# Ternary rings of operators arising from inverse semigroups

Robert Pluta  Department of Mathematics, University of California, Irvine, CA
92697-3875, USA, and Wenzhou-Kean University<EMAIL_ADDRESS>and  Bernard Russo
Department of Mathematics, University of California, Irvine, CA 92697-3875,
USA<EMAIL_ADDRESS>
###### Abstract.

We are interested in properties, especially injectivity (in the sense of
category theory), of the ternary rings of operators generated by certain
subsets of an inverse semigroup via the regular representation. We determine
all subsets of the extended bicyclic semigroup which are closed under the
triple product $xy^{*}z$ (called semiheaps) and show that the weakly closed
ternary rings of operators generated by them (W*-TROs) are injective operator
spaces.

###### Key words and phrases:

Ternary ring of operators, inverse semigroup, bicyclic semigroup, semiheap,
injective operator space

###### 2020 Mathematics Subject Classification:

Primary 20N10, 46L99; Secondary 20M18

###### Contents

  1. 1 Introduction and preliminaries
    1. 1.1 Introduction
    2. 1.2 Inverse semigroups
    3. 1.3 The extended bicyclic semigroup
  2. 2 Diagrams and examples
  3. 3 Subsemiheaps of the extended bicyclic semigroup
  4. 4 Injectivity of W*-TROs

##  1\. Introduction and preliminaries

###  1.1. Introduction

Ternary rings of operators originated in the work of M. R. Hestenes in 1962
[Hestenes62]. These are linear spaces of operators from one Hilbert space to
another which are stable under the triple product $XY^{*}Z$, which he called
ternary algebras. By their nature, these spaces satisfied an associativity
condition involving five elements, namely,

(1.1) $(XY^{*}Z)U^{*}W=XY^{*}(ZU^{*}W)=X(UZ^{*}Y)^{*}W.$

These were subsequently axiomatized and named associative triple systems
[Loos72]. A milestone in their development in the realm of functional analysis
was a Gelfand-Naimark type representation theorem for associative triple
systems equipped with an operator type norm [Zettl83].

At around the same time as Hestenes’ work, unbeknownst to the researchers in
the west due partially to the Cold War [Hollings2014], the concept of semiheap
was introduced in the Soviet Union [Lawson2017]. A semiheap is a set together
with a single three variable operation satisfying an abstract version of
(1.1), and akin to the known concepts of ternary group and inverse semigroup.
Since the concept of semiheap is central to this paper, we provide the formal
definition, as stated in [Lawson2017, p. 56].

######  Definition 1.1.

By a semiheap, we mean a set $K$ together with a singled-valued, everywhere
defined ternary operation $[\cdot\cdot\cdot]$, satisfying the condition

$[[k_{1}k_{2}k_{3}]k_{4}k_{5}]=[k_{1}[k_{4}k_{3}k_{2}]k_{5}]=[k_{1}k_{2}[k_{3}k_{4}k_{5}]].$

Semiheaps and their associated structures are closely related to inverse
semigroups. In turn, inverse semigroups, together with groupoids, give rise to
operator algebras [Paterson1999]. A ubiquitous example of an inverse semigroup
is the bicyclic semigroup [Paterson1999, p. 188].

In this paper, we generalize the bicyclic semigroup in such a way that
exhibits its semiheap structure.111After completing the final draft of this
paper, the authors found out that this extended bicyclic semigroup has already
been defined, see for example [Warne1968]. However, only binary structures are
considered in this and other papers. In Theorem 3.19, we classify all of the
subsemiheaps of this extended bicyclic semigroup. We then show in Corollary
4.3, via a general result applying to all inverse semigroups [Paterson1999,
Theorem 4.5.2], that each of the examples resulting from this classification
has the property that the weakly closed ternary ring of operators it generates
is an injective operator space. It is worth pointing out that, although the
injectivity of the W*-TROs generated by the classification of subsemiheaps
uses deep results in functional analysis ([EffOzaRua01, Theorem 2.5],
[Paterson1999, Theorem 4.5.2]), the classification itself is self-contained
using only elementary arguments.

All of the subsemigroups of the bicyclic semigroup have been determined in
[Descalco2005], while the sub-inverse semigroups were determined earlier in
[Schein2003] and recently in [Hovsepyan2020]. Since inverse subsemigroups are
semiheaps, our results give a new approach to the result in the latter two
papers.

###  1.2. Inverse semigroups

An inverse semigroup is a semigroup $S$ in which for every element $x$ there
exists a unique element $x^{*}$, called the inverse or generalized inverse of
$x$, such that $x=xx^{*}x$ and $x^{*}=x^{*}xx^{*}$. The function that sends
$x$ to $x^{*}$ is an antiautomorphism of order two. An idempotent in a
semigroup is an element that multiplied by itself results in the same element.
An inverse semigroup may contain many idempotents, but when it contains
precisely one idempotent, it is a group. In fact, the existence of a unique
idempotent characterizes groups among inverse semigroups.

######  Remark 1.2.

Inverse semigroups are precisely the semigroups in which every element has at
least one generalized inverse and in which any two idempotents commute.

######  Example 1.3.

The symmetric inverse semigroup $I(X)$, also denoted by $I_{X}$, which is the
monoid under composition of all partial bijections (or partial symmetries) on
a set $X$ is an inverse semigroup. The idempotents of $I(X)$ correspond to
subsets $A\subseteq X$ via $A\leftrightarrow 1_{A}$, where $1_{A}$ is the
identity function on $A$. In particular, $1_{\emptyset}$ is the empty function
(zero element) and $1_{X}$ is the identity element of $I(X)$. The symmetric
group on $X$ is a subgroup of $I(X)$.

A classical result originating in the work of Wagner and Preston, known as the
Wagner-Preston Representation Theorem, states that every inverse semigroup $S$
may be embedded in the symmetric inverse semigroup $I(S)$. In fact, for any
$a$ in $S$, the mapping

$a^{*}S\to aS,\qquad x\mapsto ax\qquad\text{ ($x\in a^{*}S$)}$

is a bijection from $a^{*}S$ onto $aS$ that represents the element $a$.

###  1.3. The extended bicyclic semigroup

We use the following notation: $\mathbb{N}=\\{1,2,\ldots\\}$;
$\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$;
$\mathbb{Z}=\mathbb{N}_{0}\cup-\mathbb{N}$.

######  Example 1.4.

Let $E_{22}$ be the bicyclic semigroup, as realized by the unilateral shift
([Paterson1999, p. 188]). Thus $E_{22}=\\{a_{ij}:i,j\in\mathbb{N}_{0}\\}$,
where for any $i,j\in\mathbb{Z}$, $e_{ij}$ is the matrix over $\mathbb{Z}$
with 1 in the $i,j$ position and zeros elsewhere, and $a_{ij}$ is the bounded
operator on the Hilbert space $\ell^{2}(\mathbb{Z})$,

$a_{ij}=\sum_{k\geq 0}e_{i+k,j+k}\in B(\ell^{2}(\mathbb{Z})).$

Set

$E=E_{11}\cup E_{12}\cup E_{21}\cup E_{22}$

where $E_{21}=\\{a_{ij}:i\in\mathbb{N}_{0},j\in-\mathbb{N}\\}$,
$E_{11}=\\{a_{ij}:i,j\in-\mathbb{N}\\}$ and
$E_{12}=\\{a_{ij}:i\in-\mathbb{N},j\in\mathbb{N}_{0}\\}$.

We note that for $i,j,p,q\in\mathbb{Z}$, $a_{ij}^{*}=a_{ji}$, $a_{ij}$ is a
partial isometry in $B(H)$ ($H=\ell^{2}(\mathbb{Z})$) and

(1.2) $a_{ij}a_{pq}=\left\\{\begin{array}[]{cc}a_{i,q+j-p},&p\leq j\\\
a_{i+p-j,q},&p\geq j\end{array},\right.$

equivalently

$a_{ij}a_{pq}=a_{i+p-\min(j,p),j+q-\min(j,p)}.$

In particular, $a_{ij}a_{pq}\neq 0$, $a_{ij}a_{jq}=a_{iq}$, and
$a_{ii}a_{pp}=a_{mm}$ with $m=\max(i,p)$.

Thus $E$ is an inverse semigroup consisting of partial isometries with
“inverse” $a_{ij}^{*}$ equal to the adjoint of $a_{ij}$ and the idempotents
coincide with the elements $a_{ii}$. We shall call $E$ the extended bicyclic
semigroup222Ibid., and when convenient notationally, represent $a_{ij}$ in
formulas and diagrams simply by $(i,j)\in\mathbb{Z}\times\mathbb{Z}$.

We shall analyze the extended bicyclic semigroup $E$ of Example 1.4 toward the
aims of finding all of the subsemiheaps of $E$, and showing that the
associated W*-TROs are injective operator spaces.

The idempotents of $E$ are the elements $a_{ii}$ with $i\in\mathbb{Z}$ and
$a_{ii}\leq a_{jj}$, that is, $a_{ii}a_{jj}=a_{ii}$, if and only if $j\leq i$.
From (1.2), we calculate and find that for $p,q\in\mathbb{Z}$,

$a_{ii}a_{pq}=\left\\{\begin{array}[]{ll}a_{i,q+i-p}\quad,&p\leq i\\\
a_{pq}\quad,&p\geq i\end{array}\right.,$
$a_{pq}a_{jj}=\left\\{\begin{array}[]{ll}a_{pq}\quad,&j\leq q\\\
a_{p+j-q,j}\quad,&j\geq q\end{array}\right.,$

and

$a_{ii}a_{pq}a_{jj}=\left\\{\begin{array}[]{ll}a_{i,q+i-p}\quad,&p\leq i\hbox{
and }j\leq q+i-p\\\ a_{j-q+p,q}\quad,&p\leq i\hbox{ and }j\geq q+i-p\\\
a_{pq}\quad,&p\geq i\hbox{ and }j\leq q\\\ a_{p+j-q,j}\quad,&p\geq i\hbox{ and
}j\geq q\end{array}\right.$

In particular, $a_{00}E=E_{21}\cup E_{22}$ and $Ea_{00}=E_{12}\cup E_{22}$.
Also, $a_{ii}E$, $Ea_{jj}$ and $a_{ii}Ea_{jj}$ are subsemigroups and
semiheaps, and $a_{ii}Ea_{jj}$ is an inverse semigroup if $i=j$. More
generally,

$Ea_{jj}=\\{a_{pq}:j\leq q\\},\quad a_{ii}E=\\{a_{pq}:p\geq i\\},$

and

$a_{ii}E\cap Ea_{jj}=a_{ii}Ea_{jj}=\\{a_{pq}:p\geq i,q\geq j\\}.$

##  2\. Diagrams and examples

In order to analyze the subsemiheaps of the extended bicyclic semigroup $E$ in
Example 1.4, we prepare some material. From (1.2), we have the following
lemma.

######  Lemma 2.1.

For any $a_{ij},a_{pq},a_{rs}$ in $E$, we have

$a_{ij}a_{pq}^{*}a_{rs}=\left\\{\begin{array}[]{lrl}\hbox{(i)
}a_{i,s+p+j-q-r}&\quad r\leq p+j-q,&q\leq j\\\ \hbox{(ii)
}a_{i+r-p-j+q,s}&\quad r\geq p+j-q,&q\leq j\\\ \hbox{(iii)
}a_{i+q-j+r-p,s}&\quad r\geq p,&q\geq j\\\ \hbox{(iv) }a_{i+q-j,s+p-r}&\quad
r\leq p,&q\geq j\\\ \end{array}\right.$

It is worth noting, as will be evident in the ten diagrams that follow, all
triple products in $E$ which involve only two elements, produce new elements
which propagate only to the right of, and down from, the two elements, and not
to the left or up.

Lemma 2.2 and Diagrams 1-5 describe the case in which the slope of the line
connecting the two points is negative (or zero or infinite). Lemma 2.3 and
Diagrams 6-10 describe the case in which the slope of the line connecting the
two points is positive (or zero or infinite).

######  Lemma 2.2.

If $K$ is a subsemiheap of $E$, and if $a_{\alpha\beta},a_{\gamma\delta}\in K$
with $\gamma\geq\alpha$ and $\delta\geq\beta$, then the following elements
belong to $K$:

  * •

$x_{1}=a_{\alpha+\delta-\beta,\beta+\gamma-\alpha}$

  * •

$x_{2}=a_{\alpha+\delta-\beta,\delta}$

  * •

$x_{3}=a_{\gamma,\beta+\gamma-\alpha}$

  * •

$x_{4}=a_{\gamma,\delta+(\delta-\beta)-(\gamma-\alpha)}$ if
$\gamma-\alpha\leq\delta-\beta$

  * •

$x_{5}=a_{\gamma+(\gamma-\alpha)-(\delta-\beta),\delta}$ if
$\gamma-\alpha\geq\delta-\beta$

###### Proof.

The following are the eight possible triple products containing two distinct
elements, and thus belong to $K$. They are calculated using Lemma 2.1.

  * •

$a_{\alpha\beta}a_{\alpha\beta}^{*}a_{\alpha\beta}=a_{\alpha\beta}$

  * •

$a_{\alpha\beta}a_{\gamma\delta}^{*}a_{\alpha\beta}=a_{\alpha+\delta-\beta,\beta+\gamma-\alpha}=x_{1}$

  * •

$a_{\alpha\beta}a_{\alpha\beta}^{*}a_{\gamma\delta}=a_{\gamma\delta}$

  * •

$a_{\alpha\beta}a_{\gamma\delta}^{*}a_{\gamma\delta}=a_{\alpha+\delta-\beta,\delta}=x_{2}$

  * •

$a_{\gamma\delta}a_{\alpha\beta}^{*}a_{\alpha\beta}=a_{\gamma\delta}$

  * •

$a_{\gamma\delta}a_{\gamma\delta}^{*}a_{\alpha\beta}=a_{\gamma,\beta+\gamma-\alpha}=x_{3}$

  * •

$a_{\gamma\delta}a_{\alpha\beta}^{*}a_{\gamma\delta}=\left\\{\begin{array}[]{ll}a_{\gamma,\delta+\alpha+\delta-\beta-\gamma}=x_{4},&\quad\gamma-\alpha\leq\delta-\beta\\\
a_{\gamma+\gamma-\alpha-\delta+\beta,\delta}=x_{5},&\quad\gamma-\alpha\geq\delta-\beta\end{array}\right.$

  * •

$a_{\gamma\delta}a_{\gamma\delta}^{*}a_{\gamma\delta}=a_{\gamma\delta}$

∎

Special cases of Lemma 2.2 and their diagrams, which serve as useful
references, are as follows:

######  Diagram 1.

$\delta-\beta>\gamma-\alpha>0$ ($b=\delta-\beta,h=\gamma-\alpha$)

$\beta$ $\delta$

$\alpha$ $\bullet$

$\vdots$ $h$ b$-$h

$\gamma$ $\cdots$ o $\cdots\ \cdots$ $\bullet$ o

$x_{3}$ $x_{4}$

o o

$x_{1}$ $x_{2}$

######  Diagram 2.

$\gamma=\alpha,\delta-\beta>0$

$\beta$ $\delta$

$\alpha,\gamma$ $\bullet$ $\cdots\ \cdots$ $\bullet$ o

$x_{3}$ $x_{4}$

o o

$x_{1}$ $x_{2}$

######  Diagram 3.

$\gamma-\alpha>\delta-\beta>0$ ($b=\delta-\beta,h=\gamma-\alpha$)

$\beta$ $\delta$

$\alpha$ $\bullet$

$\vdots$

$\vdots$ o $x_{2}$ o $x_{1}$

$\vdots$

$\vdots$

$\gamma$ $\cdots\ \cdots$ $\bullet$ h$-$b o $x_{3}$

o $x_{5}$

######  Diagram 4.

$\delta=\beta,\gamma-\alpha>0$

$\beta,\delta$

$\alpha$ $\bullet$ $x_{2}$ o $x_{1}$

$\vdots$

$\gamma$ $\bullet$ o $x_{3}$

o $x_{5}$

######  Diagram 5.

$\delta-\beta=\gamma-\alpha>0$

$\beta$ $\delta$

$\alpha$ $\bullet$

$\ddots$

$\gamma$ $\bullet_{\circ\circ}^{\circ\circ\circ}$
$x_{1},x_{2},x_{3},x_{4},x_{5}$

######  Lemma 2.3.

If $K$ is a subsemiheap of $E$, and if $a_{\alpha\beta},a_{\gamma\delta}\in K$
with $\gamma\geq\alpha,\delta\leq\beta$, then the following elements belong to
$K$:

  * •

$x_{1}=a_{\alpha,\beta+(\gamma-\alpha)+(\beta-\delta)}$

  * •

$x_{2}=a_{\gamma+\beta-\delta,\beta}$

  * •

$x_{3}=a_{\gamma,\beta+\gamma-\alpha}$

  * •

$x_{4}=a_{\gamma+(\beta-\delta)+(\gamma-\alpha),\delta}$

Proof. The following eight products belong to $K$ and can be calculated using
Lemma 2.1.

  * •

$a_{\alpha\beta}a_{\alpha\beta}^{*}a_{\alpha\beta}=a_{\alpha\beta}$

  * •

$a_{\alpha\beta}a_{\gamma\delta}^{*}a_{\alpha\beta}=a_{\alpha,\beta+(\gamma-\alpha)+(\beta-\delta)}=x_{1}$

  * •

$a_{\alpha\beta}a_{\alpha\beta}^{*}a_{\gamma\delta}=a_{\gamma\delta}$

  * •

$a_{\alpha\beta}a_{\gamma\delta}^{*}a_{\gamma\delta}=a_{\alpha\beta}$

  * •

$a_{\gamma\delta}a_{\alpha\beta}^{*}a_{\alpha\beta}=a_{\gamma+\beta-\delta,\beta}=x_{2}$

  * •

$a_{\gamma\delta}a_{\gamma\delta}^{*}a_{\alpha\beta}=a_{\gamma,\beta+\gamma-\alpha}=x_{3}$

  * •

$a_{\gamma\delta}a_{\alpha\beta}^{*}a_{\gamma\delta}=a_{\gamma+\beta-\delta+\gamma-\alpha,\delta}=x_{4}$

  * •

$a_{\gamma\delta}a_{\gamma\delta}^{*}a_{\gamma\delta}=a_{\gamma\delta}$ $\Box$

Special cases of Lemma 2.3 and their diagrams are as follows:

######  Diagram 6.

$\beta-\delta>\gamma-\alpha>0$ ($b=\beta-\delta,h=\gamma-\alpha$)

$\delta$ $\beta$

$\alpha$ $\bullet$ o $x_{1}$

$\vdots$ $h$ $\vdots$

$\gamma$ $\bullet$ $\cdots$ $\cdots$ $\cdots$ $\circ$ $\cdots$ $\cdots$
$\cdots$

$x_{3}$

$b+h$

o $x_{2}$

$\vdots$

o $\cdots\ \cdots\ \cdots$

$x_{4}$

######  Diagram 7.

$\gamma=\alpha,\beta-\delta>0$

$\delta$ $\beta$

$\alpha,\gamma$ $\bullet$ $\cdots\ \cdots$ $\bullet$ o

$x_{3}$ $x_{1}$

o o

$x_{4}$ $x_{2}$

######  Diagram 8.

$\gamma-\alpha>\beta-\delta>0$ ($b=\beta-\delta,h=\gamma-\alpha$)

$\delta$ $\beta$ $b+h$

$\alpha$ $\bullet$ o $x_{1}$

$\vdots$ $\vdots$

$\vdots$ $\vdots$

$\vdots$ $\vdots$

$\gamma$ $\bullet$ $\cdots$ o $\cdots$

$x_{3}$

o $x_{2}$

$\vdots$

$\vdots$

$\vdots$

$x_{4}$ o $\cdots$

######  Diagram 9.

$\delta=\beta,\gamma-\alpha>0$

$\beta,\delta$

$\alpha$ $\bullet$ o $x_{1}$

$\vdots$

$\gamma$ $\bullet$ $x_{2}$ o $x_{3}$

o $x_{4}$

######  Diagram 10.

$\delta-\beta=\gamma-\alpha>0$

$\delta$ $\beta$

$\alpha$ $\bullet$ o $x_{1}$

$\vdots$ $\vdots$

$\gamma$ $\bullet$ $\cdots$ o $\cdots$

$x_{3}$

o $x_{2}$

$\vdots$

o $\cdots$

$x_{4}$

######  Example 2.4.

For $\alpha,\beta\in\mathbb{Z}$, and $J\subset\mathbb{N}_{0}$,
$D_{\alpha,\beta}(J):=\\{a_{\alpha+j,\beta+j}:j\in J\\}$ is a subsemiheap of
$E$.

######  Example 2.5.

For $\alpha,\beta\in\mathbb{Z}$, and $\sigma\in\mathbb{N}$,
$K_{\alpha,\beta}=\\{a_{\alpha+\ell,\beta+m}:\ell,m\in\mathbb{N}_{0}\\}$, and
more generally,
$K_{\alpha,\beta}^{\sigma}:=\\{a_{\alpha+\ell\sigma,\beta+m\sigma}:\ell,m\in\mathbb{N}_{0}\\}$
are subsemiheaps of $E$.

######  Example 2.6.

$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j=1,\ldots\ell_{0}-1\\}\cup
K_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}^{\sigma}$ is a subsemiheap
of $E$, where $\alpha_{0},\beta_{0}\in\mathbb{Z}$,
$\ell_{0},\sigma,k_{i}\in\mathbb{N}$, $k_{1}<k_{2}<\cdots<k_{\ell_{0}-1}$, and

$K_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}^{\sigma}=\\{a_{\alpha_{0}+k_{\ell_{0}}+m\sigma,\beta_{0}+k_{\ell_{0}}+n\sigma}:m,n\in\mathbb{N}_{0}\\}.$

###### Proof.

Let $x_{j}=a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ for $1\leq k_{j}<\ell_{0}$
and $y_{mn}=a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+n\sigma}$ for
$m,n\in\mathbb{N}_{0}$.

The following eight products belong to $K$, as calculated by Lemma 2.1.

  1. (1)

$x_{j}y_{mn}^{*}y_{pq}=\left\\{\begin{array}[]{ll}y_{n+p-m,q}&\hbox{ if }p\geq
m\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(iii))}\\\ y_{n,q+m-p}&\hbox{ if
}p\leq m\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(iv))}\\\
\end{array}\right.$

  2. (2)

$y_{mn}x_{j}^{*}y_{pq}=\left\\{\begin{array}[]{ll}y_{m,n+q-p}&\hbox{ if }p\leq
n\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(i))}\\\ y_{m+p-n,q}&\hbox{ if
}p\geq n\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(ii))}\\\
\end{array}\right.$

  3. (3)

$y_{mn}y_{pq}^{*}x_{j}=\left\\{\begin{array}[]{ll}y_{m,p+n-q}&\hbox{ if }q\leq
n\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(i))}\\\ y_{m+q-n,p}&\hbox{ if
}q\geq n\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(iv)}\\\
\end{array}\right.$

  4. (4)

$y_{mn}x_{i}^{*}x_{j}=y_{mn}\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(i))}$

  5. (5)

$x_{i}y_{mn}^{*}x_{j}=y_{nm}\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(iv))}$

  6. (6)

$x_{i}x_{j}^{*}y_{mn}=y_{mn}\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(ii))}$

  7. (7)

$x_{i}x_{j}^{*}x_{\ell}=x_{\max(i,j,\ell)}\quad\hbox{
(Lemma~{}\ref{lem:0925211}(i)-(iv))}$

  8. (8)

$y_{mn}y_{pq}^{*}y_{rs}=\left\\{\begin{array}[]{ll}y_{m,s+p+n-r-q}&\hbox{ if
}q\leq n\hbox{ and }r+q\leq p+n\quad\hbox{ (Lemma~{}\ref{lem:0925211}(i))}\\\
y_{m+r+q-p-n,s}&\hbox{ if }q\leq n\hbox{ and }r+q\geq p+n\quad\hbox{
(Lemma~{}\ref{lem:0925211}(ii))}\\\ y_{m+q-n+r-p,s}&\hbox{ if }q\geq n\hbox{
and }r\geq p\quad\quad\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(iii))}\\\
y_{m+q-n+p-r,s}&\hbox{ if }q\geq n\hbox{ and }r\leq
p\quad\quad\quad\quad\hbox{ (Lemma~{}\ref{lem:0925211}(iv))}\\\
\end{array}\right.$

We provide some details for cases (1) and (8). For case (1), by Lemma
2.1(iii),

$x_{j}y_{mn}^{*}y_{pq}=a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+n\sigma}^{*}a_{\alpha_{0}+\ell_{0}+p\sigma,\beta_{0}+\ell_{0}+q\sigma}=a_{\alpha_{0}+\ell_{0}+(n+p-m)\sigma,\beta_{0}+\ell_{0}+q\sigma},$

if $p\geq m$ and $\ell_{0}+n\sigma\geq k_{j}$, and by Lemma 2.1(iv),

$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+n\sigma}^{*}a_{\alpha_{0}+\ell_{0}+p\sigma,\beta_{0}+\ell_{0}+q\sigma}=a_{\alpha_{0}+\ell_{0}+n\sigma,\beta_{0}+\ell_{0}+(q+m-p)\sigma},$

if $p\leq m$ and $\ell_{0}+n\sigma\geq k_{j}$.

For case (8), with$q\leq n$, by Lemma 2.1(i),

$a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+n\sigma}a_{\alpha_{0}+\ell_{0}+p\sigma,\beta_{0}+\ell_{0}+q\sigma}^{*}a_{\alpha_{0}+\ell_{0}+r\sigma,\beta_{0}+\ell_{0}+s\sigma}=a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+(s+p+n-r-q)\sigma},$

if $q\leq n$ and $r+q\leq p+n$, so that $s+p+n-r-q\in\mathbb{N}_{0}$; and by
Lemma 2.1(ii),

$a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+n\sigma}a_{\alpha_{0}+\ell_{0}+p\sigma,\beta_{0}+\ell_{0}+q\sigma}^{*}a_{\alpha_{0}+\ell_{0}+r\sigma,\beta_{0}+\ell_{0}+s\sigma}=a_{\alpha_{0}+\ell_{0}+(m+r+q-p-n)\sigma,\beta_{0}+\ell_{0}+s\sigma},$

if $q\leq n$ and $r+q\geq p+n$, so that $m+r+q-p-n\in\mathbb{N}_{0}$.

The subcases of (8) for which $q\geq n$ are as follows. By Lemma 2.1(iii),

$a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+n\sigma}a_{\alpha_{0}+\ell_{0}+p\sigma,\beta_{0}+\ell_{0}+q\sigma}^{*}a_{\alpha_{0}+\ell_{0}+r\sigma,\beta_{0}+\ell_{0}+s\sigma}=a_{\alpha_{0}+\ell_{0}+(m+q-n+r-p)\sigma,\beta_{0}+\ell_{0}+s\sigma},$

if $q\geq n$ and $r\leq p$, so that $m+q-n\in\mathbb{N}_{0}$; and by Lemma
2.1(iv),

$a_{\alpha_{0}+\ell_{0}+m\sigma,\beta_{0}+\ell_{0}+n\sigma}a_{\alpha_{0}+\ell_{0}+p\sigma,\beta_{0}+\ell_{0}+q\sigma}^{*}a_{\alpha_{0}+\ell_{0}+r\sigma,\beta_{0}+\ell_{0}+s\sigma}=a_{\alpha_{0}+\ell_{0}+(m+q-n)\sigma,\beta_{0}+\ell_{0}+(s+p-r)\sigma},$

if $q\geq n$ and $r\leq p$, so that $m+q-n\in\mathbb{N}_{0}$ and
$s+p-r\in\mathbb{N}_{0}$. ∎

######  Example 2.7.

Let $K=\bigcup_{k\in A}K_{\alpha_{0}+k,\beta_{0}+k}^{p}$, where
$\alpha_{0},\beta_{0}\in\mathbb{Z}$, $p>0$, $A\subset\\{0,1,\ldots,p-1\\}$ and
$a_{\alpha_{0}+k,\beta_{0}+k}$, $k\in A$, denote the elements of $K$ lying on
the diagonal with $k<p$. (See Diagram 11 and Proposition 3.15.) In fact, $K$
is a sub-inverse semigroup of $E$.

###### Proof.

We note first that (setting $\alpha_{0}=\beta_{0}=0$ for convenience, see
Remark 2.8)

$K=\\{(k+\ell p,k+mp):k\in A,\ell,m\in\mathbb{N}_{0}\\}$

and it suffices to show that

$(k_{1}+\ell_{1}p,k_{1}+m_{1}p)(k_{2}+\ell_{2}p,k_{2}+m_{2}p)^{*}(k_{3}+\ell_{3}p,k_{3}+m_{3}p)$

belongs to $K$. We calculate this triple product using the four cases in Lemma
2.1.

By Lemma 2.1(i), if $k_{2}+m_{2}p\leq k_{1}+m_{1}p$, and $k_{3}+\ell_{3}p\leq
k_{1}+(\ell_{2}+m_{1}-m_{2})p$, then

$(k_{1}+\ell_{1}p,k_{1}+m_{1}p)(k_{2}+\ell_{2}p,k_{2}+m_{2}p)^{*}(k_{3}+\ell_{3}p,k_{3}+m_{3}p)=(k_{1}+\ell_{1}p,k_{1}+(m_{3}+\ell_{2}+m_{1}-m_{2}-\ell_{3})p),$

and it is required to show that $m_{3}+\ell_{2}+m_{1}-m_{2}-\ell_{3}\geq 0$.

Following the argument in [Descalco2005, Lemma 4.5], we have

$k_{1}+(\ell_{2}+m_{1}-m_{2}-\ell_{3})p\geq k_{3}\geq 0$

so that $(\ell_{2}+m_{1}-m_{2}-\ell_{3})p\geq-k_{1}>-p$ and therefore
$\ell_{2}+m_{1}-m_{2}-\ell_{3}\geq 0$ and
$m_{3}+\ell_{2}+m_{1}-m_{2}-\ell_{3}\geq 0$, as required.

By Lemma 2.1(ii), if $k_{2}+m_{2}p\leq k_{1}+m_{1}p$, and $k_{3}+\ell_{3}p\geq
k_{1}+(\ell_{2}+m_{1}-m_{2})p$, then

$(k_{1}+\ell_{1}p,k_{1}+m_{1}p)(k_{2}+\ell_{2}p,k_{2}+m_{2}p)^{*}(k_{3}+\ell_{3}p,k_{3}+m_{3}p)=(k_{3}+(\ell_{1}+\ell_{3}-\ell_{2}-m_{1}+m_{2})p,k_{3}+m_{3}p)$

and it is required to show that $\ell_{3}-\ell_{2}-m_{1}+m_{2}\geq 0$.

Following the argument in [Descalco2005, Lemma 4.5], we have

$k_{3}+(\ell_{3}-\ell_{2}-m_{1}+m_{2})p\geq k_{1}\geq 0$

so that $(\ell_{3}-\ell_{2}-m_{1}+m_{2})p\geq-k_{3}>-p$ and therefore
$\ell_{3}-\ell_{2}-m_{1}+m_{2}\geq 0$ and
$\ell_{1}+\ell_{3}-\ell_{2}-m_{1}+m_{2}\geq 0$, as required.

By Lemma 2.1(iii), if $k_{3}+\ell_{3}p\geq k_{2}+\ell_{2}p$, and
$k_{2}+m_{2}p\geq k_{1}+m_{1}p$, then

$(k_{1}+\ell_{1}p,k_{1}+m_{1}p)(k_{2}+\ell_{2}p,k_{2}+m_{2}p)^{*}(k_{3}+\ell_{3}p,k_{3}+m_{3}p)=(k_{3}+(\ell_{1}+m_{2}-m_{1}+\ell_{3}-\ell_{2})p,k_{3}+m_{3}p)$

and it is required to show that $\ell_{1}+m_{2}-m_{1}+\ell_{3}-\ell_{2}\geq
0$.

Following the argument in [Descalco2005, Lemma 4.5], we have

$k_{3}+(\ell_{3}-m_{1}+m_{2}-\ell_{2})p\geq k_{1}\geq 0$

so that $(\ell_{3}-m_{1}+m_{2}-\ell_{2})p\geq-k_{3}>-p$ and therefore
$\ell_{3}-m_{1}+m_{2}-\ell_{2}\geq 0$ and
$\ell_{1}+\ell_{3}-m_{1}+m_{2}-\ell_{2}\geq 0$, as required.

By Lemma 2.1(iv), if $k_{3}+\ell_{3}p\leq k_{2}+\ell_{2}p$, and
$k_{2}+m_{2}p\geq k_{1}+m_{1}p$, then

$(k_{1}+\ell_{1}p,k_{1}+m_{1}p)(k_{2}+\ell_{2}p,k_{2}+m_{2}p)^{*}(k_{3}+\ell_{3}p,k_{3}+m_{3}p)=(k_{2}+(\ell_{1}+m_{2}-m_{1})p,k_{2}+(m_{3}+\ell_{2}-\ell_{3})p)$

and it is required to show that $\ell_{1}+m_{2}-m_{1}\geq 0$. and
$m_{3}+\ell_{2}-\ell_{3}\geq 0$.

Following the argument in [Descalco2005, Lemma 4.5], we have

$k_{2}+(\ell_{2}-\ell_{3})p\geq k_{3}\geq 0$

so that $(\ell_{2}-\ell_{3})p\geq-k_{3}>-p$ and therefore
$\ell_{2}-\ell_{3}\geq 0$ and $m_{3}+\ell_{2}-\ell_{3}\geq 0$, as required.

Following the argument in [Descalco2005, Lemma 4.5], we have

$k_{2}+(m_{2}-m_{3})p\geq k_{1}\geq 0$

so that $(m_{2}-m_{1})p\geq-k_{1}>-p$ and therefore $m_{2}-m_{1}\geq 0$ and
$\ell_{1}+m_{2}-m_{1}\geq 0$, as required. ∎

######  Remark 2.8.

The adjoint operation $a_{ij}\mapsto a_{ij}^{*}=a_{ji}$ on the extended
bicyclic semigroup $E$ is an anti-isomorphism of a subsemiheap $K$ of $E$ onto
the subsemiheap $K^{*}$, that is, $(ab^{*}c)^{*}=c^{*}ba^{*}$. As another
application of Lemma 2.1, the translation map on the extended bicyclic
semigroup is a triple isomorphism, that is, if
$\varphi_{\alpha,\beta}(a_{ij})=a_{i+\alpha,j+\beta}$, then

$\varphi(a_{ij}a_{pq}^{*}a_{rs})=\varphi(a_{ij})\varphi(a_{pq})^{*}\varphi(a_{rs}).$

Hence, if $K$ is a subsemiheap of $K_{\alpha,\beta}$, then
$\varphi_{-\alpha,-\beta}(K)$ is a subsemiheap of the bicyclic semigroup
$K_{0,0}$. At the very least, this fact can simplify notation in parts of this
paper. A sub-semigroup of the bicyclic semigroup $K_{0,0}$ (see
[Descalco2005]), is not necessarily a subsemiheap, unless it is a sub-inverse
semigroup (cf. [Descalco2005, Corollary 7.1]).

######  Diagram 11.

|  |  |  |  | $k_{1}$ |  | $k_{2}$ | $k_{3}$ |  | $k_{4}$ |  | $k_{5}$ |  | $k_{6}$ | $k_{7}$ |  | $k_{8}$ |  | $k_{9}$ |  |  | $k_{10}$ |  | $k_{11}$ | $k_{12}$ |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  | 0 |  |  | $q$ |  |  | $p$ |  |  |  | $q+p$ |  |  | $2p$ |  |  |  | $q+2p$ |  |  | $3p$ |  |  | $q+3p$ |  |
| 0 | $\bullet$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |
$k_{1}$ | q |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  | $\blacksquare$ | $\cdots$ |
$k_{2}$ |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |
$k_{3}$ | $p$ | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |
$k_{4}$ |  |  |  |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  |  |
$k_{5}$ | $q+p$ |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  | $\blacksquare$ | $\cdots$ |
$k_{6}$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |
$k_{7}$ | $2p$ | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |
$k_{8}$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  |  |
$k_{9}$ | $q+2p$ |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  | $\blacksquare$ | $\cdots$ |
$k_{10}$ | $3p$ | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |
$k_{11}$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\bullet$ |  |  |
$k_{12}$ | $q+3p$ |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  |  | $\blacksquare$ |  |  |  |  |  | $\blacksquare$ | $\cdots$ |
|  |  |  |  | $\vdots$ |  |  |  |  |  |  | $\vdots$ |  |  |  |  |  |  | $\vdots$ |  |  |  |  |  | $\vdots$ |  |

##  3\. Subsemiheaps of the extended bicyclic semigroup

In this section we shall determine all of the subsemiheaps of the extended
bicyclic semigroup. We shall proceed as follows. First, for an arbitrary
subsemiheap $K$ of $E$, we define

$\alpha_{0}=\inf\\{\alpha\in\mathbb{Z}:\exists\beta\in\mathbb{Z},a_{\alpha\beta}\in
K\\},$

and

$\beta_{0}=\inf\\{\beta\in\mathbb{Z}:\exists\alpha\in\mathbb{Z},a_{\alpha\beta}\in
K\\}.$

We have four mutually exclusive and exhaustive cases, namely,

  1. (1)

Quadrant $\alpha_{0}\neq-\infty,\beta_{0}\neq-\infty$

  2. (2)

Right Half Plane $\alpha_{0}=-\infty,\beta_{0}\neq-\infty$

  3. (3)

Lower Half Plane $\alpha_{0}\neq-\infty,\beta_{0}=-\infty$

  4. (4)

Full Plane $\alpha_{0}=-\infty,\beta_{0}=-\infty$

######  Remark 3.1.

We only need to find all of the subsemiheaps of $E$ which are in case (1),
since the other cases can be reduced to this case in steps, as follows, which
shows that every subsemiheap of the extended bicyclic semigroup is the
inductive limit of subsemiheaps in case (1) in the category of semiheaps and
semiheap homomorphisms333In fact, it is an elementary inductive limit since
the connecting maps are inclusions (see Theorem 3.19).

  * •

If a subsemiheap $K$ of $E$ is in case (2), then
$K\subset\\{a_{ij}:i\in\mathbb{Z},j\geq\beta_{0}\\}$ and
$K=\cup_{\alpha\in\mathbb{Z}}K^{\alpha}$, where
$K^{\alpha}=K\cap\\{a_{ij}:i\geq\alpha,j\geq\beta_{0}\\}$ (which we have
denoted by $K_{\alpha,\beta_{0}}$) is in case (1).

  * •

If a subsemiheap $K$ of $E$ is in case (3), then
$K\subset\\{a_{ij}:i\geq\alpha_{0},j\in\mathbb{Z}\\}$ and
$K=\cup_{\beta\in\mathbb{Z}}K_{\beta}$, where
$K_{\beta}=K\cap\\{a_{ij}:i\geq\alpha_{0},j\geq\beta\\}$
(=$K_{\alpha_{0},\beta}$) is in case (1).

  * •

If a subsemiheap $K$ of $E$ is in case (4), then
$K\subset\\{a_{ij}:i,j\in\mathbb{Z}\\}$ and
$K=\cup_{\alpha\in\mathbb{Z}}K_{(\alpha)}$, where
$K_{(\alpha)}=K\cap\\{a_{ij}:i\geq\alpha,j\in\mathbb{Z}\\}$ is in case (3).

Alternatively, if a subsemiheap $K$ of $E$ is in case (4), then
$K=\cup_{\beta\in\mathbb{Z}}K^{(\beta)}$, where
$K^{(\beta)}=K\cap\\{a_{ij}:i\in\mathbb{Z},j\geq\beta\\}$ is in case (2).

Therefore we shall concentrate only on case (1). Suppose that
$\alpha_{0}\neq-\infty$ and $\beta_{0}\neq-\infty$. Then $K\subset
K_{\alpha_{0},\beta_{0}}=\\{a_{pq}:p\geq\alpha_{0},q\geq\beta_{0}\\}.$ We
define three parameters as follows:

$\overline{\beta}=\sup\\{\beta\in\mathbb{Z}:a_{\alpha_{0}\beta}\in K\\}$
$\overline{\alpha}=\sup\\{\alpha\in\mathbb{Z}:a_{\alpha\beta_{0}}\in K\\}$
$\overline{\gamma}=\sup\\{k\in\mathbb{N}_{0}:a_{\alpha_{0}+k,\beta_{0}+k}\in
K\\}.$

.

We shall consider three primary cases:

${\bf 1.}\ \overline{\beta}=\beta_{0}\quad\quad{\bf 2.}\
\beta_{0}<\overline{\beta}<\infty\quad\quad{\bf 3.}\ \overline{\beta}=\infty$

Each of the cases 1, 2, 3, consists of three further subcases.

${\bf 1.1}\
\overline{\beta}=\beta_{0},\quad\overline{\alpha}=\alpha_{0}\quad\quad{\bf
1.2}\
\overline{\beta}=\beta_{0},\quad\alpha_{0}<\overline{\alpha}<\infty,\quad{\bf
1.3}\ \overline{\beta}=\beta_{0},\overline{\alpha}=\infty.$

${\bf 2.1}\
\overline{\beta}=\beta_{0}<\infty,\overline{\alpha}=\alpha_{0}\quad\quad{\bf
2.2}\
\beta_{0}<\overline{\beta}<\infty,\alpha_{0}<\overline{\alpha}<\infty,\quad\quad{\bf
2.3}\ \beta_{0}<\overline{\beta}<\infty,\overline{\alpha}=\infty.$

${\bf 3.1}\
\overline{\beta}=\infty,\overline{\alpha}=\alpha_{0}\quad\quad{\bf}\
\overline{\beta}=\infty,\alpha_{0}<\overline{\alpha}<\infty,\quad\quad{\bf
3.3}\ \overline{\beta}=\infty,\overline{\alpha}=\infty.$

Each of these nine cases consists of three further subcases. Thus, in order to
account for the quadrant case (1), and hence the other three cases, it will be
necessary to consider 27 cases. We summarize the results in the table on the
next page. It is worthy to note that by Diagram 2, if $\overline{\beta}$ is
finite, by which we mean, $\beta_{0}<\overline{\beta}<\infty$, then
$a_{\alpha_{0},\overline{\beta}}$ is the only point of $K$ of the form
$a_{\alpha_{0},\beta}$. A similar statement holds for $\overline{\alpha}$.
Also, if $\overline{\alpha}=\alpha_{0}$, or if $\overline{\beta}=\beta_{0}$,
then $a_{\alpha_{0},\beta_{0}}\in K$. Thus in cases 2.2, 2.3, 3.2, and 3.3, it
is necessary to consider the two possibilities: $a_{\alpha_{0},\beta_{0}}\in
K$, and $a_{\alpha_{0},\beta_{0}}\not\in K$.

We now proceed to analyze all 27 cases.

######  Lemma 3.2.

In Case 1.1.1 ($\overline{\beta}=\beta_{0},\ \overline{\alpha}=\alpha_{0},\
\overline{\gamma}=0$), we have $K=\\{a_{\alpha_{0}\beta_{0}}\\}$.

Proof. In this case, the diagram is the following, where the bullet represents
the element $a_{\alpha_{0}\beta_{0}}$, and the circles indicate that no
element of $K$ occupies that position. (Ignore, for the moment, the symbols
$\blacksquare,\blacktriangle,\triangle$)

$\bullet$ o o o o o

o $\triangle$ $\blacksquare$

o $\blacktriangle$ o

o o

o o

$\vdots$ $\ddots$

Suppose that $a_{\alpha_{0}+2,\beta_{0}+1}$, denoted by $\blacktriangle$,
belonged to $K$. Then by diagram 3 applied to the points
$a_{\alpha_{0}\beta_{0}}$ and $\blacktriangle$, the point
$a_{\alpha_{0}+1,\beta_{0}+1}$, denoted by $\triangle$, would belong to $K$, a
contradiction. So $\blacktriangle$ does not belong to $K$. By the same
argument, no element of $K$ resides in the second column of the diagram.

Suppose that $a_{\alpha_{0}+1,\beta_{0}+2}$, denoted by $\blacksquare$,
belonged to $K$. Then by diagram 1 applied to the points
$a_{\alpha_{0}\beta_{0}}$ and $\blacksquare$, the point
$a_{\alpha_{0}+1,\beta_{0}+1}$, denoted by $\triangle$, would belong to $K$, a
contradiction. So $\blacksquare$ does not belong to $K$. By the same argument,
no element of $K$ resides in the second row of the diagram.

Repetition of these two arguments shows that no element of $K$ resides in any
column or row of the diagram, other than the first row and column, and
therefore $K=\\{a_{\alpha_{0}\beta_{0}}\\}$ contains exactly one element.
$\Box$

Classification Scheme

| Case | subcase | $\overline{\beta}$ | $\overline{\alpha}$ | $\overline{\gamma}$ | exists? | result
---|---|---|---|---|---|---|---
|  | 1.1.1. | $\beta_{0}$ | $\alpha_{0}$ | 0 | yes | Lemma 3.2
| 1.1 | 1.1.2. | $\beta_{0}$ | $\alpha_{0}$ | finite | yes | Proposition 3.3
|  | 1.1.3. | $\beta_{0}$ | $\alpha_{0}$ | $\infty$ | yes | Proposition 3.4
|  | 1.2.1. | $\beta_{0}$ | finite | 0 | no | Lemma 3.7
1 | 1.2 | 1.2.2. | $\beta_{0}$ | finite | finite | no | Lemma 3.7
|  | 1.2.3. | $\beta_{0}$ | finite | $\infty$ | no | Lemma 3.7
|  | 1.3.1. | $\beta_{0}$ | $\infty$ | 0 | no | Lemma 3.7
| 1.3 | 1.3.2. | $\beta_{0}$ | $\infty$ | finite | no | Lemma 3.7
|  | 1.3.3. | $\beta_{0}$ | $\infty$ | $\infty$ | no | Lemma 3.7
|  | 2.1.1. | finite | $\alpha_{0}$ | 0 | no | Lemma 3.8
| 2.1 | 2.1.2. | finite | $\alpha_{0}$ | finite | no | Lemma 3.8
|  | 2.1.3. | finite | $\alpha_{0}$ | $\infty$ | no | Lemma 3.8
|  | 2.2.1. | finite | finite | 0 | no | Lemma 3.9
2 | 2.2 | 2.2.2. | finite | finite | finite | no | Lemma 3.9
|  | 2.2.3. | finite | finite | $\infty$ | no | Lemma 3.9
|  | 2.3.1. | finite | $\infty$ | 0 | no | Lemma 3.10
| 2.3 | 2.3.2. | finite | $\infty$ | finite | no | Lemma 3.10
|  | 2.3.3. | finite | $\infty$ | $\infty$ | no | Lemma 3.10
|  | 3.1.1. | $\infty$ | $\alpha_{0}$ | 0 | no | Lemma 3.11
| 3.1 | 3.1.2. | $\infty$ | $\alpha_{0}$ | finite | no | Lemma 3.11
|  | 3.1.3. | $\infty$ | $\alpha_{0}$ | $\infty$ | no | Lemma 3.11
|  | 3.2.1. | $\infty$ | finite | 0 | no | Lemma 3.11
3 | 3.2 | 3.2.2. | $\infty$ | finite | finite | no | Lemma 3.11
|  | 3.2.3. | $\infty$ | finite | $\infty$ | no | Lemma 3.11
|  | 3.3.1. | $\infty$ | $\infty$ | 0 | no | Proposition 3.12
| 3.3 | 3.3.2. | $\infty$ | $\infty$ | finite | no | Proposition 3.12
|  | 3.3.3. | $\infty$ | $\infty$ | $\infty$ | yes | Proposition 3.18

######  Proposition 3.3.

In Case 1.1.2 ($\overline{\beta}=\beta_{0},\ \alpha_{0}=\overline{\alpha},\
0<\overline{\gamma}<\infty$), we have

$K=\\{a_{\alpha_{0},\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:1\leq
j\leq n_{0}\\},$

where $1\leq k_{1}<k_{2}<\cdots<k_{n_{0}}=\overline{\gamma}$ and
$a_{\alpha_{0},\beta_{0}}$ and $a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ are the
only elements of $K$ that are in the diagonal.

Proof. In this case, in Diagram 12, the bullets represent some of the elements
of $K$ residing on the diagonal, the circles indicate that no element of $K$
occupies that position, and the dots represent both the finite number of
points of $K$ on the diagonal together with some positions on the diagonal not
containing points of $K$ (Ignore for the moment, the symbols
$\circledcirc,\boxdot$ which represent two elements of $K$ lying on the
diagonal, and the symbols $\blacksquare,\square,\blacktriangle,\triangle$).
The symbol $\clubsuit$ represents the element
$a_{\alpha_{0}+k_{n_{0}},\beta_{0}+k_{n_{0}}}$. We shall show that all off-
diagonal positions are not occupied by elements of $K$, which means that
$K=\\{a_{\alpha_{0},\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:1\leq
j\leq n_{0}\\}$.

Suppose that for $k_{j}\leq\ell<k_{j+1}$, the point
$a_{\alpha_{0}+k_{j+1},\beta_{0}+\ell}$, denoted by $\blacksquare$ in Diagram
12, belonged to $K$. Then by Diagram 2, starting with $\blacksquare$ and
$a_{\alpha_{0}+k_{j+1},\beta_{0}+k_{j+1}}$, denoted by $\boxdot$ in Diagram
12, shows that

$K\supset
K_{\alpha_{0}+k_{j+1},\beta_{0}+\ell}^{k_{j+1}-\ell}=\\{a_{\alpha_{0}+k_{j+1}+m(k_{j+1}-\ell),\beta_{0}+\ell+n(k_{j+1}-\ell)}:m,n\in\mathbb{N}_{0}\\}.$

Then choosing $m=n-1$, so that

$k_{j+1}+m(k_{j+1}-\ell)=\ell+n(k_{j+1}-\ell)$

and letting $n\rightarrow\infty$ shows that there are infinitely many points
of $K$ on the diagonal, a contradiction, so $\blacksquare\not\in K$. The same
argument applies to every point on each row determined by $\alpha_{0}+k_{j+1}$
to the left of $a_{\alpha_{0}+k_{j+1},\beta_{0}+k_{j+1}}$ for $0\leq j\leq
n_{0}-1$.

######  Diagram 12.

|  |  | $k_{j}$ | $\ell$ | $k_{j+1}$ |  | $k_{i}$ |  | $k_{n_{0}}$ |
---|---|---|---|---|---|---|---|---|---|---
| $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\ddots$ |  |  |  |  |  |  |  |
$k_{j}$ | $\circ$ |  | $\circledcirc$ | $\blacktriangle$ |  |  |  |  |  |
$\ell$ | $\circ$ |  | $\triangle$ | $\ddots$ | $\square$ |  |  |  |  |
$k_{j+1}$ | $\circ$ |  |  | $\blacksquare$ | $\boxdot$ |  |  |  |  |
| $\circ$ |  |  |  |  | $\ddots$ |  |  |  |
| $\circ$ |  |  |  |  |  | $\bullet$ |  |  |
| $\circ$ |  |  |  |  |  |  | $\ddots$ |  |
| $\circ$ |  |  |  |  |  |  |  | $\clubsuit$ |
| $\vdots$ |  |  |  |  |  |  |  |  | $\ddots$

Suppose now that for $k_{j}<\ell\leq k_{j+1}$, the point
$a_{\alpha_{0}+k_{j},\beta_{0}+\ell}$, denoted by $\blacktriangle$ in Diagram
12, belonged to $K$. Then by Diagram 2, starting with
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$, denoted by $\circledcirc$ in Diagram
12, and $\blacktriangle$, shows that

$K\supset K_{\alpha_{0}+k_{j},\beta_{0}+\ell}^{\ell-
k_{j}}=\\{a_{\alpha_{0}+k_{j}+m(\ell-k_{j}),\beta_{0}+k_{j}+n(\ell-
k_{j})}:m,n\in\mathbb{N}_{0}\\}.$

Then choosing $m=n$, and letting $n\rightarrow\infty$ show that there are
infinitely many points of $K$ on the diagonal, a contradiction, so
$\blacktriangle\not\in K$. The same argument applies to every point on each
row $\alpha_{0}+k_{j}$ to the right of $a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$
for $0\leq j\leq n_{0}$.

Thus all rows containing an element of $K$ on the diagonal do not contain any
other elements of $K$, as in the following diagram.

|  |  | $k_{j}$ | $\ell$ | $k_{j+1}$ |  | $k_{i}$ |  | $k_{n_{0}}$ |
---|---|---|---|---|---|---|---|---|---|---
| $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\ddots$ |  |  |  |  |  |  |  |
$k_{j}$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
$\ell$ | $\circ$ |  |  | $\ddots$ |  |  |  |  |  |
$k_{j+1}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ |  |  |  |  | $\ddots$ |  |  |  |
$k_{i}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ |  |  |  |  |  |  | $\ddots$ |  |
$k_{n_{0}}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\clubsuit$ | $\cdots$
| $\vdots$ |  |  |  |  |  |  |  |  | $\ddots$

A parallel argument, using Diagram 4 shows that all columns containing an
element of $K$ on the diagonal do not contain any other elements of $K$. For
completeness, we include the details.

Suppose that for $k_{j}\leq\ell<k_{j+1}$, the point
$a_{\alpha_{0}+\ell,\beta_{0}+k_{j+1}}$, denoted by $\square$ in Diagram 12,
belonged to $K$. Then by Diagram 4, starting with $\square$ and
$a_{\alpha_{0}+k_{j+1},\beta_{0}+k_{j+1}}$, denoted by $\boxdot$, shows that

$K\supset
K_{\alpha_{0}+\ell,\beta_{0}+k_{j+1}}^{k_{j+1}-\ell}=\\{a_{\alpha_{0}+\ell+m(k_{j+1}-\ell),\beta_{0}+k_{j+1}+n(k_{j+1}-\ell)}:m,n\in\mathbb{N}_{0}\\}.$

Then choosing $m=n+1$, so that

$k_{j+1}+n(k_{j+1}-\ell)=\ell+m(k_{j+1}-\ell)$

and letting $n\rightarrow\infty$ shows that there are infinitely many points
of $K$ on the diagonal, a contradiction, so $\square\not\in K$. The same
argument applies to every point on each column determined by
$\beta_{0}+k_{j+1}$ above $a_{\alpha_{0}+k_{j+1},\beta_{0}+k_{j+1}}$ for
$0\leq j\leq n_{0}-1$.

Suppose now that for $k_{j}\leq\ell<k_{j+1}$, the point
$a_{\alpha_{0}+\ell,\beta_{0}+k_{j}}$, denoted by $\triangle$ in Diagram 12,
belonged to $K$. Then by Diagram 4, starting with $\triangle$ and
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$, denoted by $\circledcirc$, shows that

$K\supset K_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}^{\ell-
k_{j}}=\\{a_{\alpha_{0}+k_{j}+m(\ell-k_{j}),\beta_{0}+k_{j}+n(\ell-
k_{j})}:m,n\in\mathbb{N}_{0}\\},$

Then choosing $m=n$, so that

$k_{j}+n(\ell-k_{j})=k_{j}+m(\ell-k_{j})$

and letting $n\rightarrow\infty$ shows that there are infinitely many points
of $K$ on the diagonal, a contradiction, so $\triangle\not\in K$. The same
argument applies to every point on each column determined by $\beta_{0}+k_{j}$
above $a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ for $0\leq j\leq n_{0}$.

To complete the proof, we now show that no point
$a_{\alpha_{0}+m,\beta_{0}+n}$, with $m\neq n$ can belong to $K$. By what was
just proved, it suffices to consider points which are not on a row or column
containing a point of $K$, that is, $m\neq k_{j}$ for all $j$ and $n\neq
k_{\ell}$ for all $\ell$.

We shall refer to the diagram below, which depicts the eight possible
locations for the element $a_{\alpha_{0}+m,\beta_{0}+n}$, reflecting the cases
$m>n$ and $m<n$, namely,

  1. (1)

$m>k_{n_{0}}\geq k_{\ell+1}>n>k_{\ell}$, denoted by $\blacksquare$

  2. (2)

$k_{\ell+1}>m>k_{\ell}\geq k_{j+1}>n>k_{j}$, denoted by $\blacktriangle$

  3. (3)

$k_{j+1}>m>n>k_{j}$, denoted by $\blacktriangledown$

  4. (4)

$m>n>k_{n_{0}}$, denoted by $\blacktriangleleft$

  5. (5)

$n>k_{n_{0}}\geq k_{\ell+1}>m>k_{\ell}$, denoted by $\square$

  6. (6)

$k_{\ell+1}>n>k_{\ell}\geq k_{j+1}>m>k_{j}$, denoted by $\triangle$

  7. (7)

$k_{j+1}>n>m>k_{j}$, denoted by $\triangledown$

  8. (8)

$n>m>k_{n_{0}}$, denoted by $\triangleleft$

Suppose first that $m>n$, for example case (2), $k_{\ell+1}>m_{2}>k_{\ell}\geq
k_{j+1}>n_{2}>k_{j}$. We consider the two points
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ and
$\blacktriangle=a_{\alpha_{0}+m_{2},\beta_{0}+n_{2}}$. These two points are
vertices of a triangle with height $h=m_{2}-k_{j}$ greater than the base
$b=n_{2}-k_{j}$, so $h-b=m_{2}-n_{2}$. Then by Lemma 2.2 (see Diagram 3), the
point
$x_{2}=a_{\alpha_{0}+k_{j}+\beta_{0}+m_{2}-\alpha_{0}-k_{j},\beta_{o}+m_{2}}=a_{\alpha_{0}+m_{2},\beta_{0}+m_{2}}$
would belong to $K$, which is a contradiction since
$k_{\ell}<m_{2}<k_{\ell+1}$.

|  |  |  |  |  |  |  |  |  |  |  |  $n_{6}$, |  |  |  | $n_{4},$ |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  |  |  | $n_{3}$ | $n_{2}$ | $n_{7}$ |  |  |  |  | $n_{1}$ |  |  |  | $n_{8}$ | $n_{5}$ |  |
|  |  |  | $k_{j}$ |  |  |  | $k_{j+1}$ |  |  | $k_{\ell}$ |  | $k_{\ell+1}$ |  | $k_{n_{0}}$ |  |  |  |
|  | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
|  | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  |  |  |  |  |
| $k_{j}$ | $\circ$ |  | $\bullet$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
$m_{7}$ |  | $\circ$ |  |  |  |  | $\triangledown$ |  |  |  |  |  |  |  |  |  |  |  |
$m_{6}$ |  | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ | $\triangle$ | $\circ$ |  |  |  |  |  |
$m_{3}$ |  | $\circ$ |  |  | $\blacktriangledown$ |  |  |  |  |  |  |  |  |  |  |  |  |  |
| $k_{j+1}$ | $\circ$ |  | $\circ$ |  |  |  | $\bullet$ |  |  | $\circ$ |  | $\circ$ |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
|  | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  |  |  |  |  |
|  | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  |  |  |  |  |
| $k_{\ell}$ | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\bullet$ |  | $\circ$ |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
$m_{2},m_{5}$ |  | $\circ$ |  | $\circ$ |  | $\blacktriangle$ |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  |  |  | $\square$ |  |
| $k_{\ell+1}$ | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\bullet$ |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
|  | $\circ$ |  | $\vdots$ |  |  |  | $\vdots$ |  |  | $\vdots$ |  | $\vdots$ | $\ddots$ | $\vdots$ |  |  |  |
| $k_{n_{0}}$ | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  | $\clubsuit$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
|  | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  | $\circ$ | $\circ$ |  |  |
$m_{1},m_{8}$ |  | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ | $\blacksquare$ | $\circ$ |  | $\circ$ | $\triangleleft$ | $\circ$ |  |
$m_{4}$ |  | $\circ$ |  | $\circ$ |  |  |  | $\circ$ |  |  | $\circ$ |  | $\circ$ |  | $\circ$ | $\blacktriangleleft$ | $\circ$ |  |
|  | $\vdots$ |  | $\vdots$ |  |  |  | $\vdots$ |  |  | $\vdots$ |  | $\vdots$ |  | $\vdots$ |  |  |  | $\ddots$

Suppose next that $m<n$, for example, case (6), $k_{j}<m_{6}<k_{j+1}\leq
k_{\ell}<n_{6}<k_{\ell+1}$ We again consider the two points
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ and
$\triangle=a_{\alpha_{0}+m_{6},\beta_{0}+n_{6}}$. These two points are
vertices of a triangle with height $h=m_{6}-k_{j}$ less than the base
$b=n_{6}-k_{j}$ and $b-h=n_{6}-m_{6}$. Then by Lemma 2.2 (see Diagram 1), the
point
$x_{1}=a_{\alpha_{0}+k_{j}+\beta_{0}+n_{6}-\beta_{0}-k_{j},\beta_{0}+n_{6}}=a_{\alpha_{0}+n_{6},\beta_{0}+n_{6}}$
would belong to $K$, which is a contradiction since
$k_{\ell}<n_{6}<k_{\ell+1}$.

The same two-part argument works in all the other cases, more precisely, as
follows:

  * •

For case (3), $k_{j}<n_{3}<m_{3}<k_{j+1}$, Diagram 3 applied to
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ and $\blacktriangledown$ yields
$x_{2}=a_{\alpha_{0}+n_{3},\beta_{0}+n_{3}}$

  * •

For case (7), $k_{j}<m_{7}<n_{7}<k_{j+1}$ , Diagram 1 applied to
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ and $\triangledown$ yields
$x_{3}=a_{\alpha_{0}+m_{7},\beta_{0}+m_{7}}$

  * •

For case (4), $m_{4}>n_{4}>k_{n_{0}}$, Diagram 3 applied to
$a_{\alpha_{0}+k_{n_{0}},\beta_{0}+k_{n_{0}}}$ and $\blacktriangleleft$ yields
$x_{2}=a_{\alpha_{0}+n_{4},\beta_{0}+n_{4}}$

  * •

For case (8), $n_{8}>m_{8}>k_{n_{0}}$, Diagram 1 applied to
$a_{\alpha_{0}+k_{n_{0}},\beta_{0}+k_{n_{0}}}$ and $\triangleleft$ yields
$x_{3}=a_{\alpha_{0}+m_{8},\beta_{0}+m_{8}}$

  * •

For case (1), $m_{1}>k_{n_{0}}\geq k_{\ell+1}>n_{1}>k_{\ell}$, Diagram 3
applied to $a_{\alpha_{0}+k_{\ell},\beta_{0}+k_{\ell}}$ and $\blacksquare$
yields $x_{2}=a_{\alpha_{0}+n_{1},\beta_{0}+n_{1}}$

  * •

For case (5), $n_{5}>k_{n_{0}}\geq k_{\ell+1}>m_{5}>k_{\ell}$, Diagram 1
applied to $a_{\alpha_{0}+k_{\ell},\beta_{0}+k_{\ell}}$ and $\square$ yields
$x_{3}=a_{\alpha_{0}+m_{5},\beta_{0}+m_{5}}$. $\Box$

This completes the analysis of case 1.1.2.

######  Proposition 3.4.

In Case 1.1.3: ($\overline{\beta}=\beta_{0},\ \overline{\alpha}=\alpha_{0},\
\overline{\gamma}=\infty$), either
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j\in\mathbb{N}\\}$,
or there exist $\sigma,\ell_{0}\in\mathbb{N}$, such that (see Examples 2.6 and
2.7)

(3.1)
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j=1,\ldots\ell_{0}-1\\}\cup
K_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}^{\sigma},\hbox{ or}$ (3.2)
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}:1\leq
i<\ell_{0}\\}\cup\left(\bigcup_{i=\ell_{0}}^{j}K^{\sigma}_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}\right)$

where $1\leq k_{1}<k_{2}<\cdots<k_{n}<\cdots<\infty$,
$a_{\alpha_{0},\beta_{0}}$ and $a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}$ are the
only elements of $K$ that are in the diagonal, and in (3.2),
$\sigma=k_{j}-k_{\ell_{0}}$ where $k_{j}$ is such that no point
$a_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{p}}$ belongs to $K$ for $1\leq p<j$.
If $\ell_{0}=1$, then the term $\\{a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}:1\leq
i<\ell_{0}\\}$ is missing in (3.1) and (3.2).

###### Proof.

In this case, the diagram is the following, where the bullet represents the
element $a_{\alpha_{0}\beta_{0}}$, the circles indicate that no element of $K$
occupies that position, and the dots indicate that infinitely many elements of
$K$ reside in the diagonal.

$\bullet$ o o $\cdots$

o $\ddots$

o $\ddots$

$\vdots$ $\ddots$

We consider a point $a_{\alpha_{0}+m,\beta_{0}+n}$, denoted by $\blacksquare$
in Diagram 13, with $m\neq n$ and $m\neq k_{j},n\neq k_{j}$ for every $j\geq
1$.

Suppose first that $m>n$, more precisely, $k_{j+1}>m>k_{j}\geq
k_{\ell+1}>n>k_{\ell}$. We consider the two points
$a_{\alpha_{0}+k_{\ell},\beta_{0}+k_{\ell}}$ (an element of $K$ on the
diagonal), denoted by $\square$ in Diagram 13, and
$a_{\alpha_{0}+m,\beta_{0}+n}=\blacksquare$. These two points are vertices of
a right triangle with height $h=m-k_{\ell}$ greater than the base
$b=n-k_{\ell}$, so $h-b=m-n$. Then by Lemma 2.2 (see Diagram 3), the point
$x_{2}=a_{\alpha_{0}+k_{\ell}+\beta_{0}+n-\beta_{0}-k_{\ell},\beta_{o}+m}=a_{\alpha_{0}+n,\beta_{0}+n}$,
denoted by $\circledcirc$, would belong to $K$, which is a contradiction since
$k_{\ell}<n<k_{\ell+1}$. Therefore all elements below the diagonal which are
neither located on a row nor on a column containing a diagonal point of $K$,
do not belong to $K$.

######  Diagram 13.

$k_{\ell}$ $k_{\ell+1}$ $k_{j}$ $k_{j+1}$

$\bullet$ o o o o o o o o o o o $\cdots$

o $\ddots$

$k_{\ell}$ o $\square$

$n(m)$ o $\circledcirc$ $\blacktriangle$

$k_{\ell+1}$ o $\bullet$

o $\ddots$

o $\ddots$

$k_{j}$ o $\bullet$

$m(n)$ o $\blacksquare$ $\circ$

$k_{j+1}$ o $\bullet$

o $\ddots$

Suppose next that $m<n$, more precisely, $k_{\ell}<m<k_{\ell+1}\leq
k_{j}<n<k_{j+1}$ We again consider the two points
$a_{\alpha_{0}+k_{\ell},\beta_{0}+k_{\ell}}$, denoted by $\square$ in Diagram
13, and $a_{\alpha_{0}+m,\beta_{0}+n}$, denoted by $\blacktriangle$ in Diagram
13. These two points are vertices of a right triangle with height
$h=m-k_{\ell}$ smaller than the base $b=n-k_{\ell}$ and $b-h=n-m$. Then by
Lemma 2.2 (see Diagram 1), the point
$x_{1}=a_{\alpha_{0}+k_{l}+\beta_{0}+n-\beta_{0}-k_{\ell},\beta_{0}+n}=a_{\alpha_{0}+n,\beta_{0}+n}$,
denoted by $\circledcirc$, would belong to $K$, which is a contradiction since
$k_{j}<n<k_{j+1}$.

The same two part argument works in the cases $k_{j}<m<n<k_{j+1}$ and
$k_{j}<n<m<k_{j+1}$. Therefore all elements above or below the diagonal which
are not located on a row or on a column containing a diagonal point of $K$, do
not belong to $K$.

We now have Diagram 14. Thus the only off-diagonal points that can possibly
belong to $K$ are those that are located either on a row containing a diagonal
point of $K$, indicated by $\cdots$, or on a column containing a diagonal
point of $K$, indicated by vertical dots.

We next show that points which lie on a row or column, but not both, cannot
belong to $K$.

Consider first, for any row determined by $k_{j}$ and any $n$ with $n\neq
k_{m}$ for all $m$, the element $a_{\alpha_{0}+k_{j},\beta_{0}+n}$, denoted by
$\blacksquare$ in Diagram 15, and suppose it belonged to $K$. Then by Lemma
2.1(iii),

$a_{\alpha_{0}\beta_{0}}a^{*}_{\alpha_{0}+k_{j},\beta_{0}+n}a_{\alpha_{0}+k_{j},\beta_{0}+n}=a_{\alpha_{0}+n,\beta_{0}+n}\hbox{
(denoted by $\square$)}$

would belong to $K$, a contradiction since $n\neq k_{m}$ for every $m$.

Next, for any column determined by $k_{j}$ and any $n$ with $n\neq k_{m}$ for
all $m$, consider the element $a_{\alpha_{0}+n,\beta_{0}+k_{j}}$, denoted by
$\blacktriangle$ in Diagram 15, and suppose it belonged to $K$. Then by Lemma
2.1(i),

$a_{\alpha_{0}+n,\beta_{0}+k_{j}}a^{*}_{\alpha_{0}+n,\beta_{0}+k_{j}}a_{\alpha_{0}\beta_{0}}=a_{\alpha_{0}+n,\beta_{0}+n}\hbox{
(denoted by $\square$)}$

would belong to $K$, a contradiction since $n\neq k_{m}$ for every $m$.

We now have Diagram 16. Thus the only off-diagonal points that can possibly
belong to $K$ are those, denoted by $\blacksquare$, that are located
simultaneously on a row containing a diagonal point of $K$ and a column
containing a diagonal point of $K$.

If none of the off-diagonal elements $\blacksquare$ belong to $K$, then
obviously
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j\in\mathbb{N}\\}$.
We next consider the case that some of the elements $\blacksquare$ in Diagram
16 belong to $K$.

######  Diagram 14.

|  |  | $k_{\ell}$ |  | $k_{\ell+1}$ |  |  | $k_{j}$ |  | $k_{j+1}$ |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---
| $\bullet$ | o | o | o | o | o | o | o | o | o | o | o | $\cdots$
| o | $\ddots$ | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
$k_{\ell}$ | o | $\cdots$ | $\bullet$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\cdots$
| o | o | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
$k_{\ell+1}$ | o | $\cdots$ | $\vdots$ | $\cdots$ | $\bullet$ | $\cdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\cdots$
| o | o | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
| o |  |  |  |  |  | $\ddots$ |  |  |  |  |  |
$k_{j}$ | o | $\cdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\bullet$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\cdots$
| o | o | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
$k_{j+1}$ | o | $\cdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\bullet$ | $\cdots$ | $\cdots$ | $\cdots$
| o | o | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
| $\vdots$ |  |  |  |  |  |  |  |  |  | $\ddots$ |  |

A row determined by $k_{j}$ for which no element other than
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ belongs to $K$, that is,
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}\in K$, and
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{p}}\not\in K$ for all
$p\in\mathbb{N}-\\{k_{j}\\}$, will be called a null  row. More precisely, a
right-null (respectively left-null) row determined by $k_{j}$ is one that
satisfies $a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}\in K$, and
$a_{\alpha_{0}+k_{j},\beta_{0}+k_{p}}\not\in K$ for all $p>k_{j}$
(respectively $p<k_{j}$). The row determined by $\alpha_{0}$ is a null-row.

######  Diagram 15.

|  |  | $k_{\ell}$ | $n$ | $k_{\ell+1}$ |  |  | $k_{j}$ |  | $k_{j+1}$ |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---
| $\bullet$ | o | o | o | o | o | o | o | o | o | o | o | $\cdots$
| o | $\ddots$ | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
$k_{\ell}$ | o | $\cdots$ | $\bullet$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\cdots$
$n$ | o | o | $\vdots$ | $\square$ | $\vdots$ | o | o | $\blacktriangle$ | o | $\vdots$ | o | o | $\cdots$
$k_{\ell+1}$ | o | $\cdots$ | $\vdots$ | $\cdots$ | $\bullet$ | $\cdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\cdots$
| o | o | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
| o |  |  |  |  |  | $\ddots$ |  |  |  |  |  |
$k_{j}$ | o | $\cdots$ | $\vdots$ | $\blacksquare$ | $\vdots$ | $\cdots$ | $\cdots$ | $\bullet$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\cdots$
| o | o | $\vdots$ | $\circ$ | $\vdots$ | o | o | $\vdots$ | $\circ$ | $\vdots$ | o | o | $\cdots$
$k_{j+1}$ | o | $\cdots$ | $\vdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\cdots$ | $\vdots$ | $\cdots$ | $\bullet$ | $\cdots$ | $\cdots$ | $\cdots$
| o | o | $\vdots$ | o | $\vdots$ | o | o | $\vdots$ | o | $\vdots$ | o | o | $\cdots$
| $\vdots$ |  |  |  |  |  |  |  |  |  | $\ddots$ |  |

######  Diagram 16.

|  |  | $k_{\ell}$ |  | $k_{\ell+1}$ |  |  | $k_{j}$ |  |  | $k_{j+1}$ |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
| $\bullet$ | o | o | o | o | o | o | o | o | o | o | o | o | $\cdots$
| o | $\ddots$ | o | o | o | o | o | o | o | o | o | o | $\cdots$ |
$k_{\ell}$ | o | o | $\bullet$ | o | $\blacksquare$ | o | o | $\blacksquare$ | o | o | $\blacksquare$ | o | o | $\cdots$
| o | o | o | o | o | o | o | o | o | o | o | o | $\cdots$ |
$k_{\ell+1}$ | o | o | $\blacksquare$ | o | $\bullet$ | o | o | $\blacksquare$ | o | o | $\blacksquare$ | o | o | $\cdots$
| o | o | o | o | o | o | o | o | o | o | o | o | o | $\cdots$
| o |  |  |  |  |  | $\ddots$ |  |  |  |  |  |  |
$k_{j}$ | o | o | $\blacksquare$ | o | $\blacksquare$ | o | o | $\bullet$ | o | o | $\blacksquare$ | o | o | $\cdots$
| o | o | o | o | o | o | o | o | o | o | o | o | o | $\cdots$
| o | o | o | o | o | o | o | o | o | o | o | o | o | $\cdots$
$k_{j+1}$ | o | o | $\blacksquare$ | o | $\blacksquare$ | o | o | $\blacksquare$ | o | o | $\bullet$ | o | o | $\cdots$
| o | o | o | o | o | o | o | o | o | o | o | o | o | $\cdots$
| $\vdots$ |  |  |  |  |  |  |  |  |  | $\ddots$ |  |  |

Similarly, a row determined by $k_{j}$ which contains an element of $K$ other
than $a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$, that is, there exists $\ell>j$
(respectively $\ell<j$), such that $a_{\alpha_{0}+k_{j},\beta_{0}+\ell}\in K$,
will be called a right-ample (respectively left-ample) row. A row which is
either left-ample or right-ample (or both), will be called simply ample. By
Diagram 2, a left-ample row is also right-ample, but not conversely (see the
sentence following Lemma 2.1). For the same reason, a right-null row is also
left-null. As noted above, if all rows of $K$ are null, then
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j\in\mathbb{N}\\}$.
Thus we have the following lemma.

######  Lemma 3.5.

If all the rows of $K$ are right-null, then
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j\in\mathbb{N}\\}.$

######  Lemma 3.6.

All right-null rows lie above all ample rows. Hence, if there is at least one
ample row, then there are only finitely many right-null rows.

###### Proof.

Suppose $k_{j}$ determines a right-null row, and $k_{\ell}$ determines an
ample row, which we may assume to be right-ample, say containing the element
$a_{\alpha_{0}+k_{\ell},\beta_{0}+k_{m}}$, and suppose by way of contradiction
that $\ell<j$. Since $\ell<j<m$, the diagram is the following, where
$\blacksquare$ denotes the element $a_{\alpha_{0}+k_{j},\beta_{0}+k_{m}}$.

|  | $k_{\ell}$ |  |  | $k_{j}$ |  | $k_{m}$ |  |
---|---|---|---|---|---|---|---|---|---
| $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$
$k_{\ell}$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\blacksquare$ | $\circ$ | $\cdots$
| $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$
$k_{j}$ | $\circ$ | $\cdots$ | $\circ$ | $\cdots$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$ | $\cdots$
$k_{m}$ | $\cdots$ | $\cdots$ | $\circ$ | $\cdots$ | $\cdots$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$

The two points $\blacksquare$ and $a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ are
vertices of a triangle with base $b=k_{m}-k_{j}$ and height
$h=k_{j}-k_{\ell}$. Then by Diagrams 6, 8, or 10, depending on the relative
sizes of $b$ and $h$, the point
$x_{3}=a_{\alpha_{0}+k_{j},\beta_{0}+k_{m}+k_{j}-k_{\ell}}$ would belong to
$K$, which is a contradiction since $k_{m}+k_{j}-k_{\ell}>k_{j}$. ∎

Let $\ell_{0}\in\mathbb{N}$ be such that the first ample row is determined by
$k_{\ell_{0}}$, and assume without loss of generality, that this row is right-
ample. Assume also, temporarily, that $\ell_{0}>1$. The rows lying above the
row determined by $\alpha_{0}+k_{\ell_{0}}$ do not contain any elements of $K$
above the diagonal. It follows from Diagram 4 that the columns lying to the
left of the column determined by $\beta_{0}+k_{\ell_{0}}$ do not contain any
elements of $K$ below the diagonal. By Diagram 2, $K$ contains infinitely many
elements to the right of $a_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}$,
and then by Diagrams 2 and 4, $K$ contains infinitely many elements below
$a_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}$. Since
$\overline{\gamma}=\infty$, the subsemiheap $K\cap
K_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}$ falls into subcase 3.3.3
below (see Propositions 3.15 and 3.16, and Diagram 17), and therefore

(3.3)
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}:1\leq
i<\ell_{0}\\}\cup
K_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}^{k_{j}-k_{\ell_{0}}},$

or

(3.4)
$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}:1\leq
i<\ell_{0}\\}\cup\left(\bigcup_{i=\ell_{0}}^{j}K^{k_{j}-k_{\ell_{0}}}_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}\right)$

where $k_{j}$ is such that no point
$a_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{p}}$ belongs to $K$ for $1\leq p<j$.
If $\ell_{0}=1$, then the term $\\{a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}:1\leq
i<\ell_{0}\\}$ is missing in (3.3) and (3.4). (In Diagram 17,
$\sigma=k_{\ell_{0}+1}+(k_{j}-k_{\ell_{0}})$). This completes the proof of
Proposition 3.4. ∎

This completes the analysis of case 1.1.3 and hence of case 1.1.

######  Lemma 3.7.

Cases 1.2 and 1.3 do not occur.

###### Proof.

In each of the six subcases of cases 1.2 and 1.3, the diagram is the
following:

| $\beta_{0}$ |  |  |  |  |  |
---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\ddots$ |  |  |  |  |
| $\circ$ |  | $\ddots$ |  |  |  |
| $\circ$ |  |  | $\ddots$ |  |  |
$r$ | $\bullet$ |  |  |  | $\ddots$ |  |
| $\vdots$ |  |  |  |  | $\ddots$ |
| $\vdots$ |  |  |  |  |  | $\ddots$

Then applying Diagram 4 to the points $a_{\alpha_{0}\beta_{0}}$ and
$a_{\alpha_{0}+r,\beta_{0}}$ shows that $x_{1}=a_{\alpha_{0},\beta_{0}+r}\in
K$, a contradiction. ∎

######  Diagram 17.

| $\beta_{0}$ | $\cdots$ | $k_{1}$ | $\cdots$ | $k_{2}$ | $\cdots$ | $k_{\ell_{0}-1}$ | $\cdots$ | $k_{\ell_{0}}$ | $\cdots$ | $k_{\ell_{0}+1}$ | $\cdots$ | $k_{j}$ | $\cdots$ | $\sigma$ | $\cdots$ | $\cdots$ |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$k_{1}$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$k_{2}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$\vdots$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$k_{\ell_{0}-1}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |  |
$\vdots$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$k_{\ell_{0}}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\cdots$ |  |
$\vdots$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$k_{\ell_{0}+1}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$ |
$\vdots$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$k_{j}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$\vdots$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$\sigma$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$ |
$\vdots$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$ |
$\vdots$ |  |  | $\vdots$ |  | $\vdots$ |  | $\vdots$ |  | $\vdots$ |  | $\vdots$ |  | $\vdots$ |  | $\vdots$ |  |  |

######  Lemma 3.8.

Case 2.1 does not occur.

###### Proof.

In each of the three subcases of case 2.1, the diagram is the following:

| $\beta_{0}$ |  |  |  | $p$ |  |
---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
| $\circ$ | $\ddots$ |  |  |  |  |
| $\circ$ |  | $\ddots$ |  |  |  |
| $\circ$ |  |  | $\ddots$ |  |  |
| $\circ$ |  |  |  | $\ddots$ |  |
| $\vdots$ |  |  |  |  | $\ddots$ |
| $\vdots$ |  |  |  |  |  | $\ddots$

Then applying Diagram 2 to the points $a_{\alpha_{0}\beta_{0}}$ and
$a_{\alpha_{0},\beta_{0}+p}$ shows that $x_{1}=a_{\alpha_{0}+p,\beta_{0}}\in
K$, a contradiction. ∎

######  Lemma 3.9.

Case 2.2 does not occur.

###### Proof.

Note first that by Diagrams 2 and 4, only one element of $K$ can reside on the
row determined by $\alpha_{0}$ or on the column determined by $\beta_{0}$.
Thus, the diagram for this case (if $r<p$) is the following.

| $\beta_{0}$ |  | $r$ |  | $p$ |  |
---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
| $\circ$ | $\ddots$ |  |  |  |  |
$r$ | $\bullet$ |  | $\ddots$ |  |  |  |
| $\circ$ |  |  | $\ddots$ |  |  |
$p$ | $\circ$ |  |  |  | $\ddots$ |  |
| $\vdots$ |  |  |  |  | $\ddots$ |
| $\vdots$ |  |  |  |  |  | $\ddots$

Then by Diagrams 6 and 8 (depending on whether $r<p$ or $p\geq r$, $K$ would
contain elements $a_{\alpha_{0},\beta_{0}+\ell}$ with $\ell>p$, a
contradiction. ∎

######  Lemma 3.10.

Case 2.3 does not occur, hence case 2 does not occur.

###### Proof.

In Case 2.3.1
($\beta_{0}<\overline{\beta}<\infty,\overline{\alpha}=\infty,\overline{\gamma}=0$),
note that $a_{\alpha_{0},\beta_{0}+p}$ is the only point of $K$ on the row
determined by $\alpha_{0}$. From the following diagram we see that if $p<r$,
we get a contradiction using Diagram 8, and if $r<p$ we get a contradiction
using Diagram 6, whereas if $p=r$, we get a contradiction using Diagram 10.

| $\beta_{0}$ |  |  |  | $p$ |  |  |  |
---|---|---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\cdots$ |
| $\circ$ | $\circ$ |  |  |  |  |  |  |
| $\circ$ |  | $\circ$ |  |  |  |  |  |
| $\circ$ |  |  | $\circ$ |  |  |  |  |
| $\circ$ |  |  |  | $\circ$ |  |  |  |
| $\circ$ |  |  |  |  | $\circ$ |  |  |
$r$ | $\bullet$ |  |  |  |  |  | $\circ$ |  |
| $\vdots$ |  |  |  |  |  |  | $\ddots$ |

The same proof applies to cases 2.3.2
($\beta_{0}<\overline{\beta}<\infty,\overline{\alpha}=\infty,0<\overline{\gamma}<\infty$)
and 2.3.3
($\beta_{0}<\overline{\beta}<\infty,\overline{\alpha}=\infty,\overline{\gamma}=\infty$).∎

######  Lemma 3.11.

Cases 3.1 and 3.2 do not occur.

###### Proof.

Cases 3.1.1, 3.1.2, and 3.1.3 do not occur by Diagram 2. Cases 3.2.1, 3.2.2,
and 3.2.3 do not occur by Diagram 6. ∎

######  Proposition 3.12.

Cases 3.3.1 and 3.3.2 do not occur.

###### Proof.

By Diagram 2 or 4, we may assume that $a_{\alpha_{0},\beta_{0}}\not\in K$. The
subsets $K\cap K_{\alpha_{0},\beta_{0}+k_{1}}$ and $K\cap
K_{\alpha_{0}+\ell_{1},\beta_{0}}$ are subsemiheaps of $K$ which fall into
case 3.3.3, which is described below in Proposition 3.18. However, as shown in
the proof of Lemma 3.17 below, the four possible situations each lead to
$a_{\alpha_{0},\beta_{0}}\in K$. ∎

Case 3.3.3
($\overline{\beta}=\infty,\overline{\alpha}=\infty,\overline{\gamma}=\infty$)

> Let $a_{\alpha_{0},\beta_{0}+p}\in K$ with $p\geq 1$ and
> $a_{\alpha_{0},\beta_{0}+p^{\prime}}\not\in K$ for $1\leq p^{\prime}<p$.
> Similarly, let $a_{\alpha_{0}+r,\beta_{0}}\in K$ with $r\geq 1$ and
> $a_{\alpha_{0}+r^{\prime},\beta_{0}}\not\in K$ for $1\leq r^{\prime}<r$ and
> let $a_{\alpha_{0}+q,\beta_{0}+q}\in K$ with $q\geq 1$ and
> $a_{\alpha_{0}+q^{\prime},\beta_{0}+q^{\prime}}\not\in K$ for $1\leq
> q^{\prime}<q$
>
> In the diagram below, the bullets represent the three points of $K$ which
> were just defined, the symbol $\circledcirc$ means that the element
> $a_{\alpha_{0}\beta_{0}}$ may or may not belong to $K$, and the circles
> indicate that no element of $K$ occupies that position. The diagram
> represents just one of 13 possible cases (namely case (5) below), and is for
> illustration purposes only.
>
>
>  | $\beta_{0}$ |  |  | $q$ |  |  |  | $r$ | $p$ |  |
> ---|---|---|---|---|---|---|---|---|---|---|---
> $\alpha_{0}$ | $\circledcirc$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\cdots$ |
> | $\circ$ | $\circ$ |  |  |  |  |  |  |  |  |
> | $\circ$ |  | $\circ$ |  |  |  |  |  |  |  |
> $q$ | $\circ$ |  |  | $\bullet$ |  |  |  |  |  |  |
> | $\circ$ |  |  |  | $\ddots$ |  |  |  |  |  |
> | $\circ$ |  |  |  |  | $\ddots$ |  |  |  |  |
> | $\circ$ |  |  |  |  |  | $\ddots$ |  |  |  |
> $r$ | $\bullet$ |  |  |  |  |  |  | $\ddots$ |  |  |
> $p$ | $\vdots$ |  |  |  |  |  |  |  | $\ddots$ |  |
> | $\vdots$ |  |  |  |  |  |  |  |  | $\ddots$ |
>
> Of course, we must consider the various relations between the three elements
> $p,q,r\in\mathbb{N}$, some of which can be equal, of which there are six,
> namely
>
>   * •
>
> $r\leq p\leq q$
>
>   * •
>
> $p\leq r\leq q$
>
>   * •
>
> $p\leq q\leq r$
>
>   * •
>
> $r\leq q\leq p$
>
>   * •
>
> $q\leq r\leq p$
>
>   * •
>
> $q\leq p\leq r$
>
>

>
> But for our purposes, it is necessary to distinguish 13 more refined cases,
> namely
>
>   1. (1)
>
> $r<p<q$
>
>   2. (2)
>
> $p<r<q$
>
>   3. (3)
>
> $p<q<r$
>
>   4. (4)
>
> $r<q<p$
>
>   5. (5)
>
> $q<r<p$
>
>   6. (6)
>
> $q<p<r$
>
>   7. (7)
>
> $r=p<q$
>
>   8. (8)
>
> $p=q<r$
>
>   9. (9)
>
> $r=q<p$
>
>   10. (10)
>
> $r<p=q$
>
>   11. (11)
>
> $p<r=q$
>
>   12. (12)
>
> $q<r=p$
>
>   13. (13)
>
> $r=p=q$
>
>

######  Lemma 3.13.

Cases (3) to (11) do not occur. If $a_{\alpha_{0},\beta_{0}}\in K$, then cases
(1) and (2) do not occur. If $a_{\alpha_{0}\beta_{0}}\not\in K$, then case
(12) does not occur. In case (13), $a_{\alpha_{0}\beta_{0}}\in K$

###### Proof.

By Lemma 2.1, we have

(3.5)
$a_{\alpha_{0},\beta_{0}+p}a^{*}_{\alpha_{0}+q,\beta_{0}+q}a_{\alpha_{0}+r,\beta_{0}}=\left\\{\begin{array}[]{ll}\hbox{(i)
}a_{\alpha_{0},\beta_{0}+p-r}&\hbox{ if }r\leq p\hbox{ and }q\leq p\\\
\hbox{(ii) }a_{\alpha_{0}+r-p,\beta_{0}}&\hbox{ if }r\geq p\hbox{ and }q\leq
p\\\ \hbox{(iii) }a_{\alpha_{0}+r-p,\beta_{0}}&\hbox{ if }r\geq q\hbox{ and
}q\geq p\\\ \hbox{(iv) }a_{\alpha_{0}+q-p,\beta_{0}+q-r}&\hbox{ if }r\leq
q\hbox{ and }q\geq p.\end{array}\right.$

In case (1) with $a_{\alpha_{0}\beta_{0}}\in K$, we obtain a contradiction by
Diagram 4.

In case (2) with $a_{\alpha_{0}\beta_{0}}\in K$, we obtain a contradiction by
Diagram 2.

In case (3), we obtain a contradiction by (3.5(iii)).

In case (4), we obtain a contradiction by (3.5(i)).

In case (5), we obtain a contradiction by (3.5(i)).

In case (6), we obtain a contradiction by (3.5(ii)).

In case (7), we obtain a contradiction by (3.5(iv)).

In case (8), we obtain a contradiction by (3.5(ii)).

In case (9), we obtain a contradiction by (3.5(i)).

In case (10), we obtain a contradiction by (3.5(i)).

In case (11), we obtain a contradiction by (3.5(iii)).

In case (12) with $a_{\alpha_{0}\beta_{0}}\not\in K$, we obtain a
contradiction by (3.5(i)) or (ii).

In case (13), $a_{\alpha_{0}\beta_{0}}\in K$ by (3.5(iii)). ∎

It remains to consider cases (1) and (2), with $a_{\alpha_{0}\beta_{0}}\not\in
K$, and the cases (12) and (13), with $a_{\alpha_{0}\beta_{0}}\in K$. The
latter two are resolved in Propositions 3.15 and 3.16 and the former two in
Lemma 3.17.

We start with some properties in case (12). The basic diagram for case (12) is
the following.

|  | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---
|  | $\beta_{0}$ |  |  | $q$ |  |  |  | $p$ |  |  |
0 | $\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\cdots$ |  |
1 |  | $\circ$ | $\circ$ |  |  |  |  |  |  |  |  |
2 |  | $\circ$ |  | $\circ$ |  |  |  |  |  |  |  |
3 | $q$ | $\circ$ |  |  | $\bullet$ |  |  |  |  |  |  |
4 |  | $\circ$ |  |  |  | $\ddots$ |  |  |  |  |  |
5 |  | $\circ$ |  |  |  |  | $\ddots$ |  |  |  |  |
6 |  | $\circ$ |  |  |  |  |  | $\ddots$ |  |  |  |
7 | $r=p$ | $\bullet$ |  |  |  |  |  |  | $\ddots$ |  |  |
|  | $\vdots$ |  |  |  |  |  |  |  |  |  |

######  Lemma 3.14.

In case (12), with (necessarily) $a_{\alpha_{0}\beta_{0}}\in K$,

(a):

The rows $1,2,\ldots,q-1$ contain no elements of $K$ above the diagonal

The columns $1,2,\ldots,q-1$ contain no elements of $K$ below the diagonal

(b):

The points $a_{\alpha_{0}+q,\beta_{0}+q+i}$, for $1\leq i\leq p-q$ do not
belong to $K$.

The points $a_{\alpha_{0}+q+i,\beta_{0}+q}$, for $1\leq i\leq p-q$ do not
belong to $K$.

(c):

The points $a_{\alpha_{0}+q,\beta_{0}+p}$, and $a_{\alpha_{0}+p,\beta_{0}+q}$
do not belong to $K$.

(d):

The points $a_{\alpha_{0}+q,\beta_{0}+j}$, for $mp<j<mp+q$, with
$m\in\mathbb{N}$ do not belong to $K$.

The points $a_{\alpha_{0}+i,\beta_{0}+q}$, for $mp<i<mp+q$, with
$m\in\mathbb{N}$ do not belong to $K$.

(e):

$a_{\alpha_{0}+q,\beta_{0}+p+q},a_{\alpha_{0}+p+q,\beta_{0}+q}\in K$.

(f):

The points $a_{\alpha_{0}+q,\beta_{0}+j}$, for $mp+q<j\leq(m+1)p$, with
$m\in\mathbb{N}$ do not belong to $K$.

The points $a_{\alpha_{0}+i,\beta_{0}+q}$, for $mp+q<i\leq(m+1)p$, with
$m\in\mathbb{N}$ do not belong to $K$.

(g):

The points $a_{\alpha_{0},\beta_{0}+j}$, for $mp<j<(m+1)p$, with
$m\in\mathbb{N}$ do not belong to $K$.

The points $a_{\alpha_{0}+i,\beta_{0}}$, for $mp<i<(m+1)p$, with
$m\in\mathbb{N}$ do not belong to $K$.

###### Proof.

In what follows, we shall elaborate on the above diagram. (In Diagram 18, the
symbols
$\blacksquare,\blacktriangle,\blacktriangledown,\blacktriangleleft,\blacktriangleright,\blacklozenge,\spadesuit,\clubsuit$,
and their blank versions, represent points which, a priori, do not belong to
$K$. They should be temporarily ignored. Also, Diagram 19 indicates the
locations of (a)-(g).)

(a) Consider the two points $a_{\alpha_{0}\beta_{0}}$ and
$a_{\alpha_{0}+i,\beta_{0}+j}$, the latter indicated by $\blacksquare$ in
Diagram 18, with $1\leq i<q$, $2\leq j<\infty$ and $i<j$, and suppose that
$a_{\alpha_{0}+i,\beta_{0}+j}$ belongs to $K$. Then by Diagram 1,
$x_{3}=a_{\alpha+i,\beta_{0}+i}$, indicated by $\square$, belongs to $K$, a
contradiction. Therefore the rows $1,2,3,\ldots q-1$ contain no elements of
$K$ above the diagonal.

######  Diagram 18.

|  | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  | $\beta_{0}$ |  |  |  | $q$ |  |  | $p$ |  |  |  | $p+q$ |  |  | 2p |  |
0 | $\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\triangleleft$ | $\triangledown$ | $\circ$ | $\bullet$ |  |  | $\clubsuit$ |  |  |  |  |  | $\cdots$
1 |  | $\circ$ | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
2 |  | $\circ$ |  | $\square$ |  |  |  |  | $\lozenge$ |  |  | $\blacksquare$ |  |  |  |  |  | $\cdots$
3 |  | $\circ$ |  |  | $\triangle$ |  |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
4 | $q$ | $\circ$ |  | $\boxminus$ |  | $\bullet$ |  | $\blacktriangledown$ | $\blacktriangleleft$ |  | $\blacklozenge$ |  | $\bullet$ |  |  |  |  | $\cdots$
5 |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
6 |  | $\circ$ |  |  |  | $\spadesuit$ |  | $\ddots$ |  |  |  |  |  |  |  |  |  | $\cdots$
7 | $r=p$ | $\bullet$ |  |  |  | $\blacktriangleright$ |  |  | $\bullet$ |  |  |  |  |  |  |  |  | $\cdots$
8 |  |  |  |  | $\blacktriangle$ |  |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
9 |  |  |  |  |  | $\boxplus$ |  |  |  |  | $\ddots$ |  |  |  |  |  |  | $\cdots$
10 |  | $\bigstar$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
11 | $p+q$ |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  |  |  |  | $\cdots$
12 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
13 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
14 | $2p$ |  |  |  |  |  |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  | $\cdots$

Consider the two points $a_{\alpha_{0}\beta_{0}}$ and
$a_{\alpha_{0}+i,\beta_{0}+j}$, the latter indicated by $\blacktriangle$ in
Diagram 18, with $1\leq j<q$, $2\leq i<\infty$ and $i>j$, and suppose that
$a_{\alpha_{0}+i,\beta_{0}+j}$ belongs to $K$. Then by Diagram 3,
$x_{2}=a_{\alpha+j,\beta_{0}+j}$, indicated by $\triangle$, belongs to $K$, a
contradiction. Therefore columns $1,2,3,\ldots q-1$ contain no elements of $K$
below the diagonal.

(b) Assuming that $a_{\alpha_{0}+q,\beta_{0}+q+j}$, indicated by
$\blacktriangledown$ in Diagram 18, with $1\leq j<p-q$, belongs to $K$, we
have that

$K\supset K_{\alpha_{0}+q,\beta_{0}+q}^{j}=\\{a_{\alpha_{0}+q+\ell
j,\beta_{0}+q+mj}:\ell,m\geq 0\\}.$

Then by Lemma 2.1(i),

$a_{\alpha_{0},\beta_{0}+p}a^{*}_{\alpha_{0}+q+\ell
j,\beta_{0}+q+mj}a_{\alpha_{0}\beta_{0}}=a_{\alpha_{0},\beta_{0}+(\ell-m)j+p}\in
K,$

provided that $0\leq(\ell-m)j+p$ and $q+mj\leq p$. Then with $\ell=0$ and
$m=1$, we have that $a_{\alpha_{0},\beta_{0}-j+p}$, indicated by
$\triangledown$, belongs to $K$, which is a contradiction.

For the second statement of (b), the proof is the same, namely, assuming that
$a_{\alpha_{0}+q+i,\beta_{0}+q}$, indicated by $\spadesuit$, with $1\leq
i<p-q$, belongs to $K$, we have that

$K\supset K_{\alpha_{0}+q,\beta_{0}+q}^{i}=\\{a_{\alpha_{0}+q+\ell
i,\beta_{0}+q+mi}:\ell,m\geq 0\\}.$

Then by Lemma 2.1(i),

$a_{\alpha_{0},\beta_{0}+p}a^{*}_{\alpha_{0}+q+\ell
i,\beta_{0}+q+mi}a_{\alpha_{0}\beta_{0}}=a_{\alpha_{0},\beta_{0}+(\ell-m)i+p}\in
K,$

provided that $0\leq(\ell-m)i+p$ and $q+mi\leq p$. Then with $\ell=0$ and
$m=1$, we have that $a_{\alpha_{0},\beta_{0}-i+p}$, indicated by
$\triangledown$, belongs to $K$, which is a contradiction.

(c) Assuming that $a_{\alpha_{0}+q,\beta_{0}+p}$, indicated by
$\blacktriangleleft$, belongs to $K$, then by Lemma 2.1(iii),

$a_{\alpha_{0},\beta_{0}+p}a^{*}_{\alpha_{0}+q,\beta_{0}+p}a_{\alpha_{0}+q,\beta_{0}+q}=a_{\alpha_{0},\beta_{0}+q},$

indicated by $\triangleleft$, belongs to $K$, a contradiction. Assuming that
$a_{\alpha_{0}+p,\beta_{0}+q}$, indicated by $\blacktriangleright$, belongs to
$K$, then applying Diagram 4 to the two points $a_{\alpha_{0}+q,\beta_{0}+q}$
and $\blacktriangleright$ we have $x_{1}=a_{\alpha_{0}+q,\beta_{0}+p}\in K$, a
contradiction to the previous paragraph.

(d) Suppose $mp<j<mp+q$ and assume that $a_{\alpha_{0}+q,\beta_{0}+j}$,
indicated by $\blacklozenge$ (with $m=1$), belongs to $K$. Then by Lemma
2.1(iii),

$a_{\alpha_{0},\beta_{0}+mp}a^{*}_{\alpha_{0}+q,\beta_{0}+j}a_{\alpha_{0}+q,\beta_{0}+q}=a_{\alpha_{0}+j-mp,\beta_{0}+q},$

indicated by $\lozenge$, belongs to $K$, a contradiction to (i), since
$j-mp<q$.

Suppose $mp<i<mp+q$ and assume that $a_{\alpha_{0}+i,\beta_{0}+q}$, indicated
by $\boxplus$ (with $m=1$), belongs to $K$. Then by Lemma 2.1(iv),

$a_{\alpha_{0}+q,\beta_{0}+q}a^{*}_{\alpha_{0}+i,\beta_{0}+q}a_{\alpha_{0}+mp,\beta_{0}}=a_{\alpha_{0}+q,\beta_{0}+i-mp},$

indicated by $\boxminus$, belongs to $K$, a contradiction to (i′), since
$i-mp<q$.

(e) By Diagram 6 or 8, applied to the vertices $a_{\alpha_{0}+q,\beta_{0}+q}$
and $a_{\alpha_{0}+p,\beta_{0}}$, $x_{1}=a_{\alpha_{0}+q,\beta_{0}+p+q}\in K$,
and then by Diagram 7, $a_{\alpha_{0}+p+q,\beta_{0}+q}\in K$.

In the proofs of (f) and (g), we assume with no loss of generality, that
$\alpha_{0}=\beta_{0}=0$.

(f) Suppose that $(q,j)\in K$ with $mp+q<j\leq(m+1)p$. By Lemma 2.1(iii),
$(0,mp)(q,j)^{*}(q,q)=(j-mp,q)$. This is a contradiction to (b) since $p\geq
j-mp>q$.

Suppose that $(i,q)\in K$ with $mp+q<i\leq(m+1)p$. By Lemma 2.1(iv),
$(q,q)(i,q)^{*}(mp,0)=(q,i-mp)$. This is a contradiction to (b) since $p\geq
i-mp>q$.

(g) Supposing that $a_{\alpha_{0},\beta_{0}+j}$, with $p<j<\infty$ and
$j\not\in\\{2p,3p,\ldots\\}$, denoted by $\clubsuit$ in Diagram 18, belongs to
$K$, we apply Diagram 2 to $a_{\alpha_{0},\beta_{0}+mp}$ and
$a_{\alpha_{0},\beta_{0}+j}$, where $mp<j<(m+1)p$ to get
$x_{2}=a_{\alpha_{0}+j-mp,\beta_{0}+j}\in K$, a contradiction since $j-mp<p$.
Hence no element of $K$ occupies any position in the row determined by
$\alpha_{0}$ except for the points $a_{\alpha_{0},\beta_{0}+mp}$ for
$m\in\mathbb{N}_{0}$.

Supposing that $a_{\alpha_{0}+i,\beta_{0}}$, with $p<i<\infty$ and
$i\not\in\\{2p,3p,\ldots\\}$, denoted by $\bigstar$, belongs to $K$, we apply
Diagram 4 to $a_{\alpha_{0}+mp,\beta_{0}}$ and $a_{\alpha_{0}+i,\beta_{0}}$,
where $mp<i<(m+1)p$ to get a contradiction. Hence no element of $K$ occupies
any position in the column determined by $\beta_{0}$ except for the points
$a_{\alpha_{0}+\ell p,\beta_{0}}$ for $\ell\in\mathbb{N}_{0}$.∎

We now have Diagram 19 for case (12) with $a_{\alpha_{0},\beta_{0}}\in K$, and
it is clear that $K\cap K_{q,q}$ is also in subcase (12), so it follows that
$K=\bigcup_{i=0}^{\infty}K_{q_{i},q_{i}}^{p}$, where
$a_{\alpha_{0}+q_{i},\beta_{0}+q_{i}}$ are the points of $K$ lying on the
diagonal with

$q=q_{0}<q_{1}<q_{2}<\cdots<q_{n}<q_{n+1}<\cdots.$

######  Proposition 3.15.

In case (12), with (necessarily) $a_{\alpha_{0},\beta_{0}}\in K$, let
$a_{\alpha_{0}+q_{i},\beta_{0}+q_{i}}$, $0\leq i<\infty$,be the points of $K$
lying on the diagonal, such that

$q=q_{0}<q_{1}<q_{2}<\cdots<q_{n}<p\quad\hbox{ and }\quad
p<q_{n+1}<q_{n+2}<\cdots.$

Then

$K=\bigcup_{i=0}^{n}K_{q_{i},q_{i}}^{p}.$

###### Proof.

We know that $K=\bigcup_{i=0}^{\infty}K_{q_{i},q_{i}}^{p}$. We need to show
that
$\bigcup_{i=n+1}^{\infty}K_{q_{i},q_{i}}^{p}\subset\bigcup_{i=0}^{n}K_{q_{i},q_{i}}^{p}$.
For this it suffices to show that each $q_{n+j}$ with $j\geq 1$ is congruent
to some element of $\\{q_{0},q_{1},\ldots q_{n}\\}$, modulo $p$.

Let $(q_{k}+\ell p,q_{k}+mp)\in K_{q_{k},q_{k}}^{p}$ for some $k\geq n+1$ with
fixed $\ell,m$, and let $(\ell^{\prime}p,m^{\prime}p)\in K_{0,0}^{p}$ with
variable $\ell^{\prime},m^{\prime}$. By Lemma 2.1(i),

$(q_{k}+\ell
p,q_{k}+mp)(\alpha_{0},\beta_{0})^{*}(\ell^{\prime}p,m^{\prime}p)=(q_{k}+\ell
p,q_{k}+(m^{\prime}+m-\ell^{\prime})p)\in K$

as long as $q_{k}+mp\geq 0$ and $\ell^{\prime}p\leq q_{k}+mp$, We now choose
$\ell^{\prime}$ such that $q_{k}=(\ell^{\prime}-m)p+d$, where
$\ell^{\prime}-m\geq 1$ and $0\leq d<p$. To check that $\ell^{\prime}p\leq
q_{k}+mp$, we note that $(\ell^{\prime}-m)p=q_{k}-d\leq q_{k}$. We now have

$(q_{k}+\ell
p,q_{k}+(m^{\prime}+m-\ell^{\prime})p)=(d+(\ell+\ell^{\prime}-m)p,d).$

Thus $(d+tp,d)\in K=\bigcup_{i=0}^{\infty}K_{q_{i},q_{i}}^{p}$ for some $t\geq
0$, so that $(d+tp,d)=(q_{i}+rp,q_{i}+sp)$ for some $i,r,s\geq 0$. Hence
$d+tp=q_{i}+rp$ and $d=q_{i}+sp$, so by subtraction $tp=(r-s)p$ and
$d+(r-s)p=q_{i}+rp$ so that $d=q_{i}+sp$. Since $d<p$, $s=0$ and $d=q_{i}$
with $i\leq n$ ∎

######  Diagram 19.

|  | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  | $\beta_{0}$ |  |  |  | $q$ |  |  | $p$ |  |  |  | $p+q$ |  |  | 2p |  |
0 | $\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $(g)$ | $\circ$ | $\circ$ | $(g)$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
1 |  | $\circ$ | $\circ$ | $(a)$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
2 |  | $\circ$ | $(a)$ | $\circ$ | $(a)$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
3 |  | $\circ$ | $\circ$ | $(a)$ | $\circ$ | $(a)$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
4 | $q$ | $\circ$ | $\circ$ | $\circ$ | $(a)$ | $\bullet$ | $(b)$ | $\circ$ | $(c)$ | $\circ$ | $(d)$ | $\circ$ | $\bullet(e)$ | $\circ$ | $(f)$ | $\circ$ | $\circ$ | $\cdots$
5 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $(b)$ |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
6 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  | $\ddots$ |  |  |  |  |  |  |  |  |  | $\cdots$
7 | $r=p$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $(c)$ |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  | $\cdots$
8 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
9 |  | $(g)$ | $\circ$ | $\circ$ | $\circ$ | $(d)$ |  |  |  |  | $\ddots$ |  |  |  |  |  |  | $\cdots$
10 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
11 | $p+q$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\bullet(e)$ |  |  |  |  |  |  | $\bullet$ |  |  |  |  | $\cdots$
12 |  | $(g)$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  |  |  |  | $\ddots$ |  |  |  | $\cdots$
13 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $(f)$ |  |  |  |  |  |  |  |  |  |  |  | $\cdots$
14 | $2p$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  | $\bullet$ |  |  |  |  |  |  | $\bullet$ |  | $\cdots$
15 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  |  |  |  |  |  |  | $\ddots$ | $\cdots$
16 |  | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |  | $\vdots$ |  |  |  | $\vdots$ |  |  | $\vdots$ |  | $\cdots$

######  Proposition 3.16.

In case (13), $K=K_{\alpha_{0},\beta_{0}}^{p}$

###### Proof.

The diagram for case (13) is the following. (Temporarily ignore the symbols
$\blacksquare,\blacktriangle,\blacktriangledown,\blacktriangleleft$)

######  Diagram 20.

| $\beta_{0}$ |  |  |  | $p$ |  |  |  |
---|---|---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\cdots$ | $\blacktriangledown$ | $\cdots$ |
| $\circ$ | $\circ$ |  |  |  |  |  |  |
| $\circ$ |  | $\circ$ |  |  | $\blacksquare$ |  |  |
| $\circ$ |  |  | $\circ$ |  |  |  |  |
$r=p=q$ | $\bullet$ |  |  |  | $\bullet$ |  |  |  |
| $\vdots$ |  |  |  |  | $\ddots$ |  |  |
| $\blacktriangleleft$ |  |  |  |  |  | $\ddots$ |  |
| $\vdots$ |  | $\blacktriangle$ |  |  |  |  | $\ddots$ |

In the first place, we notice that by Diagrams 2 and 4,

$K\supset K_{\alpha_{0},\beta_{0}}^{p}=\\{a_{\alpha_{0}+\ell
p,\beta_{0}+mp}:\ell,m\in\mathbb{N}_{0}\\}.$

The next four paragraphs refer to Diagram 20.

Supposing that $a_{\alpha_{0}+i,\beta_{0}+j}$ for $1\leq i<p$ and $2\leq
j<\infty$, denoted by $\blacksquare$, belongs to $K$, we apply Diagram 1 to
$a_{\alpha_{0}\beta_{0}}$ and $a_{\alpha_{0}+i,\beta_{0}+j}$ to get
$x_{3}=a_{\alpha_{0}+i,\beta_{0}+i}\in K$, a contradiction. Hence no element
of $K$ occupies any position above the diagonal in the rows determined by
$\alpha_{0}+i$, for $1\leq i<p$.

Supposing that $a_{\alpha_{0}+i,\beta_{0}+j}$ for $2\leq i<\infty$ and $1\leq
j<p$, denoted by $\blacktriangle$, belongs to $K$, we apply Diagram 3 to
$a_{\alpha_{0}\beta_{0}}$ and $a_{\alpha_{0}+i,\beta_{0}+j}$ to get
$x_{2}=a_{\alpha_{0}+j,\beta_{0}+j}\in K$, a contradiction. Hence no element
of $K$ occupies any position below the diagonal in the columns determined by
$\beta_{0}+j$, for $1\leq j<p$

Supposing that $a_{\alpha_{0},\beta_{0}+j}$, with $p<j<\infty$ and
$j\not\in\\{2p,3p,\ldots\\}$, denoted by $\blacktriangledown$, belongs to $K$,
we apply Diagram 2 to $a_{\alpha_{0},\beta_{0}+kp}$ and
$a_{\alpha_{0},\beta_{0}+j}$, where $kp<j<(k+1)p$ to get
$x_{2}=a_{\alpha_{0}+j-kp,\beta_{0}+j}\in K$, a contradiction since $j-kp<p$.
Hence no element of $K$ occupies any position in the row determined by
$\alpha_{0}$ except for the points $a_{\alpha_{0},\beta_{0}+mp}$ for
$m\in\mathbb{N}_{0}$.

Supposing that $a_{\alpha_{0}+i,\beta_{0}}$, with $p<i<\infty$ and
$i\not\in\\{2p,3p,\ldots\\}$, denoted by $\blacktriangleleft$, belongs to $K$,
we apply Diagram 4 to $a_{\alpha_{0}+kp,\beta_{0}}$ and
$a_{\alpha_{0}+i,\beta_{0}}$, where $kp<i<(k+1)p$ to get a contradiction.
Hence no element of $K$ occupies any position in the column determined by
$\beta_{0}$ except for the points $a_{\alpha_{0}+\ell p,\beta_{0}}$ for
$\ell\in\mathbb{N}_{0}$.

We now have

| $\beta_{0}$ |  |  |  | $p$ |  |  |  | $2p$ |  |
---|---|---|---|---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
$r=p=q$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ |  |  |  | $\bullet$ |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ |  | $\ddots$ |  |  |  |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  | $\ddots$ |  |  |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  | $\ddots$ |  |  |
$2p$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ |  |  |  | $\bullet$ |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  |  | $\ddots$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |  |  |  |  |  | $\ddots$

We next consider what happens in the row defined by $\alpha_{0}+p$.

Supposing that $a_{\alpha_{0}+p,\beta_{0}+p+i}$ belongs to $K$, with $1\leq
i<p$, then applying Diagram 3 to $a_{\alpha_{0},\beta_{0}+p}$ and
$a_{\alpha_{0}+p,\beta_{0}+p+i}$ we obtain
$x_{2}=a_{\alpha_{0}+i,\beta_{0}+p+i}\in K$, which is a contradiction, and
repeating this argument shows that no element of $K$ occupies any position in
the row determined by $\alpha_{0}+p$ except for the points
$a_{\alpha_{0}+p,\beta_{0}+mp}$ for $m\in\mathbb{N}_{0}$.

We next consider what happens in the column defined by $\beta_{0}+p$.

Supposing that $a_{\alpha_{0}+p+i,\beta_{0}+p}$ belongs to $K$, with $1\leq
i<p$, then applying Diagram 1 to $a_{\alpha_{0}+p,\beta_{0}}$ and
$a_{\alpha_{0}+p+i,\beta_{0}+p}$ we obtain
$x_{3}=a_{\alpha_{0}+p+i,\beta_{0}+i}\in K$, which is a contradiction, and
repeating this argument shows that no element of $K$ occupies any position in
the column determined by $\beta_{0}+p$ except for the points
$a_{\alpha_{0}+\ell p,\beta_{0}+p}$ for $\ell\in\mathbb{N}_{0}$.

We now have (ignore temporarily the symbol $\blacksquare$)

| $\beta_{0}$ |  |  |  | $p$ |  |  |  | $2p$ |  |
---|---|---|---|---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
$r=p=q$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\ddots$ |  |  |  |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  | $\blacksquare$ |  |  |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  | $\ddots$ |  |  |
$2p$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ |  |  |  | $\bullet$ |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  | $\ddots$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |  |  |  |  | $\ddots$

Finally, we consider what happens along the diagonal.

Supposing that $a_{\alpha_{0}+p+i,\beta_{0}+p+i}$, denoted by $\blacksquare$,
with $1\leq i<p$, belongs to $K$, we apply Diagram 1 to
$a_{\alpha_{0}+p,\beta_{0}}$ and $a_{\alpha_{0}+p+i,\beta_{0}+p+i}$, we obtain
$x_{3}=a_{\alpha_{0}+p+i,\beta_{0}+i}\in K$, which is a contradiction, and
repeating this argument shows that no element of $K$ occupies any position in
the diagonal except for the points $a_{\alpha_{0}+\ell p,\beta_{0}+\ell p}$
for $\ell\in\mathbb{N}_{0}$.

We now have

| $\beta_{0}$ |  |  |  | $p$ |  |  |  | $2p$ |  |
---|---|---|---|---|---|---|---|---|---|---|---
$\alpha_{0}$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\cdots$
$r=p=q$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ | $\circ$ | $\cdots$
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  | $\circ$ |  |  |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  | $\circ$ |  |  |
$2p$ | $\bullet$ | $\circ$ | $\circ$ | $\circ$ | $\bullet$ |  |  |  | $\bullet$ |  |
| $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ |  |  |  |  | $\circ$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |  |  |  |  | $\ddots$

We are now in the position at the beginning of the proof, namely, the semiheap
$K\cap K_{p,p}$ is in subcase (13) of case 3.3.3, and the result follows by
applying successively what has already been proved. ∎

We shall now consider cases (1) and (2) with $a_{\alpha_{0},\beta_{0}}\not\in
K$ (See Lemma 3.13), and assume with no loss of generality, that
$\alpha_{0}=\beta_{0}=0$. We consider Diagram 21 for case (1) and establish
the following notation. The points of $K$ on the row determined by
$\alpha_{0}$, indicated by $\blacktriangle$, are
$a_{\alpha_{0},\beta_{0}+m_{i}}$, with $1\leq m_{1}<m_{2}<\cdots$, and the
points on the column determined by $\beta_{0}$, indicated by $\boxtimes$, are
$a_{\alpha_{0}+\ell_{i},\beta_{0}}$, with $1\leq\ell_{1}<\ell_{2}<\cdots$. We
denote $\sigma=m_{2}-m_{1}$ and $\rho=\ell_{2}-\ell_{1}$. For example, in
Diagram 21, $\sigma=6$ and $\rho=r=4$.

######  Diagram 21.

|  |  |  |  |  |  |  |  |  |  | $m_{1}$ |  |  |  |  |  | $m_{2}$ |  |  |  |  |  |  |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24
|  |  |  |  |  | $r$ |  |  |  |  | $p$ |  |  | $q$ |  |  |  |  |  |  |  |  |  |  |  |
0 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\blacktriangle$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |
1 |  | $\circ$ | $\circ$ |  |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
2 |  | $\circ$ |  | $\circ$ |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
3 |  | $\circ$ |  |  | $\circ$ |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
4 | $\ell_{1}=r$ | $\boxtimes$ | $\circledcirc$ | $\circledcirc$ | $\circledcirc$ | $\square$ |  |  |  | $\square$ | $\circ$ |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$
5 |  | $\circ$ | $\circleddash$ |  |  |  | $\circ$ |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
6 |  | $\circ$ |  | $\circleddash$ |  |  |  | $\circ$ |  |  | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |
7 |  | $\circ$ |  |  | $\circleddash$ |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
8 | $\ell_{2}$ | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$
9 | $p$ |  |  |  |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
10 |  |  |  |  |  |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |
11 |  |  |  |  |  |  |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |
12 | $q$ | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square\bullet$ |  |  | $\blacktriangle$ | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square$
13 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
14 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
15 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
16 |  | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$
17 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
18 |  |  |  |  |  |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |
19 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
20 |  | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$
21 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
22 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
23 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
24 |  | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square$ |  |  | $\blacktriangle$ | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square$
25 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |

We consider first $K_{r,0}$. By Diagram 2, the points $(r,i)$, $1<i<\rho$,
indicated by $\circledcirc$, do not belong to $K$. By Diagram 4, $K\supset
K_{r,0}^{\rho}$. The semiheap $K\cap K_{r,0}$ falls into case 3.3.3, more
precisely, either cases (7), (12) or (13), but case (7) does not occur. In
case (13), the points $(r+j,j)$, $1\leq j<\rho$, indicated by $\circleddash$,
do not belong to $K$, so by Proposition 3.16, $K\cap K_{r,0}=K_{r,0}^{\rho}$,
and therefore in this case,

$K_{1}:=K_{r,0}^{\rho}\cap
K_{r,p}=\\{(\alpha_{0}+r+\ell\rho,\beta_{0}+m\rho:\ell\in\mathbb{N}_{0},m\rho\geq
p\\}.$

.

By the same argument applied to $K_{0,p}$, assuming that $K\cap K_{0,p}$ is
also in case (13), we have

$K_{2}:=K_{0,p}^{\sigma}\cap
K_{r,p}=\\{(\alpha_{0}+\ell^{\prime}\sigma,\beta_{0}+p+m^{\prime}\sigma):\ell^{\prime}\sigma\geq
r,m^{\prime}\in\mathbb{N}_{0}\\}.$

$K_{1}$ is depicted by the symbols $\square$ in $K_{r,p}$ and $K_{2}$ is
depicted by the symbols $\blacktriangle$ in $K_{r,p}$, and we must have
$K_{1}=K\cap K_{r,p}=K_{2}$.

As suggested by the diagram, we now show that $\sigma=\rho$, and that $p$ and
$r$ are divisible by $\sigma$.

  * •

Taking $m^{\prime}=0$ and $\ell^{\prime}\sigma\geq r$,
$(\ell^{\prime}\sigma,p)\in K_{2}$ so that
$(\ell^{\prime}\sigma,p)=(r+\ell\rho,m\rho)$ for some
$\ell,m\in\mathbb{N}_{0}$ with $m\rho\geq p$. Therefore $p=m\rho$ and
$r=\ell^{\prime}\sigma-\ell\rho$.

  * •

Taking $\ell=0$ and $m\rho\geq p$, $(r,m\rho)\in K_{1}$ so that
$(r,m\rho)=(\ell^{\prime}\sigma,p+m^{\prime}\rho)$ for some
$\ell^{\prime},m^{\prime}\in\mathbb{N}_{0}$ with $\ell^{\prime}\sigma\geq r$.
Therefore $r=\ell^{\prime}\sigma$ and $m\rho=p+m^{\prime}\sigma$.

Thus $p$ is divisible by $\rho$, say $p=m_{0}\rho$, and $r$ is divisible by
$\sigma$, say $r=\ell_{0}\sigma$.

  * •

Taking $\ell=0$ and $(m_{0}+1)\rho=p+\rho>p$, $(r,(m_{0}+1)\rho)\in K_{1}$ so
that $(r,(m_{0}+1)\rho)=(\ell^{\prime\prime}\sigma,p+m^{\prime\prime}\rho)$
for some $\ell^{\prime\prime},m^{\prime\prime}\in\mathbb{N}_{0}$ with
$\ell^{\prime\prime}\sigma\geq r$. Therefore $r=\ell^{\prime\prime}\sigma$ and
$(m_{0}+1)\rho=p+m^{\prime\prime}\sigma$. So
$p+\rho=p+m^{\prime\prime}\sigma$, and $\rho=m^{\prime\prime}\sigma$.

  * •

Taking $m^{\prime}=0$ and $(\ell_{0}+1)\sigma=r+\sigma>r$,
$((\ell_{0}+1)\sigma,p)\in K_{2}$ so that
$((\ell_{0}+1)\sigma,p)=(r+\ell\rho,m\rho)$ for some $\ell,m\in\mathbb{N}_{0}$
with $m\rho\geq p$. Therefore $r+\sigma=r+\ell\rho$, so that $\sigma=\ell\rho$

Thus $\rho$ is divisible by $\sigma$ and $\sigma$ is divisible by $\rho$,
hence $\sigma=\rho$.

Since $p$ and $r$ are each a multiple of $\rho$, it follows that $(r,p)\in K$,
so that $(0,p)(r,p)^{*}(r,0)=(0,0)\in K$, which is a contradiction. We
conclude that if both semiheaps $K\cap K_{r,0}$ and $K\cap K_{0,p}$ are in
case (13), then case (1) does not occur.

It remains to show that case (1) does not occur in the three other possible
cases, namely,

  * •

$K\cap K_{r,0}$ is in case (12) and $K\cap K_{0,p}$ is in case (13)

  * •

$K\cap K_{r,0}$ is in case (13) and $K\cap K_{0,p}$ is in case (12)

  * •

$K\cap K_{r,0}$ is in case (12) and $K\cap K_{0,p}$ is in case (12)

Let us now suppose that $K\cap K_{r,0}$ is in case (12), and $K\cap K_{0,p}$
is in case (13) and refer to Diagram 22. Recall that the points of $K$ on the
row determined by $\alpha_{0}$, indicated by $\blacktriangle$, are
$a_{\alpha_{0},\beta_{0}+m_{i}}$, with $1\leq m_{1}<m_{2}<\cdots$, and the
points on the column determined by $\beta_{0}$, indicated by $\boxtimes$, are
$a_{\alpha_{0}+\ell_{i},\beta_{0}}$, with $1\leq\ell_{1}<\ell_{2}<\cdots$. We
denote $\sigma=m_{2}-m_{1}$ and $\rho=\ell_{2}-\ell_{1}$. For example, in
Diagram 22, $\sigma=6$ and $\rho=r=4$.

Since $K\cap K_{r,0}$ is assumed in case (12), by Proposition 3.15, there
exist $0=j_{0}<1\leq j_{1}<j_{2}<\cdots<j_{n}<\rho$ such that

$K\cap K_{r,0}=\bigcup_{i=0}^{n}K_{r+j_{i},j_{i}}^{\rho},$

and therefore in this case,

$K_{1}:=K_{r,0}^{\rho}\cap
K_{r,p}=\bigcup_{i=0}^{n}\left(K_{r+j_{i},j_{i}}^{\rho}\cap
K_{r,p}\right)=\bigcup_{i=0}^{n}\\{(r+j_{i}+\ell\rho,j_{i}+m\rho):\ell,m\in\mathbb{N}_{0},j_{i}+m\rho\geq
p\\}.$

In Diagram 22, we indicate the points of
$K_{r+j_{1},j_{1}}^{\rho}=K_{r+3,3}^{4}$ with the symbols $\heartsuit$, and
the points of $K_{r+j_{2},j_{2}}^{\rho}=K_{r+6,6}^{4}$ with the symbols
$\oplus$.

As before, assumng that $K\cap K_{0,p}$ is in case (13), we have

$K_{2}:=K_{0,p}^{\sigma}\cap
K_{r,p}=\\{(\alpha_{0}+\ell^{\prime}\sigma,\beta_{0}+p+m^{\prime}\sigma):\ell^{\prime}\sigma\geq
r,m^{\prime}\in\mathbb{N}_{0}\\}$

In Diagram 22 below, $K_{1}$ is depicted by the symbols
$\square,\heartsuit,\oplus$ in $K_{r,p}$ and $K_{2}$ is depicted by the
symbols $\blacktriangle$ in $K_{r,p}$, and we must have $K_{1}=K\cap
K_{r,p}=K_{2}$.

######  Diagram 22.

|  |  |  |  | $j_{1}$ |  |  | $j_{2}$ |  |  | $m_{1}$ |  |  |  |  |  | $m_{2}$ |  |  |  |  |  |  |  |  |  |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25
|  |  |  |  |  | $r$ |  |  |  |  | $p$ |  |  | $q$ |  |  |  |  |  |  |  |  |  |  |  |  |
0 |  | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\blacktriangle$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\circ$ | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |
1 |  | $\circ$ | $\circ$ |  |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
2 |  | $\circ$ |  | $\circ$ |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
3 |  | $\circ$ |  |  | $\circ$ |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
4 | $\ell_{1}=r$ | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ | $\circ$ |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |
5 |  | $\circ$ |  |  |  |  | $\circ$ |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
6 |  | $\circ$ |  |  |  |  |  | $\circ$ |  |  | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |  |  | $\blacktriangle$ |  |  |  |
7 | $r+j_{1}$ | $\circ$ |  |  | $\heartsuit$ |  |  |  | $\heartsuit\circ$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |
8 | $\ell_{2}$ | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |
9 | $p$ |  |  |  |  |  |  |  |  |  | $\circ$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
10 | $r+j_{2}$ |  |  |  |  |  |  | $\oplus$ |  |  |  | $\oplus\circ$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |
11 |  |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit\circ$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |
12 | $q$ | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square\bullet$ |  |  | $\blacktriangle$ | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square$ |
13 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
14 |  |  |  |  |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |
15 |  |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |
16 |  | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |
17 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
18 |  |  |  |  |  |  |  | $\oplus$ |  |  | $\blacktriangle$ | $\oplus$ |  |  |  | $\oplus$ | $\blacktriangle$ |  |  | $\oplus$ |  |  | $\blacktriangle$ | $\oplus$ |  |  |
19 |  |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |
20 |  | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |  |  |  | $\square$ |
21 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
22 |  |  |  |  |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |  | $\oplus$ |  |  |
23 |  |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |  |  | $\heartsuit$ |  |
24 |  | $\boxtimes$ |  |  |  | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square$ |  |  | $\blacktriangle$ | $\square$ |  |  |  | $\square$ | $\blacktriangle$ |  |  | $\square$ |
25 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |

We show first that $\rho=\sigma$. Since $K_{1}\subset K_{2}$, for
$\ell,m,i\in\mathbb{N}_{0}$, with $j_{i}+m\rho\geq p$, there exist
$\ell^{\prime},m^{\prime}\in\mathbb{N}_{0}$ with $\ell^{\prime}\sigma\geq r$
and

(3.6)
$(r+j_{i}+\ell\rho,j_{i}+m\rho)=(\ell^{\prime}\sigma,p+m^{\prime}\sigma).$

Fix $i$ such that $j_{i}\geq p$. Then for all $\ell,m\in\mathbb{N}_{0}$, there
exist $\ell^{\prime},m^{\prime}\in\mathbb{N}_{0}$ with
$\ell^{\prime}\sigma\geq r$ such that

$r+j_{i}+\ell\rho=\ell^{\prime}\rho\hbox{ and
}j_{i}+m\rho=p+m^{\prime}\sigma.$

Eliminating $j_{i}$ from these two equations results in

(3.7) $r+p=(m-\ell)\rho+(\ell^{\prime}-m^{\prime})\sigma,$

with $(\ell^{\prime},m^{\prime})$ depending on $(\ell,m)$ and satisfying
$\ell^{\prime}\sigma\geq r$.

Since $K_{2}\subset K_{1}$, for $\ell,m\in\mathbb{N}_{0}$, with
$\ell\sigma\geq r$, there exist $\ell^{\prime},m^{\prime},i\in\mathbb{N}_{0}$
with $j_{i}+m^{\prime}\rho\geq p$ such that

$r+j_{i}+\ell^{\prime}\rho=\ell\sigma\hbox{ and
}j_{i}+m^{\prime}\rho=p+m\sigma.$

Eliminating $j_{i}$ from these two equations results in

(3.8) $r+p=(m^{\prime}-\ell^{\prime})\rho+(\ell-m)\sigma,$

with $(\ell^{\prime},m^{\prime})$ depending on $(\ell,m)$, provided
$\ell\sigma\geq r$.

With $\ell\geq 0$ and $m\geq 0$, from (3.7), there exist $\ell_{1},m_{1}$ such
that

$r+p=(m-\ell)\rho+(\ell_{1}-m_{1})\sigma$

and there exist $\ell_{2},m_{2}$ such that

$r+p=(m+1-\ell)\rho+(\ell_{2}-M_{2})\sigma,$

so by subtraction, $0=\rho+[(\ell_{2}-m_{2})+(\ell_{1}-m_{1})]\sigma$ and
$\sigma$ divides $\rho$.

With $\ell\sigma\geq r$ and $m\geq 0$, from (3.8), there exist
$\ell_{3},m_{3}$ such that

$r+p=(m_{3}-\ell_{3})\rho+(\ell-m)\sigma$

and there exist $\ell_{4},m_{4}$ such that

$r+p=(m_{4}-\ell_{4})\rho+(\ell+1-m)\sigma,$

so by subtraction, $0=[(m_{4}-\ell_{4})-(m_{3}-\ell_{3})]\rho+\sigma$ and
$\rho$ divides $\sigma$.

Hence $\rho=\sigma$ and from (3.7) or (3.8), $\rho$ divides $r+p$. Now, taking
$i=0,\ell=0$ in (3.6), $r=\ell^{\prime}\sigma$, so that also $\rho$ divides
$p$. Hence, as in the previous case, $(r,p)\in K$, so that
$(0,p)(r,p)^{*}(r,0)=(0,0)\in K$, which is a contradiction. We conclude that
if the semiheap $K\cap K_{r,0}$ is in case (12) and the semiheap $K\cap
K_{0,p}$ is in case (13), then case (1) does not occur.

Since the adjoint mapping is an anti-isomorphism of the extended bicyclic
semigroup (See Remark 2.8), it follows that if the semiheap $K\cap K_{r,0}$ is
in case (13) and the semiheap $K\cap K_{0,p}$ is in case (12), then case (1)
does not occur.

It remains to consider the case when both semiheaps $K\cap K_{r,0}$ and $K\cap
K_{0,p}$ are in case (12). After this, again since the adjoint mapping is an
anti-isomorphism, and case (1) has been shown to not occur , it will follow
that case (2) also does not occur, so we will have the following lemma.

######  Lemma 3.17.

Cases (1) and (2) with (necessarily) $(0,0)\not\in K$, do not occur.

###### Proof.

It suffices to show that if both semiheaps $K\cap K_{r,0}$ and $K\cap K_{0,p}$
are in case (12), then $(0,0)\in K$ and therefore case (1) does not occur. We
have

$K_{1}:=K\cap K_{r,0}\cap
K_{r,p}=\bigcup_{i=0}^{n}\\{(r+j_{i}+\ell\rho,j_{i}+m\rho):\ell,m\in\mathbb{N}_{0},j_{i}+m\rho\geq
p\\}$

and

$K_{2}:=K\cap K_{0,p}\cap
K_{r,p}=\bigcup_{i=0}^{n^{\prime}}\\{(k_{i}+\ell\sigma,p+k_{i}+m\sigma):\ell,m\in\mathbb{N}_{0},k_{i}+\ell\sigma\geq
r\\},$

where $0=k_{0}<1\leq k_{1}<k_{2}<\cdots<k_{n^{\prime}}<\sigma$.

Since $K_{1}\subset K_{2}$, for $i,\ell,m\in\mathbb{N}_{0}$ with
$j_{i}+m\rho\geq p$, there exist
$i^{\prime},\ell^{\prime},m^{\prime}\in\mathbb{N}_{0}$ satisfying
$k_{i^{\prime}}+\ell^{\prime}\sigma\geq r$, such that

$(r+j_{i}+\ell\rho,j_{i}+m\rho)=(k_{i^{\prime}}+\ell^{\prime}\sigma,p+k_{i^{\prime}}+m^{\prime}\sigma)$

so that

$r+j_{i}+\ell\rho=k_{i^{\prime}}+\ell^{\prime}\sigma\hbox{ and
}j_{i}+m\rho=p+k_{i^{\prime}}+m^{\prime}\sigma$

Fix $i$ such that $j_{i}\geq p$. Then for every $\ell,m\in\mathbb{N}_{0}$, by
subtraction, we have

(3.9) $r+p=(m-\ell)\rho+(\ell^{\prime}-m^{\prime})\sigma$

with $\ell^{\prime},m^{\prime}$ depending only on $\ell,m\in\mathbb{N}_{0}$
(and $\ell^{\prime}$ satisfying $k_{i^{\prime}}+\ell^{\prime}\sigma\geq r$ for
some $i^{\prime}$).

Since $K_{2}\subset K_{1}$, for $i,\ell,m\in\mathbb{N}_{0}$ with
$k_{i}+\ell\sigma\geq r$, there exist
$i^{\prime},\ell^{\prime},m^{\prime}\in\mathbb{N}_{0}$ satisfying
$j_{i^{\prime}}+m^{\prime}\rho\geq p$, such that

$(k_{i}+\ell\sigma,p+k_{i}+m\sigma)=(r+j_{i^{\prime}}+\ell^{\prime}\rho,j_{i^{\prime}}+m^{\prime}\rho)$

so that

$k_{i}+\ell\sigma=r+j_{i^{\prime}}+\ell^{\prime}\rho\hbox{ and
}p+k_{i}+m\sigma=j_{i^{\prime}}+m^{\prime}\rho$

Fix $i$ such that $k_{i}\geq r$. Then for every $\ell,m\in\mathbb{N}_{0}$, by
subtraction, we have

(3.10) $r+p=(\ell-m)\sigma+(m^{\prime}-\ell^{\prime})\rho$

with $\ell^{\prime},m^{\prime}$ depending only on $\ell,m\in\mathbb{N}_{0}$
(and $m^{\prime}$ satisfying $j_{i^{\prime}}+m^{\prime}\rho\geq p$ for some
$i^{\prime}$).

With $\ell\geq 0$ and $m\geq 0$, from (3.9), there exist $\ell_{1},m_{1}$ such
that

$r+p=(m-\ell)\rho+(\ell_{1}-m_{1})\sigma$

and there exist $\ell_{2},m_{2}$ such that

$r+p=(m+1-\ell)\rho+(\ell_{2}-M_{2})\sigma,$

so by subtraction, $0=\rho+[(\ell_{2}-m_{2})+(\ell_{1}-m_{1})]\sigma$ and
$\sigma$ divides $\rho$.

With $\ell\geq 0$ and $m\geq 0$, from (3.10), there exist $\ell_{1},m_{1}$
such that

$r+p=(\ell-m)\sigma+(m_{1}-\ell_{1})\rho$

and there exist $\ell_{2},m_{2}$ such that

$r+p=(\ell+1-m)\sigma+(m_{2}-\ell_{2})\rho,$

so by subtraction, $0=\sigma+[(m_{2}-\ell_{2})+(m_{1}-\ell_{1})]\rho$ and
$\rho$ divides $\sigma$.

Hence $\rho=\sigma$ and from (3.9) or (3.10), $\sigma$ divides $p+r$. In fact,
$\sigma$ divides both $p$ and $r$. Indeed, since $(q,q)\in K_{1}$ and
$K_{1}=K_{2}$, there exist $\ell,m,i$ and
$\ell^{\prime},m^{\prime},i^{\prime}$, such that

$(q,q)=(r+j_{i}+\ell\sigma,j_{i}+\sigma)=(k_{i^{\prime}}+\ell^{\prime}\sigma,p+k_{i^{\prime}}+m^{\prime}\sigma),$

so that $r+\ell\sigma=m\sigma$ and $p+m^{\prime}\sigma=\ell^{\prime}\sigma$.
We now have $(r,p)\in K$, so that $(0,0)=(0,p)(r,p)^{*}(r,0)\in K$, a
contradiction. ∎

We summarize the results of Lemma 3.13 to Lemma 3.17 in the following
proposition.

######  Proposition 3.18.

If the semiheap $K$ is in case 3.3.3, then either
$K=K_{\alpha_{0},\beta_{0}}^{p}$ for some $p>0$, or there exists $p>0$ such
that

$K=\bigcup_{i=0}^{n}K_{q_{i},q_{i}}^{p}.$

where $a_{\alpha_{0}+q_{i},\beta_{0}+q_{i}}$, $0\leq i<\infty$, are the points
of $K$ lying on the diagonal, such that

$q=q_{0}<q_{1}<q_{2}<\cdots<q_{n}<p\quad\hbox{ and }\quad
p<q_{n+1}<q_{n+2}<\cdots.$

###### Proof.

By Lemma 3.17, cases (1) and (2) do not occur. By Lemma 3.13, cases (3)-(11)
do not occur. Cases (12) and (13) are described in Propositions 3.15 and 3.16.
∎

The following theorem is the main result of this paper, listing all of the
subsemiheaps of the extended bicyclic semigroup.

######  Theorem 3.19.

The subsemiheaps of the extended bicyclic semigroup are the inductive limits
(see Remark 3.1) of sequences of the following semiheaps $K$:

  * •

$K$ is a single point $\\{a_{pq}\\}$ (Lemmal 3.2)

  * •

$K=\\{a_{\alpha_{0},\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:1\leq
j\leq n_{0}\\}$, where $1\leq k_{1}<k_{2}<\cdots<k_{n_{0}}$ and
$a_{\alpha_{0},\beta_{0}}$ and $a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}$ are the
only elements of $K$ that are in the diagonal. (Proposition 3.3)

  * •

$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j\in\mathbb{N}\\}$
(Proposition 3.4)

  * •

There exist $\sigma,\ell_{0}\in\mathbb{N}$, such that

$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{j},\beta_{0}+k_{j}}:j=1,\ldots\ell_{0}-1\\}\cup
K_{\alpha_{0}+k_{\ell_{0}},\beta_{0}+k_{\ell_{0}}}^{\sigma}$

or

$K=\\{a_{\alpha_{0}\beta_{0}}\\}\cup\\{a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}:1\leq
i<\ell_{0}\\}\cup\left(\bigcup_{i=\ell_{0}}^{j}K^{\sigma}_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}\right)$

where $1\leq k_{1}<k_{2}<\cdots<k_{n}<\cdots<\infty$,
$a_{\alpha_{0},\beta_{0}}$ and $a_{\alpha_{0}+k_{i},\beta_{0}+k_{i}}$ are the
only elements of $K$ that are in the diagonal. (Proposition 3.4)

  * •

$K=K_{\alpha_{0},\beta_{0}}^{p}$ for some $p>0$ (Proposition 3.18)

  * •

There exists $p>0$ such that

$K=\bigcup_{i=0}^{n}K_{q_{i},q_{i}}^{p}.$

where $a_{\alpha_{0}+q_{i},\beta_{0}+q_{i}}$, $0\leq i<\infty$, are the points
of $K$ lying on the diagonal, such that

$q=q_{0}<q_{1}<q_{2}<\cdots<q_{n}<p\quad\hbox{ and }\quad
p<q_{n+1}<q_{n+2}<\cdots.$

(Proposition 3.18)

##  4\. Injectivity of W*-TROs

The notation for this section is the following.

$\mathbb{N}=1,2,\ldots$; $\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$;
$\mathbb{Z}=\mathbb{N}_{0}\cup-\mathbb{N}$

$S$ is an inverse semigroup with generalized inverse $x^{*}$

$K$ is a subset of $S$ closed under the triple product $xy^{*}z$ (semiheap).

$\pi$ is the left regular representation of $S$ on $H:=\ell^{2}(S)$ so that
$S$ is an orthonormal basis for $H$ and $\pi(x)$ is the partial isometry
defined by $\pi(x)y=xy$ if $yy^{*}\leq x^{*}x$ and $\pi(x)y=0$ otherwise.

$C^{*}_{\hbox{red}}(S)$ is the C*-algebra generated by $\\{\pi(x):x\in S\\}$
and is the norm closure of $\hbox{span}\,\pi(S)$.

$\hbox{TRO}(K)$ is the TRO generated by $\pi(K)$ and is the norm closure of
$\hbox{span}\,\pi(K)$.

$\hbox{VN}(S)$ is the von Neumann algebra generated by $\pi(S)$ and is the
weak closure of $C^{*}_{\hbox{red}}(S)$

$\hbox{VNTRO}(K)$ is the W*-TRO generated by $\pi(K)$ and is the weak closure
of $\hbox{TRO}(K)$.

Details of the left regular representation are as follows ([Paterson1999, pp.
25–27]). We have

$\pi(a_{ij})a_{pq}=\left\\{\begin{array}[]{cc}a_{ij}a_{pq},&a_{pq}a_{pq}^{*}\leq
a_{ij}^{*}a_{ij}\\\ 0,&\hbox{otherwise}\end{array}\right.,$

that is,

$\pi(a_{ij})a_{pq}=\left\\{\begin{array}[]{cc}a_{ij}a_{pq},&a_{pp}\leq
a_{jj}\\\ 0,&\hbox{otherwise}\end{array}\right.,$

or

$\pi(a_{ij})a_{pq}=\left\\{\begin{array}[]{cc}a_{ij}a_{pq},&p\geq j\\\
0,&\hbox{otherwise}\end{array}\right.,$

or

$\pi(a_{ij})a_{pq}=\left\\{\begin{array}[]{cc}a_{i+p-j,q},&p\geq j\\\
0,&\hbox{otherwise}\end{array}\right.$

Define provisionally a linear map
$\Phi_{0}:\hbox{span}\,\pi(S)\rightarrow\hbox{span}\,\pi(K)$ as follows:
$\Phi_{0}(0)=0$, and for $x_{1},\ldots x_{n}\in S$,

$\Phi_{0}\left(\sum_{i=1}^{n}\lambda_{i}\pi(x_{i})\right)=\sum_{x_{i}\in
K}\lambda_{i}\pi(x_{i}).$

######  Proposition 4.1.

The idempotent map $\Phi_{0}$ is well-defined and contractive, and therefore
extends to a contractive projection $\Phi$ on $C^{*}_{\hbox{red}}(S)$ with
range $\hbox{TRO}(K)$. Moreover, $\Phi$ extends to a completely contractive
projection on $\hbox{VN}(S)$ with range $\hbox{VNTRO}(K)$. Hence, if
$\hbox{VN}(S)$ in an injective von Neumann algebra, then $\hbox{VNTRO}(K)$ is
an injective operator space.

###### Proof.

Let $a=\left\|\sum_{i=1}^{n}\lambda_{i}\pi(x_{i})\right\|$ and
$b=\left\|\sum_{x_{i}\in K}\lambda_{i}\pi(x_{i})\right\|$. With
$\xi=\sum_{z\in S}(\xi,z)z\in\ell^{2}(S)$,

$\pi(x_{i})\xi=\sum_{zz^{*}\leq x_{i}^{*}x_{i}}(\xi,z)x_{i}z$

so that

$b^{2}=\sup_{\|\xi\|\leq 1}\left\|\sum_{x_{i}\in
K}\lambda_{i}\pi(x_{i})\xi\right\|^{2}=\sup_{\|\xi\|\leq 1}\sum_{x_{i}\in
K,z\in S,zz^{*}\leq x_{i}^{*}x_{i}}|\lambda_{i}(\xi,z)|^{2}$

and by the same calculation

$a^{2}=\sup_{\|\xi\|\leq 1}\left\|\sum_{x_{i}\in
S}\lambda_{i}\pi(x_{i})\xi\right\|^{2}=\sup_{\|\xi\|\leq 1}\sum_{x_{i}\in
S,z\in S,zz^{*}\leq x_{i}^{*}x_{i}}|\lambda_{i}(\xi,z)|^{2}.$

Therefore $\Phi_{0}$ is contractive and extends to a contractive projection on
$C^{*}_{\hbox{red}}(S)$ with range $\hbox{TRO}(K)$.

Let $A=C^{*}_{\hbox{red}}(S)$, $U=TRO(K)$, so that $\Phi^{**}$ is a
contractive projection on the von Neumann algebra $A^{**}$ with range
$U^{**}$. By [LanRus83, Lemma], $U^{**}$ is isomorphic to $\hbox{VNTRO}(K)$,
and by [EffOzaRua01, Theorem 2.5], $\Phi^{**}$ is a completely contractive
projection with range $\hbox{VNTRO}(K)$. Therefore,
$\overline{\Phi}:=\Phi^{**}|_{VN(S)}$ is a completely contractive projection
of $VN(S)$ onto $VNTRO(K)$. If $VN(S)$ is injective, then there is a
completely contractive projection $P$ of $B(H)$ onto $VN(S)$, so that
$\overline{\Phi}\circ P$ is a completely contractive projection with range
$VNTRO(K)$. ∎

######  Example 4.2.

Suppose that $e$ and $f$ are idempotents in the inverse semigroup $S$ and that
$K=eSf$, which is a subsemiheap of $S$. The corresponding induced map takes
the form $\pi(x)\mapsto\pi(exf)$ (with $0\rightarrow 0$) and is contractive
since

$\left\|\sum_{i}\lambda_{i}\pi(ex_{i}f)\right\|\leq\|\pi(e)\|\left\|\sum_{i}\lambda_{i}\pi(x_{i})\right\|\|\pi(f)\|=\left\|\sum_{i}\lambda_{i}\pi(x_{i})\right\|.$

Hence Proposition 4.1 applies. This also applies to maps of the form $x\mapsto
ex$ and $x\mapsto xf$.

The maximal subgroups of any inverse semigroup $S$ are of the form

$S_{e}^{e}=\\{s\in S:ss^{*}=s^{*}s=e\\}$

for some idempotent $e$ (See [Paterson1999, p. 198]). Thus, the maximal
subgroups of $E$ in Example 1.4 reduce to one-element groups, so are trivially
amenable and hence by [Paterson1999, Theorem 4.5.2], $\hbox{VN}(E)$ is
injective.444As pointed out to the authors by Alan Paterson, the proof of
[Paterson1999, Theorem 4.5.2] required the assumption that the universal
groupoid of the inverse semigroup be Hausdorff. This assumption holds for the
extended bicyclic semigroup $E$, because it is an E-unitary semigroup (see
[SkandalisEtAlJRAM02, Corollary 3.7] and [Lawson2017, p.57]). In addition, it
appears from [SkandalisEtAlJRAM02] that the Hausdorff assumption can actually
be dropped.

Since $VN(E)$ is injective, where $E$ is the extended bicyclic semigroup, it
follows from Proposition 4.1 and Example 4.2, that $VNTRO(a_{ii}Ea_{jj})$ is
an injective operator space, as are $VNTRO(Ea_{jj})$ and $VNTRO(a_{ii}E)$.
More generally, we have

######  Corollary 4.3.

All of the subsemiheaps of the extended bicyclic semigroup $E$ (which were
determined in Theorem 3.19) give rise to injective W*-TROs.

## References