(x,0)}\cup A_{(x,1)}\subseteq T_{x}$.

  2. $(b)$

If $y\not\leqslant x$ then $A_{(y,0)}\cup
A_{(y,1)}\subseteq^{*}T_{y}\backslash T_{x}$. In other words, $(A_{(y,0)}\cup
A_{(y,1)})\cap T_{x}=^{*}\emptyset$.

  3. $(c)$

$T_{x}\cap T_{y}=^{*}T_{x\wedge y}.$

In particular, properties (b) and (c) imply that $\mathcal{T}=\langle
T_{x}\rangle_{x\in X}$ is a $*$-lower semi-lattice isomorphic to $X$. We claim
that $|G(\mathcal{T})|=2^{\omega_{1}}$. Trivially $|G(\mathcal{T})|\leqslant
2^{\omega_{1}}$, so we will only prove the other inequality. For this, let
$S\in\mathscr{P}(X)$. . We will build recursively (using that $X$ is well-
founded) a coherent subsystem
$g_{S}:\mathcal{T}\longrightarrow\mathscr{P}(\omega)$ such that
$A_{(x,0)}\subseteq g_{S}(T_{x})$ for any $x\in X$ and:

(A)

If $x\in S$, then $A_{(x,1)}\cap g_{S}(T_{x})=^{*}\emptyset.$

(B)

If $x\not\in S$, then $A_{(x,1)}\subseteq g_{S}(T_{x})$.

Suppose that $y\in X$ and we have defined $g_{S}(T_{x})$ for any $x<y$. By
means of Lemma 3.1.58 we know that
$\mathcal{G}(\mathcal{T}|_{y},g_{S}|_{\mathcal{T}|_{y}})$ is pregap where
$\mathcal{T}|_{y}=\langle T_{x}\rangle_{x<y}$. As there are no countable
pregaps and $|(-\infty,y)|\leqslant\omega$, we conclude that there is
$C\in\mathscr{P}(\omega)$ separating it. Since $g_{S}(T_{x})\subseteq
T_{x}\subseteq^{*}T_{y}$ for any $x<y$, we may assume without loss of
generality that $C\subseteq T_{y}$. Note that $C\cap T_{x}=^{*}g_{S}(T_{x})$
for any $x<y$. We now define $g_{S}(T_{y})$ by cases. If $y\in S$ define
$g_{S}(T_{y})$ as $(C\cup A_{(y,0)})\backslash A_{(y,1)}$. Otherwise define
$g_{S}(T_{y})$ as $C\cup A_{(y,0)}\cup A_{(y,1)}$. Since both $A_{(y,0)}$ and
$A_{(y,1)}$ are almost disjoint with $T_{x}$ for all $x<y$ due to the point
(b), it follows that $g_{S}(T_{y})\cap T_{x}=^{*}C\cap T_{x}=^{*}g_{S}(T_{x})$
for any such $x$. This finishes the recursion.

Now consider $\mathcal{S}\subseteq[X]^{\omega_{1}}$ a family of cardinality
$2^{\omega_{1}}$ so that $S\Delta S^{\prime}$ is uncountable for any two
distinct $S,S^{\prime}\in\mathcal{S}$. The key fact needed to finish the proof
is that if $x\in S\Delta S^{\prime}$ then $A_{(x,1)}\subseteq g_{S}\cdot
g_{S^{\prime}}(T_{x})$ and $A_{(x,0)}\subseteq T_{x}\backslash(g_{S}\cdot
g_{S^{\prime}}(T_{x}))$. In this way, any separation of the pregap
$\mathcal{G}(\mathcal{T},g_{S}\cdot g_{S^{\prime}})$ would also separate the
pregap $(A_{(x,1)},A_{(x,0)})_{x\in S\Delta S^{\prime}}$ which is impossible
since $\mathcal{A}$ is a Luzin family and $S\Delta S^{\prime}$ is uncountable.
This shows that $g_{S}$ and $g_{S^{\prime}}$ are not cohomologous. Since
$|S|=2^{\omega_{1}}$ then $|G(\mathcal{T})|=2^{\omega_{1}}$. ∎

####  3.1.3 Destructibility of gaps

In this subsection we will study $(\omega_{1},\omega_{1})$-gaps in the context
of forcing. The reader can find fairly complete treatments of this subject in
[13], [91] and [113].

The following fundamental lemma is attributed to Kenneth Kunen. We will prove
it for the sake of completeness. For others proofs see [37], [91] and [113].

######  Lemma 3.1.70.

Let $(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ be an
$(\omega_{1},\omega_{1})$-pregap with $L_{\alpha}\cap R_{\alpha}=\emptyset$
for any $\alpha$. Then $(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is a
gap if and only if for any $S\in[\omega_{1}]^{\omega_{1}}$ there are
$\alpha<\beta\in S$ such that $(L_{\alpha}\cap R_{\beta})\cup(L_{\beta}\cap
R_{\alpha})\not=\emptyset$.

###### Proof.

###### Proof of $\Rightarrow$.

Assume that $(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is a gap and let
$S\in[\omega_{1}]^{\omega_{1}}$. Suppose towards a contradiction that
$(L_{\alpha}\cap R_{\beta})\cup(L_{\beta}\cap R_{\alpha})=\emptyset$ for all
$\alpha,\beta\in S$. Let $C=\bigcup\limits_{\beta\in S}L_{\beta}$. Given
$\alpha\in\omega_{1}$ there is $\beta\in S$ such that $\alpha<\beta$. Note
that $L_{\alpha}\subseteq^{*}L_{\beta}\subseteq C$ and $R_{\alpha}\cap
C\subseteq^{*}R_{\beta}\cap C=\emptyset$. In this way $C$ separates
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ which is a contradiction.
Thus, the proof of this implication is over. ∎

###### Proof of $\Leftarrow$.

Again, by contradiction. Suppose that
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is not a gap and let $C$ be a
set which separates it. By the pigeonhole principle we can find
$S\in[\omega_{1}]^{\omega_{1}}$ and $a,b\in[\omega]^{<\omega}$ such that
$L_{\alpha}\backslash C=a$ and $R_{\alpha}\cap C=b$ for any $\alpha\in S$.
Note that if $\alpha\in S$ then $L_{\alpha}\cap b=\emptyset$ because
$b\subseteq R_{\alpha}$ and $L_{\alpha}\cap R_{\alpha}=\emptyset$. Therefore,
$L_{\alpha}\cap R_{\beta}\subseteq((C\cup a)\backslash b)\cap
R_{\beta}=\emptyset$ whenever $\alpha,\beta\in S$. With this contradiction we
finish the proof. ∎

∎

######  Remark 3.1.71.

For the rest of this section we will assume that if
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is an
$(\omega_{1},\omega_{1})$-pregap then it is normal. That is, $L_{\alpha}\cap
R_{\alpha}=\emptyset$ any $\alpha.$

######  Definition 3.1.72 (Destructible gaps).

Let $(\mathcal{L},\mathcal{R})$ be an $(\omega_{1},\omega_{1})$-pregap. We say
that $(\mathcal{L},\mathcal{R})$ is destructible if there is a forcing notion
$\mathbb{P}$ which preserves $\omega_{1}$ in such way that
$(\mathcal{L},\mathcal{R})$ is not a gap in some generic extension through
$\mathbb{P}$. If this does not happen, the gap is said to be indestructible.

######  Remark 3.1.73.

The Hausdorff condition is absolute. Hence any Hausdorff gap is
indestructible. Note that under $PID$ any $(\omega_{1},\omega_{1})$-gap is
undestructible. This is due to the Remark 3.1.8.

In Definition 3.1.32 we defined the forcing
$\mathbb{P}(\mathcal{D},\mathcal{E})$ where $(\mathcal{D},\mathcal{E})$ is a
pregap (not necesarily of type $(\omega_{1},\omega_{1})$) with
$|\mathcal{D}|=|\mathcal{E}|=\omega_{1}$. According to Proposition 3.1.34,
this forcing is $ccc$ if and only if the pregap $(\mathcal{D},\mathcal{E})$
can be separated in some $\omega_{1}$-preserving extension of the universe.
Forcings with such properties have already been studied for the case of
$(\omega_{1},\omega_{1})$-gaps. In the following definition we present some
well-known reincarnations of them (see [25], [91], [121] or [133]).

######  Definition 3.1.74.

Let $(\mathcal{L},\mathcal{R})$ be an $(\omega_{1},\omega_{1})$-pregap indexed
as $(L_{\alpha},R_{\alpha})_{\alpha\in X}$ where $X$ is an uncountable set of
ordinals. We define the following forcing notions:

  * $\blacksquare$

$\chi_{0}(\mathcal{L},\mathcal{R})=\\{p\in[X]^{<\omega}\,:\,\big{(}\bigcup\limits_{\alpha\in
p}L_{\alpha}\big{)}\cap\big{(}\bigcup\limits_{\alpha\in
p}R_{\alpha}\big{)}=\emptyset\\}.$

  * $\blacksquare$

$\chi_{1}(\mathcal{L},\mathcal{R})=\\{p\in[X]^{<\omega}\,:\,\forall\alpha\not=\beta\in
p\,\big{(}(L_{\alpha}\cap R_{\beta})\cup(L_{\beta}\cap
R_{\alpha})\not=\emptyset\big{)}\\}.$

both ordered by reverse inclusion.

######  Lemma 3.1.75.

Let $(\mathcal{L},\mathcal{R})$ be an $(\omega_{1},\omega_{1})$-pregap indexed
as $(L_{\alpha},R_{\alpha})_{\alpha\in X}$ where $X$ is an uncountable set of
ordinals. Then $(\mathcal{L},\mathcal{R})$ is a gap if and only if
$\mathbb{P}=\chi_{1}(\mathcal{L},\mathcal{R})$ is $ccc$.

###### Proof.

First suppose that $(\mathcal{L},\mathcal{R})$ is a gap and consider
$\mathcal{A}\in[\mathbb{P}]^{\omega_{1}}$. By refining $\mathcal{A}$ we can
assume that the following conditions hold:

  * $\blacksquare$

There is $n\in\omega$ such that $|p|=n$ for any $p\in\mathcal{A}$.

  * $\blacksquare$

There are $\sigma,\tau:n\times n\longrightarrow[\omega]^{<\omega}$ so that for
any $i\leqslant j\in n$, $L_{p(i)}\backslash L_{p(j)}=\sigma(i,j)=\sigma(j,i)$
and $R_{p(i)}\backslash R_{p(j)}=\tau(i,j)=\tau(j,i)$.

  * $\blacksquare$

There is $\mu:n\times n\longrightarrow[\omega]^{<\omega}$ so that for any
$(i,j)\in n\times n$, $L_{p(i)}\cap R_{q(j)}=\mu(i,j)$.

  * $\blacksquare$

$\mathcal{A}$ forms a root-tail-tail $\Delta$-system with a root $R$ of
cardinality $r<n$.

By means of Lemma 3.1.70 we know that there are distinct $p,q\in\mathcal{A}$
so that

$(L_{p(r)}\cap R_{q(r)})\cup(L_{q(r)}\cap R_{p(r)})\not=\emptyset.$

Claim: $p\cup q\in\mathbb{P}$.

###### Proof of claim.

It is enough to take $i,j<n$ and show that $(L_{p(i)}\cap
R_{q(j)})\cup(L_{q(j)}\cap R_{p(i)})\not=\emptyset.$ If $i<r$ then $p(i)=q(i)$
so the previous set is equal to $(L_{q(i)}\cap R_{q(j)})\cup(L_{q(j)}\cap
R_{q(i)})$ which by hypothesis is non-empty because $q\in\mathbb{P}$. An
analogous argument applies when $j<r$ so let us assume that $r\leqslant i,j$.
But if $i\not=j$ then

$\displaystyle\emptyset$ $\displaystyle\not=(L_{p(i)}\cap
R_{p(j)})\cup(L_{p(j)}\cap R_{p(i)})=\mu(i,j)\cup\mu(j,i)$
$\displaystyle\subseteq(L_{p(i)}\cap R_{q(j)})\cup(L_{q(j)}\cap R_{p(i)}).$

So we might as well assume that $i=j$. In this case we have that

$(L_{p(r)}\cap R_{q(r)})\cap\tau(r,i)\subseteq(L_{p(r)}\cap R_{q(r)})\cap
R_{p(r)}\subseteq R_{p(r)}\cap L_{p(r)}=\emptyset,$ $(L_{p(r)}\cap
R_{q(r)})\cap\sigma(r,i)\subseteq R_{q(r)}\cap L_{q(r)}=\emptyset.$

Therefore $L_{p(r)}\cap R_{q(r)}\subseteq R_{q(r)}\backslash\tau(r,i)\subseteq
R_{q(i)}$ and $L_{p(r)}\cap R_{q(r)}\subseteq
L_{p(r)}\backslash\sigma(r,i)\subseteq L_{p(i)}$. That is, $L_{p(r)}\cap
R_{q(r)}\subseteq L_{p(i)}\cap R_{q(j)}$. In the same way we can show that
$L_{p(r)}\cap R_{q(r)}\subseteq L_{q(i)}\cap R_{p(i)}$ so

$\emptyset\not=(L_{p(r)}\cap R_{q(r)})\cup(L_{q(r)}\cap
R_{p(r)})\subseteq(L_{p(i)}\cap R_{q(i)})\cup(L_{q(i)}\cap R_{p(i)}).$

∎

This finishes the proof of this implication. The other one is direct from
Lemma 3.1.70. ∎

######  Lemma 3.1.76.

Let $(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ be an
$(\omega_{1},\omega_{1})$-pregap indexed as
$(L_{\alpha},R_{\alpha})_{\alpha\in X}$ where $X$ is an uncountable set of
ordinal. Then $\mathbb{P}=\chi_{0}(\mathcal{L},\mathcal{R})$ is $ccc$ if and
only if for any $S\in[X]^{\omega_{1}}$ there are $\alpha<\beta$ such that
$(L_{\alpha}\cap R_{\beta})\cup(L_{\beta}\cap R_{\alpha})=\emptyset$.

###### Proof.

###### Proof of $\Leftarrow$.

Take $S\in[X]^{\omega_{1}}$. Then $\mathcal{A}=\\{\\{\alpha\\}\,:\,\alpha\in
S\\}$ is an uncountable subset of $\mathbb{P}$. By hypothesis there are
distinct $\alpha,\beta\in S$ so that $\\{\alpha\\}$ and $\\{\beta\\}$ are
compatible. It is easy to see that $(L_{\alpha}\cap
R_{\beta})\cup(L_{\beta}\cap R_{\alpha})=\emptyset$. ∎

###### Proof of $\Rightarrow$.

Let $\mathcal{A}\in[\mathbb{P}]^{\omega_{1}}$. We can suppose without loss of
generality that $\mathcal{A}$ satisfies the same conditions as in Lemma 3.1.75
for some $\sigma,\tau,\mu$ and $R$. Note that the set
$S=\\{p(n-1)\,:\,p\in\mathcal{A}\\}$ is uncountable so by the hypotheses we
can find distinct $p$ and $q$ so that

$(L_{p(n-1)}\cap R_{q(n-1)})\cup(L_{q(n-1)}\cap R_{p(n-1)})=\emptyset.$

Again, we claim that $p\cup q\in\mathbb{P}$. For this it suffices to prove
that if $i,j<n$ then $L_{p(i)}\cap R_{q(j)}=\emptyset.$ Indeed, for any such
$i,j$ it happens that

$L_{p(i)}\subseteq L_{p(n-1)}\cup\sigma(i,n-1)\subseteq L_{p(n-1)}\cup
L_{q(i)}.$

Therefore $L_{p(i)}\cap(R_{q(n-1)}\subseteq L_{p(n-1)})\cap
R_{q(n-1)}\cup(L_{q(i)}\cap R_{q(n-1)})=\emptyset$. In an analogous way
$R_{q(j)}\subseteq R_{q(n-1)}\cup R_{p(j)}$. Hence, we conclude that

$L_{p(i)}\cap R_{q(j)}\subseteq(L_{p(i)}\cap R_{q(n-1)})\cup(L_{p(i)}\cap
L_{p(j)})=\emptyset.$

∎

∎

The following theorem can be found in [25], [91], [121] and [133].

######  Theorem 3.1.77.

Let $(\mathcal{A},\mathcal{B})$ be an $(\omega_{1},\omega_{1})$-pregap:

  * $\blacksquare$

$\chi_{1}(\mathcal{L},\mathcal{R})$ is $ccc$ if and only if
$(\mathcal{L},\mathcal{R})$ is a gap. In this case, there is some condition in
$\chi_{1}(\mathcal{L},\mathcal{R})$ forcing $(\mathcal{L},\mathcal{R})$ to be
indestructible.

  * $\blacksquare$

$\chi_{0}(\mathcal{L},\mathcal{R})$ is $ccc$ if and only if
$(\mathcal{L},\mathcal{R})$ is destructible. In this case, there is some
condition in $\chi_{0}(\mathcal{L},\mathcal{R})$ forcing
$(\mathcal{L},\mathcal{R})$ to be separated.

A particular instance of destrucible gaps are the ones satifying the following
condition.

######  Definition 3.1.78 (Todorčević condition).

Let $(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ be an
$(\omega_{1},\omega_{1})$-pregap. We say that
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ is Todorčević if for any
$S\in[\omega_{1}]^{\omega_{1}}$ there are $\alpha<\beta$ such that
$L_{\alpha}\subseteq L_{\beta}$ and $R_{\alpha}\subseteq R_{\beta}$.

Fulgencio Lopez and Stevo Todorčević showed that the existence of Todorčević
gaps follows from $CA_{3}$ and $CA_{2}(part)$. Here we give another proof of
such result based on the construction given in Theorem 3.1.10.

######  Theorem 3.1.79 (Under $CA_{2}(part)$).

Let $\mathcal{F}$ be a $\mathcal{P}$-$2$-capturing $2$-construction scheme
with $\mathcal{P}=\\{P_{0},P_{1}\\}$. Also, consider
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$ the Hausdorff gap constructed
in Theorem 3.1.10. Let

$C=\bigcup\\{\,\\{2k,2k+1\\}\,:\,k\in P_{0}\backslash 1\\}$

and define

$L^{C}_{\alpha}=L_{\alpha}\cap C=\\{2k+\Xi_{\alpha}(k)\,:\,k\in
P_{0}\backslash 1\textit{ and }\Xi_{\alpha}(k)\geqslant 0\\},$
$R^{C}_{\alpha}=R_{\alpha}\cap C=\\{2k+(1-\Xi_{\alpha}(k))\,:\,k\in
P_{0}\backslash 1,\textit{ and }\Xi_{\alpha}(k)\geqslant 0\\}$

for each $\alpha\in\omega_{1}$. Then
$(\mathcal{L}|_{C},\mathcal{R}|_{C})=(L^{C}_{\alpha},R^{C}_{\alpha})_{\alpha\in\omega_{1}}$
is a Todorčević gap.

###### Proof.

$(L^{C}_{\alpha},R^{C}_{\alpha})_{\alpha\in\omega_{1}}$ is a pregap by similar
reasons as the ones in the proof of Theorem 3.1.10. That is,
$L^{C}_{\alpha}\cap R^{C}_{\alpha}=\emptyset$ for each $\alpha\in\omega_{1}$
and if $\alpha<\beta$ then:

  1. $(a)$

$L^{C}_{\alpha}\backslash
L^{C}_{\beta}\subseteq\\{2k+\Xi_{\alpha}(k)\,:\,k\leqslant\rho(\alpha,\beta)\textit{
and }\Xi_{\alpha}(k)\not=\Xi_{\beta}(k)\\},$

  2. $(b)$

$R^{C}_{\alpha}\backslash
R^{C}_{\beta}\subseteq\\{2k+(1-\Xi_{\alpha}(k))\,:\,k\leqslant\rho(\alpha,\beta)\textit{
and }\Xi_{\alpha}(k)\not=\Xi_{\beta}(k)\\}.$

In particular, as a consequence of (a) we have:

  1. $(c)$

$L^{C}_{\alpha}\cap
R^{C}_{\beta}\subseteq\\{2k+\Xi_{\alpha}(k)\,:\,k\leqslant\rho(\alpha,\beta)\textit{
and }\Xi_{\alpha}(k)\not=\Xi_{\beta}(k)\\}.$

In order to prove that $(L^{C}_{\alpha},R^{C}_{\alpha})_{\alpha\in\omega_{1}}$
is a gap we will use the equivalence provided by Lemma 3.1.70. Let
$S\in[\omega_{1}]^{\omega_{1}}$. Since $\mathcal{F}$ is
$\mathcal{P}$-2-capturing there are $\alpha<\beta\in S$ so that
$\\{\alpha,\beta\\}$ is captured at some level $l\in P_{0}$. By Proposition
2.4.10 it follows that $\Delta(\alpha,\beta)=l=\rho(\alpha,\beta)$. Hence,
$\Xi_{\alpha}(k)=\Xi_{\beta}(k)$ for any $k<l$. From this and by the the point
$(c)$ above we deduce that $L^{C}_{\alpha}\cap
R^{C}_{\beta}\subseteq\\{2l+\Xi_{\alpha}(l)\\}$. Furthermore, since
$l=\rho(\alpha,\beta)$ and $\mathcal{F}$ is a construction scheme(or just by
point (2) in Proposition 2.4.10) we also have that $\Xi_{\alpha}(l)=0$ and
$\Xi_{\beta}(l)=1$. Therefore we conclude that $L^{C}_{\alpha}\cap
R^{C}_{\beta}=\\{2k\\}$, so in particular $(L^{C}_{\alpha}\cap
R^{C}_{\beta})\cup(L^{C}_{\beta}\cap R^{C}_{\alpha})\not=\emptyset.$ As we
said before, by Lemma 3.1.70 we are done.

Now we will prove that $(L^{C}_{\alpha},R^{C}_{\alpha})_{\alpha\in\omega_{1}}$
is Todorčević. Let $S\in[\omega_{1}]^{\omega_{1}}$. Since $\mathcal{F}$ is
$\mathcal{P}$-2-capturing there are $\alpha<\beta\in S$ such that
$\\{\alpha,\beta\\}$ is captured at some level $l\in P_{1}$. Again, by
Proposition 2.4.10 and by the points (a) and (b) written at the beginning of
the proof we deduce that $L^{C}_{\alpha}\backslash
L^{C}_{\beta}\subseteq\\{2l\\}$ and $R^{C}_{\alpha}\backslash
R^{C}_{\beta}\subseteq\\{2l+1\\}$. But $l\in P_{1}$ so $\\{2l,2l+1\\}$ has
empty intersection with both $L^{C}_{\alpha}$ and $R^{C}_{\alpha}$ by
definition of these two sets. In this way we conclude that
$L^{C}_{\alpha}\backslash L^{C}_{\beta}=\emptyset$ and
$R^{C}_{\alpha}\backslash R^{C}_{\beta}=\emptyset.$ In other words,
$L^{C}_{\alpha}\subseteq L^{C}_{\beta}$ and $R^{C}_{\alpha}\subseteq
R^{C}_{\beta}$. ∎

It is natural to ask whether destructible can be constructed only by using a
$2$-capturing construction scheme. In the next theorem we will show that this
is not the case. But first, let us note that the proofs of both Lemma 3.1.76
and Lemma 3.1.75 actually yield the two following results.

######  Lemma 3.1.80.

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap. Suppose that
$\mathcal{A}\in[\chi_{0}(\mathcal{L},\mathcal{R})]^{\omega_{1}}$. Then there
are $n\in\omega$ and
$\mathcal{B}\in[\mathcal{A}]^{\omega_{1}}\cap\mathscr{P}([\omega_{1}]^{n})$ an
uncountable root-tail-tail $\Delta$-system such that for any two distinct
$p,q\in\mathcal{B}$, the following conditions are equivalent:

  * $\blacksquare$

$p$ is compatible with $q$.

  * $\blacksquare$

$(L_{p(n-1)}\cap R_{q(n-1)})\cup(L_{q(n-1)}\cap R_{p(n-1)})=\emptyset.$ In
other words, $\\{p(n-1)\\}$ is compatible with $\\{q(n-1)\\}.$

######  Lemma 3.1.81.

Let $(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$
be an $(\omega_{1},\omega_{1})$-gap. Suppose that
$\mathcal{A}\in[\chi_{1}(\mathcal{L},\mathcal{R})]^{\omega_{1}}$. Then there
are $n,r\in\omega$ and
$\mathcal{B}\in[\mathcal{A}]^{\omega_{1}}\cap\mathscr{P}([\omega_{1}]^{n})$ a
root-tail-tail $\Delta$-system with root $R$ of cardinality $r$ such that for
any two distinct $p,q\in\mathcal{B}$, the following conditions are equivalent:

  * $\blacksquare$

$p$ is compatible with $q$.

  * $\blacksquare$

$(L_{p(r)}\cap R_{q(r)})\cup(L_{q(r)}\cap R_{p(r)})\not=\emptyset.$ In other
words, $\\{p(r)\\}$ is compatible with $\\{q(r)\\}.$

######  Theorem 3.1.82.

Let $\mathcal{F}$ be a $2$-capturing construction scheme. If
$\mathfrak{m}_{\mathcal{F}}>\omega_{1}$, then there are no destructible gaps.

###### Proof.

Let us assume towards a contradiction that there is a destructible gap
$(\mathcal{L},\mathcal{R})=(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}.$ The
forcing $\chi_{0}(\mathcal{L},\mathcal{R})$ is never Knaster. Therefore, by
Propositions 4.1.3 and 4.0.3, there are
$X^{\prime}\in[\chi_{0}(\mathcal{L},\mathcal{R})]^{\omega_{1}}$ and an
injective function $\zeta^{\prime}:X\longrightarrow\omega_{1}$ in such way
that for any two distinct $p,q\in X^{\prime}$, if
$\\{\zeta^{\prime}(p),\zeta^{\prime}(q)\\}$ is captured, then $p$ and $q$ are
incompatible in $\chi_{0}(\mathcal{L},\mathcal{R})$. By virtue of the Lemma
3.1.80, we may assume that the elements $X^{\prime}$ are singletons. In this
way, we can define $X=\\{\alpha\in\omega_{1}\,:\,\\{\alpha\\}\in
X^{\prime}\\}$ and $\zeta:X\longrightarrow\omega_{1}$ given by
$\zeta(\alpha)=\zeta^{\prime}(\\{\alpha\\})$. Note that for any two distinct
$\alpha,\beta\in X$, if $\\{\zeta(\alpha),\zeta(\beta)\\}$ is captured, then

$(L_{\alpha}\cap R_{\beta})\cup(L_{\alpha}\cap R_{\beta})\not=\emptyset.$

Claim: $\chi_{1}(L_{\alpha},R_{\alpha})_{\alpha\in X}$ is $ccc$ and
$n$-preserves $\mathcal{F}.$

###### Proof of claim.

We will prove the claim by appealing to the Lemma 4.0.3. Let $\mathcal{A}$ be
an uncountable subset of $\chi_{1}(L_{\alpha},R_{\alpha})_{\alpha\in X}$ and
$\nu:\mathcal{A}\longrightarrow\omega_{1}$ be an injective function. According
to the Lemma 3.1.81, we may suppose that there are $n,r\in\omega$ so that
$\mathcal{A}\in[X]^{n}$ and it forms a root-tail-tail $\Delta$-system with
root $R$ of cardinality $r$ so that for any two distinct $p,q\in\mathcal{A}$,
if $(L_{p(r)}\cap R_{q(r)})\cup(L_{q(r)}\cap R_{p(r)})\not=\emptyset$, then
$p$ and $q$ are compatible. Given $p\in\mathcal{A}$, let us define

$A_{p}=\ \\{\zeta(p(r)),\nu(p)\\},$ $\alpha_{p}=\max(A_{p}).$

Since both $\zeta$ and $\nu$ are injective functions and $p(r)\not=q(r)$
whenever $p\not=q$, we may assume that the $A_{p}$’s are pairwise disjoint.
Even more,we can suppose that there are $k,a\in\omega$ so that the following
conditions hold for any two distinct $p,q\in\mathcal{A}$:

  1. $(a)$

$\rho^{A_{p}}=k.$

  2. $(b)$

$\lVert\alpha_{p}\rVert_{k}=a.$

  3. $(c)$

If $h:(\alpha_{p})_{k}\longrightarrow(\alpha_{q})_{k}$ is the increasing
bijection $h(\nu(p))=\nu(q)$ and $h(\zeta(p(r)))=\zeta(q(r))$.

  4. $(d)$

$\rho(\alpha_{p},\alpha_{q})>k.$

Since $\mathcal{F}$ is assumed to be $2$-capturing, there are distinct
$p_{0},p_{1}\in\mathcal{A}$ for which $\\{\alpha_{p_{0}},\alpha_{p_{1}}\\}$ is
captured. By virtue of the Lemma 2.4.11 and the points $(a)$, $(b)$, $(c)$ and
$(d)$ above, are $\\{\nu(p_{0}),\nu(p_{1})\\}$ and
$\\{\zeta(p_{0}(r)),\zeta(p_{1}(r))\\}$ are also captured. In particular, this
means that $(L_{p_{0}(r)}\cap R_{p_{1}(r)})\cup(L_{p_{1}(r)}\cap
R_{p_{0}(r)})\not=\emptyset.$ Thus, $p_{1}$ and $p_{1}$ are compatible. This
finishes the proof.

∎

By the previous claim $\chi_{1}(L_{\alpha},R_{\alpha})_{\alpha\in X}$ is an
uncountable $ccc$ forcing which $2$-preserves $\mathcal{F}$. Since
$\mathfrak{m}_{\mathcal{F}}>\omega_{1}$, it follows that
$(L_{\alpha},R_{\alpha})_{\alpha\in X}$ is an indestructible gap due to the
Lemma 3.1.77 (hence, so is $(\mathcal{L},\mathcal{R})$). This is a
contradiction, so the proof is over. ∎

######  Remark 3.1.83.

Under the hypotheses of the Theorem 3.1.79 , define $L^{\omega\backslash
C}_{\alpha}=L_{\alpha}\cap(\omega\backslash C)$ and $R^{\omega\backslash
C}_{\alpha}=R_{\alpha}\cap(\omega\backslash C)$ for each
$\alpha\in\omega_{1}$. A similar arguing yields that
$(\mathcal{L}|_{\omega\backslash C},\mathcal{R}|_{\omega\backslash
C})=(L^{\omega\backslash C}_{\alpha},R^{\omega\backslash
C}_{\alpha})_{\alpha\in\omega_{1}}$ is also a Todorčević gap.

Recall that the product of two forcings, say $\mathbb{P}$ and $\mathbb{Q}$, is
$ccc$ if and only if $\mathbb{P}$ is $ccc$ and $\mathbb{P}\Vdash\text{\say{
$\mathbb{Q}\textit{ is ccc }$}}$. Suppose we are in the particular situation
where have $(\mathcal{L}_{0},\mathcal{R}_{0})$ and
$(\mathcal{L}_{1},\mathcal{R}_{1})$ are two gaps such that
$\chi_{i}(\mathcal{L}_{0},\mathcal{R}_{0})\times\chi_{j}(\mathcal{L}_{1},\mathcal{R}_{1})$
is $ccc$ for any $i,j\in 2$. According to the previous theorem, destroying or
making indestructible one of these two gaps by any of the previous forcings
will not destroy or or make indestructible the remaining one. This motivates
the following definition.

######  Definition 3.1.84 (Independent families of gaps).

We say that a family $\langle(\mathcal{L}^{c},\mathcal{R}^{c})\rangle_{c\in
I}$ of $(\omega_{1},\omega_{1})$-pregaps is independent if:

$\prod\limits_{c\in I}^{FS}\chi_{\phi(c)}(\mathcal{L}^{c},\mathcal{R}^{c})$

is $ccc$ for any $\phi:I\longrightarrow 2.$ Additionally, we say that a
coherent family of functions $\mathfrak{F}$ supported by an $\omega_{1}$-tower
is independent if the family
$\langle(\mathcal{L}^{c(0)},\mathcal{L}^{c(1)})\rangle_{c\in[\omega_{1}\backslash\omega]^{2}}$
is independent.

######  Remark 3.1.85.

If $\langle(\mathcal{L}^{c},\mathcal{R}^{c})\rangle_{c\in I}$ is an
independent family of $(\omega_{1},\omega_{1})$-pregaps, then
$(\mathcal{L}^{c},\mathcal{R}^{c})$ is a gap for any $c\in I.$

######  Theorem 3.1.86 (Under $CA_{2}(part)$).

There are two $(\omega_{1},\omega_{1})$-gaps which are Todorčević but not
independent.

###### Proof.

Let $\mathcal{F}$ be a $2$-capturing $2$-construction scheme and consider the
gaps $(\mathcal{L}|_{C},\mathcal{R}|_{C})$ and
$(\mathcal{L}|_{\omega\backslash C},\mathcal{R}|_{\omega\backslash C})$
defined in Theorem 3.1.79 and Remark 3.1.83 respectively. These two gaps are
Todorčević, so we only need to prove that they are not independent. This will
be done in the following claim.

Claim:
$\mathbb{P}=\chi_{0}(\mathcal{L}|_{C},\mathcal{R}|_{C})\times\chi_{0}(\mathcal{L}|_{\omega\backslash
C},\mathcal{R}|_{\omega\backslash C})$ is not $ccc$.

###### Proof of claim.

Suppose towards a contradiction that the forcing $\mathbb{P}$ is $ccc$. By
Theorem 3.1.77, it follows that there is $G$ a $\mathbb{P}$-generic filter
over $V$ so that both of $(\mathcal{L}|_{C},\mathcal{R}|_{C})$ and
$(\mathcal{L}|_{\omega\backslash C},\mathcal{R}|_{\omega\backslash C})$ are
separated in $V[G]$ by some $C_{0}$ and $C_{1}$ respetively. In order to
finish, just note that $C_{0}\cup C_{1}$ separates the Hausdorff gap
$(L_{\alpha},R_{\alpha})_{\alpha\in\omega_{1}}$. This is a contradiction since
Hausdorff gaps undestructible. Thus, the proof is over. ∎

∎

In [3], Uri Abraham and Saharon Shelah used large independent families of
Suslin trees in order to code certain sets of reals. Independent families of
gaps can be used in a similar way. In [133], Teruyuki Yorioka proved, assuming
the $\Diamond$-principle, that there is an independent family of
$2^{\omega_{1}}$ gaps. In here we will construct an independent coherent
family of functions from $FCA$. For this purpose, we will need the following
proposition (see [113]).

######  Proposition 3.1.87.

Let $\langle(\mathcal{L}^{c},\mathcal{R}^{c})\rangle_{c\in I}$ be a finite
family of $(\omega_{1},\omega_{1})$-gaps indexed as
$(L^{c}_{\alpha},R^{c}_{\alpha})_{\alpha\in X_{c}}$ for each $c\in I$. Also
let $\phi:I\longrightarrow 2$. Suppose that for any uncountable
$\mathcal{B}\subseteq\prod\limits_{i\in I}X_{i}$ there are distinct
$g,h\in\mathcal{B}$ so that for any $c\in I$, if $g(c)\not=h(c)$ then:

$(L^{c}_{g(c)}\cap R^{c}_{h(c)})\cup(L^{c}_{h(c)}\cap
R^{c}_{g(c)})=\emptyset\textit{ if }\phi(c)=0,$ $(L^{c}_{g(c)}\cap
R^{c}_{h(c)})\cup(L^{c}_{h(c)}\cap R^{c}_{g(c)})\not=\emptyset\textit{ if
}\phi(c)=1.$

Then $\mathbb{P}=\prod\limits_{c\in
I}\chi_{\phi(c)}(\mathcal{L}^{c},\mathcal{R}^{c})$ is $ccc$.

###### Proof.

Let $\mathcal{A}$ be an uncountable subset of $\mathbb{P}$. For any $c\in I$
consider $\mathcal{A}^{c}=\\{p(c)\,:\,p\in\mathcal{A}\\}$. Note that
$\mathcal{A}^{c}\subseteq\chi_{\phi(c)}(\mathcal{L}^{c},\mathcal{R}^{c})$. In
this way we can suppose that each $\mathcal{A}^{c}$ satisfies the conditions
listed in the proof of Lemma 3.1.75 for some
$n_{c},\sigma_{c},\tau_{c},\mu_{c},R_{c}$ and $r_{c}$. For any
$p\in\mathcal{A}$ define $g_{p}\in\prod\limits_{c\in I}X_{c}$ as:

$g_{p}(i)=\begin{cases}p(c)(n_{c}-1)&\textit{ if }\phi(c)=0\\\
p(c)(r_{c})&\textit{ if }\phi(i)=1\par\end{cases}$

By the hypotheses there are distinct $p,q\in\mathcal{A}$ such that for any
$c\in I$, if $g_{p}(c)\not=g_{q}(c)$ then

$(L^{c}_{g_{p}(c)}\cap R^{c}_{g_{q}(c)})\cup(L^{c}_{g_{q}(c)}\cap
R^{c}_{g_{p}(c)})=\emptyset\textit{ if }\phi(i)=0,$ $(L^{c}_{g_{p}(c)}\cap
R^{c}_{g_{q}(c)})\cup(L^{c}_{g_{q}(c)}\cap
R^{c}_{g_{p}(c)})\not=\emptyset\textit{ if }\phi(c)=1.$

By arguing on each coordinate in the same way as we did in the proof of Lemmas
3.1.76 an 3.1.75 we conclude that the function
$p^{\prime}\in\prod\limits_{c\in I}[X_{c}]^{<\omega}$ given by
$p^{\prime}(c)=p(c)\cup q(c)$ for each $c\in I$, is an element of $\mathbb{P}$
which is obviously smaller than both $p$ and $q$. This finishes the proof. ∎

######  Theorem 3.1.88 (FCA).

There is an independent coherent family of functions supported by an
$\omega_{1}$-tower.

###### Proof.

The construction here is similar to the one in Theorem 3.1.69. Fix a type
$\langle m_{k},n_{k+1},r_{k+1}\rangle_{k\in\omega}$ such that
$n_{k+1}>2^{r_{k+1}^{2}}$ for all $k\in\omega$, and let $\mathcal{F}$ be a
fully capturing construction scheme of that type. For each $k>0$, let

$N_{k}=\\{k\\}\times[r_{k}]^{2},$ $N=\bigcup\limits_{k>0}N_{k}.$

Also enumerate $\mathscr{P}([r_{k}]^{2})$ (possibly with repetitions) as
$\langle S^{k}_{i}\rangle_{i<n_{k}}$ in such way that
$S^{k}_{0}=S^{k}_{1}=[r_{k}]^{2}$. We start by defining an $\omega_{1}$-tower
over $N$. Given $\alpha\geqslant\omega$ and $k>0$ we define
$T^{k}_{\alpha}\subseteq N_{k}$ as follows:

$T^{k}_{\alpha}=\begin{cases}\emptyset&\textit{ if }\Xi_{\alpha}(k)=-1\\\
\\{k\\}\times S^{k}_{i}&\textit{ if }\Xi_{\alpha}(k)=i\geqslant
0\par\end{cases}$

Finally, let $T_{\alpha}=\bigcup_{k\in\omega\backslash 1}T^{k}_{\alpha}$. In
order to prove that $\mathcal{T}=\langle
T_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ is in fact an
$\omega_{1}$-tower just note that if
$\alpha<\beta\in\omega_{1}\backslash\omega$ and $k>\rho(\alpha,\beta)$ then
$T^{k}_{\alpha}\subseteq T^{k}_{\beta}$ by means of the point (c) of Lemma
2.3.9.

Now we will construct a coherent family of functions supported by
$\mathcal{T}$. For this, let $\beta\geqslant\omega$. Note that if $x\in
T_{\beta}$ then $x=(k,s)$ where $k>0$, $\Xi_{k}(\beta)\geqslant 0$ and $s\in
S^{k}_{\Xi_{\beta}(k)}\subseteq[r_{k}]^{2}$. In this way, we can define
$f_{\beta}:T_{\beta}\longrightarrow\beta$ as:

$f_{\beta}(k,s)=\begin{cases}(\beta)_{k}(s(0))&\textit{ if
}\Xi_{\beta}(k)=0\\\ (\beta)_{k}(s(1))&\textit{ if
}\Xi_{\beta}(k)>0\par\end{cases}$

It is easy to check that $\langle
f_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ is a coherent family
of functions supported by $\mathcal{T}$. This is done, again, by appealing to
the point (c) of Lemma 2.3.9.

The only thing left to show is that $\langle
f_{\alpha}\rangle_{\alpha\in\omega_{1}\backslash\omega}$ is independent.
Recall that a finite support product of forcings is $ccc$ if and only if each
finite subproduct is $ccc$. Let $I$ be a non-empty finite subset of
$[\omega_{1}\backslash\omega]^{2}$ and let $\phi:I\longrightarrow 2.$ We will
finish by proving the following claim.

Claim: $\mathbb{P}=\prod\limits_{c\in
I}\chi_{\phi(c)}(\mathcal{L}^{c(0)},\mathcal{L}^{c(1)})$ is $ccc$.

###### Proof of claim.

The proof of this claim will be performed by appealing to the equivalence
provided by Proposition 3.1.87. First note that for any $c\in I$,
$(\mathcal{L}^{c(0)},\mathcal{L}^{c(1)})=(L^{c(0)}_{\alpha},L^{c(1)}_{\alpha})_{\alpha\in
X_{c}}$ where $X_{c}=\omega_{1}\backslash(c(1)+1)$. Now, let $\mathcal{B}$ be
an uncountable subset of $\prod\limits_{c\in I}X_{c}$. For any
$g\in\mathcal{B}$, define $D_{g}=\bigcup I\cup im(g)$. We need to prove the
following claim.

  1. $(1)$

If $g,h\in B$ then $|im(g)|=|im(h)|$. Furthermore, if
$\psi:im(g)\longrightarrow im(h)$ is the increasing bijection then
$\psi(g(c))=h(c)$ for any $c\in I$.

  2. $(2)$

$\\{D_{g}\,:g\in\mathcal{B}\\}$ is a root-tail-tail $\Delta$-system with root
$\bigcup I$.

Fix $k>\rho^{\cup I}$. Since $\mathcal{F}$ is a fully capturing construction
scheme, there is $l>k$ and there are distinct
$g_{0},\dots,g_{n_{l}-1}\in\mathcal{B}$ so that the family
$\\{D_{g_{0}},\dots,D_{g_{n_{l}-1}}\\}$ is a captured at level $l$. According
to (2) and by Lemma 2.4.5, $\Xi_{\cup I}(l)={-1}$. In this way, the set
$S=\\{\,\\{\lVert c(0)\rVert_{l},\lVert c(1)\rVert_{l}\\}\,:\,c\in I\textit{
and }\phi(c)=1\\}$ is a subset of $[r_{l}]^{2}$. Therefore we can find
$i<n_{l}$ so that $S^{l}_{i}=g_{i}$. The proof will be over once we show the
following subclaim.

Subclaim: $g_{0}$ and $g_{i}$ satisfy the hypotheses of Proposition 3.1.87.
That is, for any $c\in I$, if $g_{0}(c)\not=g_{i}(c)$ then:333In this case
$(\mathcal{L}^{c},\mathcal{R}^{c})=(L^{c(0)}_{\alpha},L^{c(1)}_{\alpha})_{\alpha\in
X_{c}}$. Therefore $L^{c}_{g_{0}(c)}=L^{c(0)}_{g_{0}(c)}$,
$R^{c}_{g_{i}(c)}=L^{c(1)}_{g_{i}(c)}$, $L^{c}_{g_{i}(c)}=L^{c(0)}_{g_{i}(c)}$
and $R^{c}_{g_{0}(c)}=L^{c(1)}_{g_{0}(c)}$.

$(L^{c(0)}_{g_{0}(c)}\cap L^{c(1)}_{g_{i}(c)})\cup(L^{c(0)}_{g_{i}(c)}\cap
L^{c(1)}_{g_{0}(c)})=\emptyset\textit{ if }\phi(c)=0,$
$(L^{c(0)}_{g_{0}(c)}\cap L^{c(1)}_{g_{i}(c)})\cup(L^{c(0)}_{g_{i}(c)}\cap
L^{c(1)}_{g_{0}(c)})\not=\emptyset\textit{ if }\phi(c)=1.$

###### Proof of subclaim.

Let $c\in I$. Then $g(c)\not=h(c)$ due to condition (2). Furthermore,
$\rho(g(c),h(c))=l=\Delta(g(c),h(c))$ for any $c\in I.$ This follows from
condition (1) together with the fact that $D_{g_{0}}$ is strongly isomorphic
to $D_{g_{i}}$. By the definitions of $f_{g_{0}(c)}$ and $f_{g_{i}(c)}$ and
the points (a) and (c) of Lemma 2.3.9, it follows that

$L^{c(0)}_{g_{0}(c)}\cap
L^{c(1)}_{g_{i}(c)}=\big{(}f^{-1}_{g_{0}(c)}[\\{c(0)\\}]\cap
f^{-1}_{g_{i}(c)}[\\{c(1)\\}]\big{)}\cap T^{l}_{g_{0}(c)}.$

On one hand, since $\Xi_{g_{0}(c)}(l)=0$ then

$f^{-1}_{g_{0}(c)}[\\{c(0)\\}]\cap
T^{l}_{g_{0}(c)}=\\{l\\}\times\\{s\in[r_{l}]^{2}\,:\,s(0)=\lVert
c(0)\rVert_{l}\,\\}.$

On the other hand, since $\Xi_{g_{i}(c)}(l)>0$ and $S=S^{l}_{i}$ then

$f^{-1}_{g_{0}(c)}[\\{c(1)\\}]\cap T^{l}_{g_{0}(c)}=\\{l\\}\times\\{s\in
S\,:\,s(1)=\lVert c(1)\rVert_{l}\,\\}.$

Thus, $L^{c(0)}_{g_{0}(c)}\cap L^{c(1)}_{g_{i}(c)}\subseteq\\{\,\\{\lVert
c(0)\rVert_{l},\lVert c(1)\rVert_{l}\\}\,\\}$. Furthermore, such intersection
is non-empty if and only if $\phi(c)=1$ by the definition of $S$. By arguing
in a similar way and using that $\lVert c(0)\rVert_{l}<\lVert c(1)\rVert_{l}$
we conclude that $L^{c(0)}_{g_{i}(c)}\cap L^{c(1)}_{g_{0}(c)}$ is always
empty. This finishes the proof of the subclaim.

∎

∎

∎

###  3.2 Trees and lines

The purpose of this section is to study uncountable trees and lines (totally
ordered sets) from the point of view of construction schemes. Before dealing
with any specific notion, it is convenient to recall the definition of a tree.
Complete introductions to trees, lines and the duality between them can be
found in [54], [64] and [114].

######  Definition 3.2.1 (Tree).

A partial ordered set $(T,<)$ is called a tree if $(-\infty,t)_{T}$ is well-
ordered for any $t\in T$.

When dealing with trees, we apply some changes in the notation related to
partial orders. In this realm, intervals $(-\infty,t)_{T}$ and
$(t,\infty)_{T}$ are denoted as $t_{\downarrow_{T}}$ and $t_{\uparrow_{T}}$
respectively. As usual, whenever there is no risk of confusion we will write
those sets simply as $t_{\downarrow}$ and $t_{\uparrow}$ respectively.

It is easy to see each tree $T$ is well-founded. Thus, according to the
Section 1.2, there is a unique function $rank_{T}:T\longrightarrow Ord$ so
that:

$rank_{T}(t)=\sup(\,rank_{T}(s)+1\ ,:\,\,s<t\,).$

Due to uniqueness, it can be shown that $rank_{T}(t)=ot(t_{\downarrow_{T}})$
for all $t\in T$. In this way, for each ordinal $\alpha$, the level $\alpha$
of $T$ can be described as $T_{\alpha}=\\{\,t\in
T\,:\,ot(t_{\downarrow})=\alpha\,\\}.$ Given an ordinal $\beta$ and $t\in T$,
we define

$T|_{\beta}=\bigcup\limits_{\alpha<\beta}T_{\alpha},$ $T|_{t}=\\{s\in
T\,:\,s\textit{ is comparable with }t\\}.$

Finally, we say that $B\subseteq T$ is a branch if it is a maximal chain in
$T$.

####  3.2.1 Countryman lines and Aronszajn trees

Countryman lines are certain kind of linear orders whose existence was
proposed by Roger Simmons Countryman in the 1970’s. In [93], Saharon Shelah
constructed for the first time Countryman lines without appealing to any extra
axioms. Years later, Stevo Todorčević presented in [120] an easy construction
of such a line using walks on ordinals. The reader may find some variations of
Todorčević original proof in [52] and [110].

######  Definition 3.2.2 (Countryman line).

We say that a totally ordered set $(C,<)$ is a Countryman line if $C$ is
uncountable and $C^{2}$ can be covered by countably many chains444Here, we
consider the order over $C^{2}$ given by $(x,y)\leqslant(w,z)$ if and only
$x\leqslant w$ and $y\leqslant z$..

In [75], Justin Moore put Countryman lines into the realm of canonical objects
inside Mathematics. Specifically, he proved the following theorem.

######  Theorem 3.2.3 (Under $PFA$).

Let $C$ be any Countryman line and $X$ be a fixed subset of $\mathbb{R}$ of
cardinality $\omega_{1}$. Then any uncountable order contains an isomorphic
copy of one of the following five orders:

$X,\,\omega_{1},\,\omega_{1}^{*},\,C,\textit{ or }C^{*}.$

Where $\omega_{1}^{*}$ and $C^{*}$ denote $\omega_{1}$ and $C$ but with the
reverse order.

We will now define a Countryman line using construction schemes. It should be
noted that in Chapter 3 of [110], Todorčević showed that ordinal metrics can
also be used to define Countryman lines. Although working with construction
schemes is equivalent to working with their induced ordinal metric, the
definition of the line appearing in this thesis differs from the ones in
[110].
The following proposition will help by giving us a better picture of the next
definitions.

######  Proposition 3.2.4.

Let $\alpha,\beta\in\omega_{1}$ be distinct ordinals and
$k<\Delta(\alpha,\beta)$. Then $\Xi_{\delta}(\Delta(\alpha,\beta))=-1$ for
each $\delta\in(\alpha)_{k}\cap(\beta)_{k}$. In particular
$|(\alpha)_{k}\cap(\beta)_{k}|\leqslant r_{\Delta(\alpha,\beta)}.$

###### Proof.

Let $\delta\in(\alpha)_{k}\cap(\beta)_{k}$. Suppose towards a contradiction
that $\Xi_{\delta}(\Delta(\alpha,\beta))\not=-1$. Thus, according to the part
(c) of Lemma 2.3.9 and the fact that
$\rho(\alpha,\delta),\rho(\beta,\delta)\leqslant k<\Delta(\alpha,\beta)$, we
have that
$\Xi_{\alpha}(\Delta(\alpha,\beta))=\Xi_{\delta}(\Delta(\alpha,\beta))=\Xi_{\beta}(\Delta(\alpha,\beta)).$
This is impossible since
$\Xi_{\alpha}(\Delta(\alpha,\beta)\not=\Xi_{\beta}(\Delta(\alpha,\beta))$ due
to the point (d) of the same lemma. ∎

######  Definition 3.2.5.

Let $\alpha,\beta\in\omega_{1}$ be distinct ordinals and
$k=\Delta(\alpha,\beta)-1$. We define $c(\alpha,\beta)$ as:

$c^{\alpha}_{\beta}=\min(\,(\alpha)_{k}\backslash(\beta)_{k}\,)$

######  Remark 3.2.6.

Note that $c^{\alpha}_{\beta}\not=c^{\beta}_{\alpha}$ for any two distinct
$\alpha,\beta\in\omega_{1}$. Furthermore, if $k=\Delta(\alpha,\beta)-1$, then
$(c^{\alpha}_{\beta})^{-}_{k}=(\alpha)_{k}\cap(\beta)_{k}=(c^{\beta}_{\alpha})^{-}_{k}.$
In this way, $\lVert c^{\beta}_{\alpha}\rVert_{k}=\lVert
c^{\alpha}_{\beta}\rVert_{k}\leqslant r_{k+1}$ due to the Proposition 3.2.4.

In the next definition we define the relation that turns $\omega_{1}$ into a
Countryman line.

######  Definition 3.2.7.

Let $\mathcal{F}$ be a construction scheme. Given different
$\alpha,\beta\in\omega_{1}$ and $k=\Delta(\alpha,\beta)-1$, we recursively
decide whether $\alpha<_{\mathcal{F}}\beta$ when one of the following
conditions holds:

  1. $(a)$

$\lVert c^{\alpha}_{\beta}\rVert_{k}=r_{k+1}$ and
$\Xi_{\alpha}(k+1)<\Xi_{\beta}(k+1).$

  2. $(b)$

$\lVert c^{\alpha}_{\beta}\rVert_{k}<r_{k+1}$ and
$c^{\alpha}_{\beta}<_{\mathcal{F}}c^{\beta}_{\alpha}.$

Where $k=\Delta(\alpha,\beta)-1.$

In here, $|(\alpha)_{k}\cap(\beta)_{k}|=r_{\Delta(\alpha,\beta)}$. Thus, we
decide whether $\alpha<_{\mathcal{F}}\beta$ by comparing
$\Xi_{\alpha}(\Delta(\alpha,\beta))$ and $\Xi_{\beta}(\Delta(\alpha,\beta)).$

In here, $|(\alpha)_{k}\cap(\beta)_{k}|<r_{\Delta(\alpha,\beta)}$. Thus, we
decide whether $\alpha<_{\mathcal{F}}\beta$ by comparing $c^{\alpha}_{\beta}$
and $c^{\beta}_{\alpha}.$

######  Lemma 3.2.8.

Suppose $\alpha,\beta\in\omega_{1}$ are distinct and let
$k=\Delta(\alpha,\beta)-1$. Consider
$h:(\alpha)_{k}\longrightarrow(\beta)_{k}$ the increasing bijection and take
$\gamma\in(\alpha)_{k}$ such that $h(\gamma)\not=\gamma$. Then the following
statements are equivalent:

  1. $(1)$

$\alpha<_{\mathcal{F}}\beta$.

  2. $(2)$

$\gamma<_{\mathcal{F}}h(\gamma)$.

###### Proof.

The proof is carried by induction over $\alpha$ and $\beta$. So suppose that
we have proved the Lemma for each $\alpha^{\prime}<\alpha$ and
$\beta^{\prime}<\beta$. We start with some remarks about $\gamma$. First of
all, $\Delta(\gamma,h(\gamma))\geqslant\Delta(\alpha,\beta)$ due to the point
(a) of Lemma 2.3.5. Furthermore, according to Remark 2.3.6 it follows that
$\gamma\in(\alpha)_{k}\backslash(\beta)_{k}$ and
$h(\gamma)\in(\beta)_{k}\backslash(\alpha)_{k}$. This means that
$c^{\alpha}_{\beta}\in(\gamma)_{k}$ and
$c^{\beta}_{\alpha}\in(h(\gamma))_{k}$. Lastly, as
$(\alpha)_{k}\cap(\beta)_{k}$ is an initial segment of both $(\alpha)_{k}$ and
$(\beta)_{k}$, we conclude that
$(\alpha)_{k}\cap(\beta)_{k}=(\delta)_{k}\cap(h(\delta))_{k}$. We divide the
rest of the proof into two cases.

Case 1: If $\Delta(\alpha,\beta)=\Delta(\gamma,h(\gamma))$.

###### Proof of case.

In this case we have that $c^{\alpha}_{\beta}=c^{\gamma}_{h(\gamma)}$ and
$c^{\beta}_{\alpha}=c^{h(\gamma)}_{\gamma}$. Thus, if $\lVert
c^{\alpha}_{\beta}\rVert_{k}<r_{k+1}$ then $\alpha<_{\mathcal{F}}\beta$ if and
only if $c^{\alpha}_{\beta}<_{\mathcal{F}}c^{\beta}_{\alpha}$ if and only if
$\gamma<_{\mathcal{F}}h(\gamma)$. On the other hand, if $\lVert
c^{\alpha}_{\beta}\rVert_{k}=r_{k+1}$ then $\alpha<_{\mathcal{F}}\beta$ if and
only if $\Xi_{\alpha}(k+1)<\Xi_{\beta}(k+1)$, and
$\gamma<_{\mathcal{F}}h(\gamma)$ if and only if
$\Xi_{\gamma}(k+1)<\Xi_{h(\gamma)}(k+1)$. Now, as
$k+1=\Delta(\gamma,h(\gamma))$ then both $\Xi_{\gamma}(k+1)$ and
$\Xi_{h(\gamma)}(k+1)$ are non-negative. From this fact and the part (c) of
Lemma 2.3.9, we conclude that $\Xi_{\alpha}(k+1)=\Xi_{\gamma}(k+1)$ and
$\Xi_{\beta}(k+1)=\Xi_{h(\gamma)}(k+1)$. In this way,
$\alpha<_{\mathcal{F}}\beta$ if and only if $\gamma<_{\mathcal{F}}h(\gamma)$.
∎

Case 2: If $\Delta(\alpha,\beta)<\Delta(\gamma,h(\gamma)).$

###### Proof of case.

First we argue that $\alpha<_{\mathcal{F}}\beta$ if and only if
$c^{\alpha}_{\beta}<_{\mathcal{F}}c^{\beta}_{\alpha}$. For this purpose it is
enough to show that, in this case, $\lVert
c^{\alpha}_{\beta}\lVert_{k}<r_{k+1}$. Indeed, according to the point (a) of
Lemma 2.3.9, we have that $\Xi_{\gamma}(k+1)=\Xi_{h(\gamma)}(k+1)$. On the
other hand, by the part (d) of such lemma, we also know that
$\Xi_{\alpha}(k+1)\not=\Xi_{\beta}(k+1)$. Thus, $\Xi_{\gamma}(k+1)=-1$ due to
the part $(c)$ of Lemma 2.3.9. Since $c^{\alpha}_{\beta}\in(\gamma)_{k}$, it
follows that $\lVert c^{\alpha}_{\beta}\rVert_{k}<r_{k+1}$. In this way,
$\alpha<_{\mathcal{F}}\beta$ if and only if
$c^{\alpha}_{\beta}<_{\mathcal{F}}c^{\beta}_{\alpha}$ as we wanted. The proof
of this case will be over once we show that $\gamma<_{\mathcal{F}}h(\gamma)$
if and only if $c^{\alpha}_{\beta}<_{\mathcal{F}}c^{\beta}_{\alpha}$. Let
$l=\Delta(\gamma,h(\gamma))-1$ and consider
$h^{\prime}:(\gamma)_{l}\longrightarrow(h(\gamma))_{l}$ be the increasing
bijection. As $k<l$ then
$c^{\alpha}_{\beta}\in(\gamma)_{k}\subseteq(\gamma)_{l}$. Furthermore,

$h^{\prime}(c^{\alpha}_{\beta})=h^{\prime}(\,(\gamma)_{k}(\lVert
c^{\alpha}_{\beta}\rVert_{k})\,)=(h(\gamma))_{k}(\lVert
c^{\alpha}_{\beta}\rVert_{k})=(h(\gamma))_{k}(\lVert
c_{\alpha}^{\beta}\rVert_{k})=c^{\beta}_{\alpha}.$

As $c^{\alpha}_{\beta}\not=c^{\beta}_{\alpha}$, we may use the inductive
hypotheses to conclude that $\gamma<_{\mathcal{F}}h(\gamma)$ if and only if
$c^{\alpha}_{\beta}<_{\mathcal{F}}c^{\beta}_{\gamma}$. This finishes the
proof. ∎

∎

######  Corollary 3.2.9.

let $\alpha,\beta\in\omega_{1}$ be distinct ordinals. Then
$\alpha<_{\mathcal{F}}\beta$ if and only if
$c^{\alpha}_{\beta}<_{\mathcal{F}}c^{\beta}_{\alpha}$.

######  Corollary 3.2.10.

Suppose $\alpha,\beta\in\omega_{1}$ are distinct and let
$k<\Delta(\alpha,\beta)$. Consider $h:(\alpha)_{k}\longrightarrow(\beta)_{k}$
the increasing bijection and take $\gamma\in(\alpha)_{k}$ such that
$h(\gamma)\not=\gamma$. Then the following statements are equivalent:

  1. $(1)$

$\alpha<_{\mathcal{F}}\beta$.

  2. $(2)$

$\gamma<_{\mathcal{F}}h(\gamma)$.

###### Proof.

Let
$\psi:(\alpha)_{\Delta(\alpha,\beta)-1}\longrightarrow(\beta)_{\Delta(\alpha,\beta)-1}$
be the increasing bijection. To prove the corollary, just notice that
$\psi|_{(\alpha)_{z}}=\phi.$ ∎

######  Lemma 3.2.11.

Let $\alpha,\beta,\delta\in\omega_{1}$ be distinct ordinals and $k\in\omega$.
If $(\alpha)_{k}\cap(\beta)_{k}\subseteq(\alpha)_{k}\cap(\delta)_{k}$ then
$(\alpha)_{k}\cap(\beta)_{k}=(\beta)_{k}\cap(\delta)_{k}.$

###### Proof.

Since $(\alpha)_{k}\cap(\beta)_{k}\subsetneq(\alpha)_{k}\cap(\delta)_{k}$ then
$(\alpha)_{k}\cap(\beta)_{k}=(\alpha)_{k}\cap(\delta)_{k}\cap(\beta)_{k}\subseteq(\delta)_{k}\cap(\beta)_{k}$.
In order to prove that
$(\beta)_{k}\cap(\delta)_{k}\subseteq(\alpha)_{k}\cap(\beta)_{k}$, note that
both $(\alpha)_{k}\cap(\delta)_{k}$ and $(\beta)_{k}\cap(\delta)_{k}$ are
initial segments of $(\delta)_{k}$ by virtue of the point (6) in Proposition
2.1.7. In this way, either
$(\beta)_{k}\cap(\delta)_{k}\sqsubseteq(\alpha)_{k}\cap(\delta)_{k}$ or
$(\alpha)_{k}\cap(\delta)_{k}\sqsubseteq(\beta)_{k}\cap(\delta)_{k}$. The
latter alternative can not happen as it would imply that
$(\alpha)_{k}\cap(\delta)_{k}\subseteq(\alpha)_{k}\cap(\beta)_{k}$. Thus, the
first alternative holds. In this way,
$(\beta)_{k}\cap(\delta)_{k}\subseteq(\beta)_{k}\cap(\alpha)_{k}\cap(\delta)_{k}\subseteq(\alpha)_{k}\cap(\beta)_{k}.$
This finishes the proof. ∎

######  Proposition 3.2.12.

$(\omega_{1},<_{\mathcal{F}})$ is a total order.

###### Proof.

The only non-trivial part of this task is to prove transitivity. This proof
will be performed by induction. For this purpose let
$\alpha,\beta,\delta\in\omega_{1}$ be distinct ordinals. Suppose that we have
proved that for any $\alpha^{\prime}<\alpha$, $\beta^{\prime}<\beta$ and
$\delta^{\prime}<\delta$, the triplet
$\\{\alpha^{\prime},\beta^{\prime},\delta^{\prime}\\}$ do not form a cycle.
That is, neither
$\alpha^{\prime}<_{\mathcal{F}}\beta^{\prime}<_{\mathcal{F}}\delta^{\prime}<_{\mathcal{F}}<\alpha^{\prime}$
nor
$\alpha^{\prime}>_{\mathcal{F}}\beta^{\prime}>_{\mathcal{F}}>\delta^{\prime}>_{\mathcal{F}}\alpha^{\prime}$.
We will show that the same holds for $\\{\alpha,\beta,\delta\\}$. By virtue of
Lemma 2.3.4, we may assume without loss of generality that
$\Delta(\alpha,\beta)=\Delta(\alpha,\delta)\leqslant\Delta(\beta,\delta).$ Let
$k=\Delta(\alpha,\beta)-1.$ For each distinct
$x,y\in\\{\alpha,\beta,\delta\\}$ consider
$e^{x}_{y}=\min(\,(x)_{k}\backslash(y)_{k}\,)$ and
$h^{x}_{y}:(x)_{k}\longrightarrow(y)_{k}$ the increasing bijection. Observe
that $h^{z}_{x}\circ h^{y}_{z}=h^{y}_{x}$, $h^{x}_{y}(e^{x}_{y})=e^{y}_{x}$
and $e^{x}_{y}\not=e^{y}_{x}$. We divide the rest of the proof into the
following cases.

Case 1: There are distinct $x,y,z\in\\{\alpha,\beta,\delta\\}$ for which
$(x)_{k}\cap(y)_{k}\subsetneq(x)_{k}\cap(z)_{k}.$

###### Proof of case.

By means of the Lemma 3.2.11, $(x)_{k}\cap(y)_{k}=(z)_{k}\cap(y)_{k}$. It then
follows that $e^{y}_{z}=e^{y}_{x}$. Now we will now show that
$e^{x}_{y}=e^{z}_{y}$. First recall that both $(x)_{k}\cap(y)_{k}$ and
$(x)_{k}\cap(z)_{k}$ are initial segments of $(x)_{k}$. From this fact we can
deduce that $(x)_{k}\cap(y)_{k}{\mathrlap{\sqsubseteq}{\,{}_{\not}}}$
$\;(x)_{k}\cap(z)_{k}.$ By minimality, $e^{x}_{y}\in(x)_{k}\cap(z)_{k}.$ As
$e^{x}_{y}\notin(y)_{k}$, it follows that $e^{x}_{y}\geqslant e^{z}_{y}$.
Moreover, since $(x)_{k}\cap(z)_{k}\sqsubseteq(z)_{k}$ then
$e^{z}_{y}\in(x)_{k}\cap(z)_{k}$. Again, we have that $e^{z}_{y}\geqslant
e^{x}_{y}$ because $e^{z}_{y}\notin(y)_{k}$. Therefore, $e^{z}_{y}=e^{x}_{y}$.
According to the Corollary 3.2.10, $y$ and $x$ are ordered (with respect to
$<_{\mathcal{F}}$) in the same way as $e^{y}_{x}$ and
$h^{y}_{x}(e^{y}_{x})=e^{x}_{y}$, and $y$ and $z$ are ordered in the same way
as $e^{y}_{z}=e^{y}_{x}$ and $h^{y}_{z}(e^{y}_{z})=e^{z}_{y}=e^{x}_{y}$. We
conclude that either $y<_{\mathcal{F}}x,z$ or $x,z<_{\mathcal{F}}y.$ This
finishes the proof of this case. ∎

Case 2:
$(\alpha)_{k}\cap(\beta)_{k}=(\alpha)_{k}\cap(\delta)_{k}=(\beta)_{k}\cap(\delta)_{k}.$

###### Proof of case.

We will divide the proof of this case into two subcases, but first, note that
$e^{\alpha}_{\beta},e^{\beta}_{\alpha},e^{\alpha}_{\delta}$ and
$e^{\delta}_{\alpha}$ are equal to
$c^{\alpha}_{\beta},c^{\beta}_{\alpha},c^{\alpha}_{\delta}$ and
$c^{\delta}_{\alpha}$ respectively. Moreover, $e^{x}_{y}=e^{x}_{z}$ for all
distinct $x,y,z\in\\{\alpha,\beta,\delta\\}$.

Subcase 1: $|(\alpha)_{k}\cap(\beta)_{k}|<r_{k+1}$.

###### Proof of subcase.

By the remarks made at the start of this case and by Corollary 3.2.10, we have
that $\alpha,\beta$ and $\delta$ are ordered in the same way as
$e^{\alpha}_{\beta}$, $e^{\beta}_{\alpha}$ and $e^{\delta}_{\alpha}$
respectively. Thus, by virtue of the inductive hypotheses, it suffices to show
that $e^{\alpha}_{\beta}<\alpha$, $e^{\beta}_{\alpha}<\beta$ and
$e^{\delta}_{\alpha}<\delta$. Indeed, according to the hypotheses of this
subcase, $\lVert e^{x}_{y}\rVert_{k}=|(x)_{k}\cap(y)_{k}|<r_{k+1}$ for all
$x,y\in\\{\alpha,\beta,\delta\\}$. On the other hand, since
$k+1=\Delta(\alpha,\beta)=\Delta(\alpha,\delta)\leqslant\Delta(\beta,\delta)$,
then $\Xi_{x}(k+1)\geqslant 0$ for each $x\in\\{\alpha,\beta,\delta\\}$. In
other words $\lVert x\rVert_{k}\geqslant r_{k+1}$. Particularly,
$e^{\alpha}_{\beta}<\alpha$, $e^{\beta}_{\alpha}<\beta$ and
$e^{\delta}_{\alpha}<\delta$, so we are done. ∎

Subcase 2: If $|(\alpha)_{k}\cap(\beta)_{k}|=r_{k+1}.$

###### Proof of subcase.

According to the definition of the order $<_{\mathcal{F}}$ and since