# Beyond the Galvin–Hindman–Furstenberg Theorem:
Partition regularity of IP sets in arbitrary semigroups

Yonatan Gadot and Boaz Tsaban

###### Abstract.

According to Furstenberg, an _infinite parallelepiped (IP)_ set in a semigroup
is a set that contains a sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$,
together with all finite sums $a_{i_{1}}+\dotsb+a_{i_{m}}$, for natural
numbers $m$ and $i_{1}<\dotsb<i_{m}$. Using a method of Galvin, Hindman proved
that IP sets of natural numbers are _partition regular_ : For each finite
partition of an IP set, some part must be an IP set, too. Furstenberg noted
that the same proof applies to arbitrary semigroups. However, in a semigroup
with idempotents, an IP set may be a singleton.

A _proper IP set_ is one where the witnessing sequence $a_{1},\allowbreak
a_{2},\allowbreak\dots$ may be chosen to be _bijective_ (in $\mathbb{N}$,
every IP set is proper). We provide a complete characterization of semigroups
where the proper IP sets are partition regular, and show that this property is
equivalent to several other fundamental notions of additive Ramsey theory.

Using our results, we prove that if there is a monochromatic proper IP set for
each finite coloring of a semigroup, then for each finite coloring of the
semigroup there are infinitely many, pairwise disjoint, proper IP sets sharing
the same color.

###### 2020 Mathematics Subject Classification:

05D10, 20M10

##  1\. Introduction

Ramsey theory deals with the phenomenon that, whenever a rich mathematical
structure is partitioned into finitely many parts, at least one of the parts
is mathematically rich. A family of sets is _partition regular_ if, for each
finite partition of a set from the family, there is a part that is also in
this family. A _finite coloring_ of a set is a function from that set into a
finite set of colors. A subset of the colored set is _monochromatic_ if all of
its members have the same color. Thus, a family of sets is partition regular
if and only if for each finite coloring of a set in the family, there is in
the family a monochromatic subset of the colored set.

For clarity and compatibility of notation, we use additive notation for all,
_including noncommutative_ , semigroups. Thus, the results of this paper hold
for all, commutative and noncommutative, semigroups. For a sequence
$a_{1},\allowbreak a_{2},\allowbreak\dots$ in a semigroup, let

$\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots):=\\{\,a_{i_{1}}+\dotsb+a_{i_{m}}:m\geq
1,i_{1}<\dotsb<i_{m}\,\\}.$

A subset of a semigroup is an _infinite parallelepiped (IP)_ set if it
contains a set $\operatorname{FS}(a_{1},\allowbreak a_{2},\allowbreak\dots)$,
for some sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in the semigroup.
For convenience, we define that IP sets are not necessarily of the form
$\operatorname{FS}(a_{1},\allowbreak a_{2},\allowbreak\dots)$; it suffices for
them to contain a set of this form. Hindman’s celebrated Finite Sums Theorem
[5, Theorem 3.1] asserts that for each finite coloring of the natural numbers,
there is a monochromatic IP set. Using the Stone-Čech compactification, Galvin
and Glazer provided an elegant proof of Hindman’s Theorem, that applies to all
semigroups [2, Thorem 10.3]. Furstnberg [3, Proposition 8.13] attributes the
following stronger assertion to Hindman. Hindman proves this assertion in the
case of $\mathbb{N}$, and attributes his result to Galvin [6, Corrolary 2.9].

######  Theorem 1 (Furstenberg–Galvin–Hindman).

For each semigroup, the family of IP sets is partition regular.

In the semigroup $\mathbb{N}$ of natural numbers, every IP set contains a set
$\operatorname{FS}(a_{1},\allowbreak a_{2},\allowbreak\dots)$ where the
sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ is _bijective_. Thus, in
$\mathbb{N}$, the term “infinite parallelepiped” is justified. However, as
defined above, an IP set in a general semigroup may be a singleton! Indeed,
for each idempotent element $e$ in a semigroup, we have
$\operatorname{FS}(e,e,\dotsc)=\\{e\\}$. The following definition rectifies
this problem.

######  Definition 2.

A _proper IP_ set in a semigroup is a set that contains a set
$\operatorname{FS}(a_{1},\allowbreak a_{2},\allowbreak\dots)$, for some
_bijective_ sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in the
semigroup.

For a sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in a semigroup and a
natural number $n$, let

$\displaystyle\operatorname{FS}(a_{1},\dotsc,a_{n})$
$\displaystyle:=\\{\,a_{i_{1}}+\dotsm+a_{i_{m}}:1\leq m,i_{1}<\dotsb<i_{m}\leq
n\,\\}.$

######  Lemma 3.

Every infinite set of the form $\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)$ (and thus every superset of such a set) is a proper
IP set.

###### Proof.

Let $n$ be a natural number. Since

$\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)=\operatorname{FS}(a_{1},\dotsc,a_{n})\cup\operatorname{FS}(a_{n+1},a_{n+2},\dotsc)\\\
\cup(\operatorname{FS}(a_{1},\dotsc,a_{n})+\operatorname{FS}(a_{n+1},a_{n+2},\dotsc)),$

the set $\operatorname{FS}(a_{n+1},\allowbreak a_{n+2},\dotsc)$ must be
infinite. It follows that we can construct a bijective sequence
$b_{1},\allowbreak b_{2},\allowbreak\dots$ such that for each $n$ there is an
index $m_{n}$ such that $b_{n}\in\operatorname{FS}(a_{1},\dotsc,a_{m_{n}})$
and $b_{n+1}\in\operatorname{FS}(a_{m_{n}+1},a_{m_{n}+2},\dotsc)$. Clearly,
$\operatorname{FS}(b_{1},\allowbreak
b_{2},\allowbreak\dots)\subseteq\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)$. ∎

In general, the proper IP sets need not be partition regular. Consider the fan
semilattice $S$, colored as follows:

|
---|---
$\textstyle{{\color[rgb]{0.00,0.59,0.00}\bullet}}$$\textstyle{{\color[rgb]{0.00,0.59,0.00}\bullet}}$$\textstyle{{\color[rgb]{0.00,0.59,0.00}\bullet}}$$\textstyle{{\color[rgb]{0.00,0.59,0.00}\bullet}}$$\textstyle{\dotsm}$$\textstyle{{\color[rgb]{1,0,0}\bullet}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

The entire semilattice is a proper IP set, but the only monochromatic IP set
for this coloring is a singleton. It follows that, for a semigroup to have the
proper IP sets partition regular, it may not include the fan semilattice as a
subsemigroup.

For which semigroups are the proper IP sets partition regular? This problem
was raised, for example, by Andrews and Goldbring [1, Question 4.9]. We
provide a complete solution: the proper IP sets in a semigroup are partition
regular if and only if the semigroup has no subsemigroup of one of three
explicit types. We also establish the equivalence of this feature to other
fundamental notions of additive coloring theory. Finally, using our results,
we prove that if a semigroup $S$ contains a monochromatic proper IP set for
each finite coloring of the entire semigroup, then for each finite coloring it
contains _infinitely many_ pairwise disjoint monochromatic proper IP sets.

##  2\. Proper sumsequences

Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a sequence in a semigroup.
For a finite index set $F=\\{i_{1},\dotsc,i_{m}\\}$ with $i_{1}<\dotsb<i_{m}$,
let

$a_{F}:=a_{i_{1}}+\dotsb+a_{i_{m}}.$

For finite index sets $F_{1},F_{2}\subseteq\mathbb{N}$, we write $F_{1}<F_{2}$
if all elements of the set $F_{1}$ are smaller than all elements of the set
$F_{2}$. The sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ is _proper_
[10, Definition 1.3] if $a_{F_{1}}\neq a_{F_{2}}$ for all $F_{1}<F_{2}$.

######  Remark 4.

A proper IP set need not contain a set $\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)$ for a proper sequence $a_{1},\allowbreak
a_{2},\allowbreak\dots$. However, since proper sequences are bijective, an IP
set containing a set $\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)$ for a proper sequence $a_{1},\allowbreak
a_{2},\allowbreak\dots$ must be a proper IP set. These notions of properness
are only partially related.

######  Lemma 5.

Every bijective sequence in a group has a proper subsequence.

###### Proof.

Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a bijective sequence in a
group. We construct a proper subsequence
$b_{1}:=a_{m_{1}},b_{2}:=a_{m_{2}},\dotsc$ by induction.

Let $m_{1}:=1$. Assume that $m_{1}\dotsc,m_{n}$ are defined. Since the set
$A:=\operatorname{FS}(b_{1},\allowbreak\dotsc,\allowbreak b_{n})\cup\\{0\\}$
is finite, so is the set $-A+A$. Choose an index $m_{n+1}$ such that

$a_{m_{n+1}}\in\operatorname{FS}(a_{m_{n}+1},a_{m_{n}+2},\dotsc)\setminus(-A+A).$

Let $F_{1}<F_{2}$ be finite index sets, and let $n:=\max F_{2}$. Then there is
$a\in A:=\operatorname{FS}(b_{1},\dotsc\allowbreak,b_{n-1})\cup\\{0\\}$ such
that $b_{F_{2}}=a+b_{n}$, and $a+b_{n}\notin A$, whereas $b_{F_{1}}\in A$. ∎

A _sumsequence_ (traditionally called _sum subsystem_) of a sequence
$a_{1},\allowbreak a_{2},\allowbreak\dots$ is a sequence
$a_{F_{1}},a_{F_{2}},\dotsc$, for an increasing sequence $F_{1}<F_{2}<\dotsb$
of finite index sets. The proof of Lemma 3 shows that if a set
$\operatorname{FS}(a_{1},\allowbreak a_{2},\allowbreak\dots)$ is infinite,
then the sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ has a bijective
sumsequence. Among other things, our main result (Theorem 15) characterizes
the semigroups where every bijective sequence has a proper sumsequence.

######  Proposition 6.

A sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in a semigroup has a
proper sumsequence if and only if it has a sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ with
$\bigcap_{n=1}^{\infty}\operatorname{FS}(b_{n},b_{n+1},\dotsc)=\emptyset$.

###### Proof.

($\Rightarrow$) Let $b_{1},\allowbreak b_{2},\allowbreak\dots$ be a proper
sumsequence.

$(\Leftarrow)$ We construct a proper subsequence $b_{m_{1}},b_{m_{2}},\dotsc$
of the sequence $b_{1},\allowbreak b_{2},\allowbreak\dots$.

Let $m_{1}:=1$. Assume that we have defined $m_{1},\dotsc,m_{n}$. Since the
set $\operatorname{FS}(b_{m_{1}},\allowbreak\dotsc,\allowbreak b_{m_{n}})$ is
finite and
$\bigcap_{m=1}^{\infty}\operatorname{FS}(b_{m},b_{m+1},\dotsc)=\emptyset$, we
can choose an index $m_{n+1}$ with

$\operatorname{FS}(b_{m_{1}},\dotsc,b_{m_{n}})\cap\operatorname{FS}(b_{m_{n+1}},b_{m_{n+1}+1},\dotsc)=\emptyset.$

Let $c_{1}:=b_{m_{1}},c_{2}:=b_{m_{2}},\dotsc$. For each $n$, we have

$\operatorname{FS}(c_{1},\dotsc,c_{n})\cap\operatorname{FS}(c_{n+1},c_{n+2},\dotsc)=\emptyset.$

Let $F_{1}<F_{2}$ be finite index sets. Let $n:=\max F_{1}$. Then
$c_{F_{1}}\in\operatorname{FS}(c_{1},\dotsc,c_{n})$ and
$c_{F_{2}}\in\operatorname{FS}(c_{n+1},c_{n+2},\dotsc)$, and thus
$c_{F_{1}}\neq c_{F_{2}}$. ∎

In particular, if a sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ has no
proper sumsequence, then the set
$\bigcap_{n=1}^{\infty}\operatorname{FS}(a_{n},\allowbreak a_{n+1},\dotsc)$ is
nonempty.

######  Lemma 7.

Let $S$ be a semigroup and $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a
bijective sequence with no proper sumsequence. Let
$A_{\infty}:=\bigcap_{n=1}^{\infty}\operatorname{FS}(a_{n},a_{n+1},\dotsc)$.
Then:

  1. (1)

For each element $s\in A_{\infty}$ and each $n$, there is a finite index set
$F\geq n$ with such that $s=a_{F}$.

  2. (2)

The set $A_{\infty}$ is a subsemigroup of $S$.

  3. (3)

Every sequence $b_{1},\allowbreak b_{2},\allowbreak\dots$ in $A_{\infty}$ is a
sumsequence of $a_{1},\allowbreak a_{2},\allowbreak\dots$.

###### Proof.

(1) For each $n$, we have $s\in\operatorname{FS}(a_{n},a_{n+1},\dotsc)$.

(2) By Lemma 6, we have $A_{\infty}\neq\emptyset$. Let $s_{1},s_{2}\in
A_{\infty}$. For each $n$, by (1) there are index sets $n\leq F_{1}<F_{2}$
with $s_{1}=a_{F_{1}}$ and $s_{2}=a_{F_{2}}$. Then $s_{1}+s_{2}=a_{F_{1}\cup
F_{2}}\in\operatorname{FS}(a_{n},a_{n+1},\dotsc)$. Thus, $s_{1}+s_{2}\in
A_{\infty}$.

(3) This follows from (1). ∎

######  Definition 8.

A bijective sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in a semigroup
is _minimal_ if, for each bijective sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ of $a_{1},\allowbreak a_{2},\allowbreak\dots$, we have

$\bigcap_{n=1}^{\infty}\operatorname{FS}(b_{n},b_{n+1},\dotsc)=\bigcap_{n=1}^{\infty}\operatorname{FS}(a_{n},a_{n+1},\dotsc).$

######  Proposition 9.

Let $S$ be a semigroup and $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a
bijective sequence with no proper sumsequence. There is a bijective
sumsequence $b_{1},\allowbreak b_{2},\allowbreak\dots$ of $a_{1},\allowbreak
a_{2},\allowbreak\dots$ such that:

  1. (1)

The set $\bigcap_{n=1}^{\infty}\operatorname{FS}(b_{n},b_{n+1},\dotsc)$ is a
finite subsemigroup of $S$.

  2. (2)

The sequence $b_{1},\allowbreak b_{2},\allowbreak\dots$ is minimal.

###### Proof.

(1) By Lemma 7, the set
$\bigcap_{n=1}^{\infty}\operatorname{FS}(a_{n},a_{n+1},\dotsc)$ is a
subsemigroup of $S$. If it is finite, then we are done. Thus, assume that it
is infinite. By a theorem of Shevrin [8], this subsemigroup has one of the
following subsemigroups:

  1. (1)

$(\mathbb{N},+)$.

  2. (2)

An infinite periodic group.

  3. (3)

An infinite right or left zero semigroup.

  4. (4)

$(\mathbb{N},\wedge)$ where $m\wedge n=\operatorname{min}\\{m,n\\}$.

  5. (5)

$(\mathbb{N},\vee)$ where $m\vee n=\operatorname{max}\\{m,n\\}$.

  6. (6)

An infinite semigroup $S$ with $S+S$ finite.

  7. (7)

The fan semilattice $(\mathbb{N},\wedge)$ with $m\wedge n=1$ for distinct
$m,n$.

The semigroups of types (1)–(5) have proper sequences, and by Lemma 7(3),
every such sequence is a sumsequence of the initial sequence. Thus, our
subsemigroup must have a subsemigroup of type (6) or (7). Each of these, in
turn, contains a bijective sequence $b_{1},\allowbreak b_{2},\allowbreak\dots$
with $\bigcap_{n=1}^{\infty}\operatorname{FS}(b_{n},b_{n+1},\dotsc)$ finite.
By Lemma 7(3) again, the sequence $b_{1},\allowbreak b_{2},\allowbreak\dots$
is a sumsequence of the sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$.

(2) Since the semigroup
$\bigcap_{n=1}^{\infty}\operatorname{FS}(b_{n},\allowbreak b_{n+1},\dotsc)$ is
finite, we can move to a bijective sumsequence $c_{1},\allowbreak
c_{2},\allowbreak\dots$ of $b_{1},\allowbreak b_{2},\allowbreak\dots$ such
that the set $\bigcap_{n=1}^{\infty}\operatorname{FS}(c_{n},\allowbreak
c_{n+1},\dotsc)$ is minimal with respect to inclusion. ∎

######  Proposition 10.

Assume that a bijective sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in
a semigroup is minimal, and the set
$A_{\infty}:=\bigcap_{n=1}^{\infty}\operatorname{FS}\allowbreak(a_{n},a_{n+1},\dotsc)$
is a finite subsemigroup. Then there are a right ideal $R$ of the semigroup
$A_{\infty}$ and a bijective sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ of $a_{1},\allowbreak a_{2},\allowbreak\dots$ such
that $R+\langle b_{1},\allowbreak b_{2},\allowbreak\dots\rangle\subseteq R$.

###### Proof.

First, assume that there are an idempotent element $e$ in $A_{\infty}$ and a
bijective sumsequence $b_{1},\allowbreak b_{2},\allowbreak\dots$ of
$a_{1},\allowbreak a_{2},\allowbreak\dots$ such that
$e+\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)\subseteq
A_{\infty}$. Consider the right ideal $R:=e+A_{\infty}$ of $A_{\infty}$. Since
the sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ is minimal, we have

$A_{\infty}=\bigcap_{n=1}^{\infty}\operatorname{FS}(b_{n},b_{n+1},\dotsc).$

Since the semigroup $A_{\infty}$ is finite, there is a number $m$ such that
$A_{\infty}\subseteq\operatorname{FS}(b_{1},\allowbreak\dotsc,\allowbreak
b_{m})$. Fix a number $k>m$. Then

$R+b_{k}=e+A_{\infty}+b_{k}\subseteq
e+\operatorname{FS}(b_{1},\dotsc,b_{m})+b_{k}\subseteq
e+\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)\subseteq
A_{\infty}.$

As $R=e+A_{\infty}$, we have $e+R=R$, and thus

$R+b_{k}=e+R+b_{k}\subseteq e+A_{\infty}=R.$

This shows that $R+\langle b_{m+1},b_{m+2},\dotsc\rangle\subseteq R$, and the
set $R$ and the sumsequence $(b_{m+1},b_{m+2},\allowbreak\dotsc)$ are as
required.

Next, towards a contradiction, assume that for each idempotent $e\in
A_{\infty}$ and every bijective sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$, we have $e+\operatorname{FS}(b_{1},\allowbreak
b_{2},\allowbreak\dots)\nsubseteq A_{\infty}$. Let $e\in A_{\infty}$ be an
idempotent. We construct by induction a bijective sumsequence
$b_{1},\allowbreak b_{2},\allowbreak\dots\not\in A_{\infty}$ such that
$e+b_{n}=b_{n}$ for all $n$.

Assume that $b_{1}:=a_{F_{1}},\dotsc,b_{n}:=a_{F_{n}}$ are defined. Since
$b_{1},\dotsc,b_{n}\not\in A_{\infty}$, there is some natural number $m>F_{n}$
such that

$b_{1},\dotsc,b_{n}\not\in\operatorname{FS}(a_{m},a_{m+1},\dotsc).$

Since $e\in A_{\infty}\subseteq\operatorname{FS}(a_{m},a_{m+1},\dotsc)$, there
is a finite index set $G\geq m$ such that $e=a_{G}$. Denote $k:=\max G$. By
our assumption, there is a finite sum
$a_{H}\in\operatorname{FS}(a_{k+1},a_{k+2},\dotsc)$ such that $e+a_{H}\not\in
A_{\infty}$. Define $b_{n+1}:=a_{G\cup H}=e+a_{H}$. It is clear that
$b_{1},\dotsc,b_{n},b_{n+1}\not\in A_{\infty}$ is a bijective sumsequence and
that $e+b_{n+1}=b_{n+1}$.

If there is another idempotent $e^{\prime}\in A_{\infty}$, we repeat this
construction for the sumsequence $b_{1},\allowbreak b_{2},\allowbreak\dots$
and the idempotent $e^{\prime}$. Since the set $A_{\infty}$ is finite, we
eventually receive a bijective sumsequence $c_{1},\allowbreak
c_{2},\allowbreak\dots$ such that all idempotents in $A_{\infty}$ are left
identities for $c_{1},\allowbreak c_{2},\allowbreak\dots$. We consider the two
possible cases.

_Case 1._ There is an idempotent $e\in A_{\infty}$ such that for every
$m\in\mathbb{N}$, we have

$\operatorname{FS}(c_{m},c_{m+1},\dotsc)+e\nsubseteq A_{\infty}.$

As above, we construct a bijective sumsequence $d_{1},\allowbreak
d_{2},\allowbreak\dots$ such that $d_{n}+e=d_{n}$ for all $n$. The
subsemigroup $M:=\langle d_{1},\allowbreak d_{2},\allowbreak\dots\rangle$ is a
monoid with the identity $e$. Since $e\in
A_{\infty}=\bigcap_{n=1}^{\infty}\operatorname{FS}(d_{n},\allowbreak
d_{n+1},\dotsc)$, there is a subsequence $d_{n_{1}},d_{n_{2}},\dotsc$ of
elements such that, for each $i$, there is a finite index set $F>n_{i}$ with
$d_{n_{i}}+d_{F}=e$. For each $m$, let $s_{m}:=d_{n_{m}}$, an element with a
right $e$-inverse. Similarly, since
$e\in\bigcap_{n=1}^{\infty}\operatorname{FS}(s_{n},\allowbreak
s_{n+1},\dotsc)$, we have a subsequence $s_{n_{1}},s_{n_{2}},\dotsc$ of
elements with a left $e$-inverse. Since they already have a right $e$-inverse,
we have $s_{n_{1}},s_{n_{2}},\dotsc\in H(e)$ where $H(e)$ is the largest
subgroup in $M$ with identity $e$. By Lemma 5, every bijective sequence in a
group has a proper subsequence, and thus the sequence $a_{1},\allowbreak
a_{2},\allowbreak\dots$ has a proper sumsequence; a contradiction.

_Case 2._ Since the set $A_{\infty}$ is finite, in the remaining case there is
$m$ such that $\operatorname{FS}(c_{m},c_{m+1},\allowbreak\dotsc)+e\subseteq
A_{\infty}$ for every idempotent $e\in A_{\infty}$. Since every idempotent
element $e\in A_{\infty}$ is a left identity for $c_{1},\allowbreak
c_{2},\allowbreak\dots$, it is also a left identity for
$A_{\infty}\subseteq\operatorname{FS}(c_{1},\allowbreak
c_{2},\allowbreak\dots)$. It follows that

(1)
$\operatorname{FS}(c_{m},c_{m+1},\dotsc)+A_{\infty}=\operatorname{FS}(c_{m},c_{m+1},\dotsc)+e+A_{\infty}\subseteq
A_{\infty}+A_{\infty}\subseteq A_{\infty}$

We claim that for each element $s\in A_{\infty}$ and for any $n\in\mathbb{N}$,

(2) $c_{n}\in A_{\infty}\text{ whenever }s+c_{n}\in A_{\infty}.$

Indeed, let $k$ be a number such that $k\cdot s\in A_{\infty}$ is an
idempotent. Then

$c_{n}=k\cdot s+c_{n}=(k-1)\cdot s+(s+c_{n})\in A_{\infty}.$

We construct a bijective sumsequence $d_{1},\allowbreak
d_{2},\allowbreak\dots$ of $c_{1},\allowbreak c_{2},\allowbreak\dots$ such
that for each $m$ we have $d_{m}:=s_{m}+c^{\prime}_{m}$ for some element
$s_{m}\in A_{\infty}$ and some element $c^{\prime}_{m}:=c_{n_{m}}\not\in
A_{\infty}$. Let $F=\\{\,m_{1}<\dotsb<m_{k}\,\\}$ be a finite index set. Then

$d_{F}=d_{m_{1}}+\dotsb+d_{m_{k}}=s_{m_{1}}+c^{\prime}_{m_{1}}+\dotsb+c^{\prime}_{m_{k-1}}+s_{m_{k}}+c^{\prime}_{m_{k}}.$

By Equation (1), we have

$s:=s_{m_{1}}+c^{\prime}_{m_{1}}+\dotsb+c^{\prime}_{m_{k-1}}+s_{m_{k}}\in
A_{\infty}$

since $s_{m_{k}}\in A_{\infty}$. By Equation (2), we have
$d_{F}=s+c^{\prime}_{m_{k}}\not\in A_{\infty}$. This is true for every finite
index set $F$, and therefore $\operatorname{FS}(d_{1},\allowbreak
d_{2},\allowbreak\dots)\cap A_{\infty}=\emptyset$, in contradiction to the
minimality of the sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$. ∎

######  Example 11.

Consider the semigroup $\mathbb{N}\cup\\{0\\}$ with addition modulo $k$. The
sequence $1,2,3,\dotsc$ has the subsequence $k,2k,3k,\dotsc$ with
$\bigcap_{n=1}^{\infty}\operatorname{FS}(nk,(n+1)k,\dotsc)=\\{0\\}$, clearly a
minimal subsemigroup. It is also clear that $\\{0\\}+\langle
nk,(n+1)k,\dotsc\rangle=\\{0\\}$.

##  3\. Parition regularity of proper IP sets and related notions

For a sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in a semigroup, let

$\operatorname{FS}_{\geq 2}(a_{1},\allowbreak
a_{2},\allowbreak\dots):=\\{\,a_{i_{1}}+\dotsb+a_{i_{m}}:m\geq
2,i_{1}<\dotsb<i_{m}\,\\}.$

######  Proposition 12.

Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a bijective sequence in a
semigroup $S$ with no proper sumsequence. There is a bijective sumsequence
$b_{1},\allowbreak b_{2},\allowbreak\dots$ with $\operatorname{FS}_{\geq
2}(b_{1},\allowbreak b_{2},\allowbreak\dots)$ finite.

###### Proof.

The previous results imply that by moving to a bijective sumsequence, we may
assume that the sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ is
minimal, the set
$A_{\infty}:=\bigcap_{n=1}^{\infty}\operatorname{FS}(a_{n},\allowbreak
a_{n+1},\allowbreak\dotsc)$ is finite, and there is a right ideal $R\leq
A_{\infty}$ such that $R+\langle a_{1},\allowbreak
a_{2},\allowbreak\dots\rangle\subseteq R$. Since the set $A_{\infty}$ is
finite, we may assume further (by moving to a bijective sumsequence of
$a_{1},\allowbreak a_{2},\allowbreak\dots$) that this right ideal is _maximal_
with respect to inclusion, in the sense that for each right ideal $J\leq
A_{\infty}$ with $R\subseteq J$, for which there is a bijective sumsequence
$s_{1},\allowbreak s_{2},\allowbreak\dots$ of $a_{1},\allowbreak
a_{2},\allowbreak\dots$ such that $J+\langle s_{1},\allowbreak
s_{2},\allowbreak\dots\rangle\subseteq R$, we have $J=R$.

The sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ is of one of the
following types:

  1. (1)

There are a finite sum $a_{F}\in\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)\setminus A_{\infty}$ and a number $n$ such that

$(a_{F}+\operatorname{FS}(a_{n},a_{n+1},\dotsc))\cap R=\emptyset.$

  2. (2)

There are a finite sum $a_{F}\in\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)\setminus A_{\infty}$ and a number $n$ such that

$\emptyset\neq\\{\,a_{H}\in\operatorname{FS}(a_{n},a_{n+1},\dotsc):a_{F}+a_{H}\in
R\,\\}\subseteq A_{\infty}.$

  3. (3)

Otherwise, for each finite sum $a_{F}\in\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)\setminus A_{\infty}$ and each $n$, there is a finite
sum $a_{H}\in\operatorname{FS}(a_{n},a_{n+1},\dotsc)\setminus A_{\infty}$ such
that $a_{F}+a_{H}\in R$. It follows that for each finite sum
$a_{F}\in\operatorname{FS}(a_{1},\allowbreak a_{2},\allowbreak\dots)\setminus
A_{\infty}$ there is a bijective sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ such that $a_{F}+b_{n}\in R$ for all $n$. (Thus, since
$R+\langle a_{1},\allowbreak a_{2},\allowbreak\dots\rangle\subseteq R$, we
actually have $a_{F}+\langle b_{1},\allowbreak
b_{2},\allowbreak\dots\rangle\subseteq R$.)

By moving to an appropriate sumsequence, we may have $a_{1},\allowbreak
a_{2},\allowbreak\dots$ satisfying one of the following conditions:

  1. (1)

Every bijective sumsequence has a bijective sumsequence of type 1.

  2. (2)

Every bijective sumsequence has a bijective sumsequence of type 2.

  3. (3)

Every bijective sumsequence is of type 3.

We complete the proof addressing these three possible cases.

_Case 1_. We construct by induction a bijective sumsequence $c_{1},\allowbreak
c_{2},\allowbreak\dots\not\in A_{\infty}$, together with a bijective
sumsequence $b^{(n)}_{1},b^{(n)}_{2},\dotsc$ for each $n$ such that the
following conditions hold for each $n$:

  * •

The sequence $b^{(n+1)}_{1},b^{(n+1)}_{2},\dotsc$ is a sumsequence of
$b^{(n)}_{1},b^{(n)}_{2},\dotsc$.

  * •

$c_{n}\in\operatorname{FS}(b^{(n)}_{1},b^{(n)}_{2},\dotsc)$.

  * •

$c_{n}+\operatorname{FS}(b^{(n+1)}_{1},b^{(n+1)}_{2},\dotsc)\cap R=\emptyset$.

Define $b^{(0)}_{1},b^{(0)}_{2},\dotsc:=a_{1},\allowbreak
a_{2},\allowbreak\dots$. Assume we have defined $c_{1},\dotsc,c_{n-1}$ and a
sumsequence $b^{(i)}_{1},b^{(i)}_{2},\dotsc$ for all $1\leq i\leq n$. We now
define $c_{n}$ and $b^{(n+1)}_{1},b^{(n+1)}_{2},\dotsc$ as requested.

Since $c_{1},\dotsc,c_{n-1}\not\in A_{\infty}$, there exists a number $k$ such
that $c_{1},\dotsc,c_{n-1}\in\operatorname{FS}(a_{1},\dotsc,a_{k-1})$ and
$c_{1},\allowbreak
c_{2},\dotsc,c_{n-1}\not\in\operatorname{FS}(a_{k},\allowbreak
a_{k+1},\dotsc)$. There exists a number $m$ such that $b^{(n)}_{m},\allowbreak
b^{(n)}_{m+1},\dotsc\in\operatorname{FS}(a_{k},\allowbreak a_{k+1},\dotsc)$.
The sumsequence $b^{(n)}_{m},\allowbreak b^{(n)}_{m+1},\dotsc$ has a bijective
sumsequence $s_{1},\allowbreak s_{2},\allowbreak\dots$ of type 1. Therefore,
there are $s_{F}\in\operatorname{FS}(s_{1},\allowbreak
s_{2},\allowbreak\dots)\setminus A_{\infty}$ and a number $n$ such that

$(s_{F}+\operatorname{FS}(s_{n},s_{n+1},\dotsc))\cap R=\emptyset.$

Define $c_{n}:=s_{F}$ and
$b^{(n+1)}_{1},b^{(n+1)}_{2},\dotsc:=s_{n},s_{n+1},\dotsc$. The requested
conditions are satisfied.

We prove that $\operatorname{FS}(c_{1},\allowbreak c_{2},\allowbreak\dots)\cap
R=\emptyset$, in contradiction to our assumption that

$R\subseteq
A_{\infty}=\bigcap_{n=1}^{\infty}\operatorname{FS}(c_{n},c_{n+1},\dotsc),$

by the minimality of the sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$..
Indeed, for a finite index set $F:=\\{\,m_{1}<\dotsb<m_{k}\,\\}$, we have

$c_{F}=c_{m_{1}}+c_{m_{2}}+\dotsb+c_{m_{k}}\in
c_{m_{1}}+\operatorname{FS}(b^{(m_{1}+1)}_{1},b^{(m_{1}+1)}_{2},\dotsc).$

Since
$(c_{m_{1}}+\operatorname{FS}(b^{(m_{1}+1)}_{1},b^{(m_{1}+1)}_{2},\dotsc))\cap
R=\emptyset$, we have $c_{F}\not\in R$.

_Case 2_. In a similar way, we can construct a bijective sumsequence
$b_{1},\allowbreak b_{2},\allowbreak\dots$ of $a_{1},\allowbreak
a_{2},\allowbreak\dots$ such that for each $n$,

$\emptyset\neq\\{\,b_{H}\in\operatorname{FS}(b_{n+1},b_{n+2},\dotsc):b_{n}+b_{H}\in
R\,\\}\subseteq A_{\infty}.$

The ideal $A_{\infty}$ is finite, so we may assume that $b_{n}\not\in
A_{\infty}$ for any $n\in\mathbb{N}$.

Since $b_{m_{1}}+\dotsb+b_{m_{n}}\in R$ implies $b_{m_{2}}+\dotsb+b_{m_{n}}\in
A_{\infty}$ for each finite sum, we necessarily have a finite sum
$b_{m_{1}}+\dotsb+b_{m_{n}}\in R$ such that $b_{m_{2}}+\dotsb+b_{m_{n}}\in
A_{\infty}\setminus R$. (Otherwise, given a sum $b_{k_{1}}+\dotsb+b_{k_{r}}\in
R$ it must be that either $b_{k_{r}}\in A_{\infty}$ or there exists $1\leq
l\leq r-1$ such that $b_{k_{l}}+\dotsb+b_{k_{r}}\in R$ and
$b_{k_{l+1}}+\dotsb+b_{k_{r}}\not\in A_{\infty}$.)

Consider the right ideal

$R^{\prime}:=\\{\,s\in A_{\infty}:b_{m_{1}}+s\in R\,\\}\leq A_{\infty}.$

Since $R^{\prime}$ is finite and $R^{\prime}\subseteq
A_{\infty}\subseteq\operatorname{FS}(b_{m_{1}+1},b_{m_{1}+2},\dotsc)$, there
is $k$ such that
$R^{\prime}\subseteq\operatorname{FS}(b_{m_{1}+1},\allowbreak\dotsc,\allowbreak
b_{k-1})$. Clearly, $b_{m_{1}}+R^{\prime}\subseteq R$ and thus
$b_{m_{1}}+R^{\prime}+\operatorname{FS}(b_{k},b_{k+1},\dotsc)\subseteq R$.
Each element in $R^{\prime}+\operatorname{FS}(b_{k},b_{k+1},\dotsc)$ may be
presented as a finite sum $b_{F}$ for a finite index set $F>m_{1}$. Since
$b_{m_{1}}+b_{F}\in R$, we have $b_{F}\in A_{\infty}$. Therefore
$R^{\prime}+\operatorname{FS}(b_{k},b_{k+1},\dotsc)\subseteq A_{\infty}$, and
thus $R^{\prime}+\operatorname{FS}(b_{k},b_{k+1},\dotsc)\subseteq R^{\prime}$.

It follows that $R\cup R^{\prime}$ is also a right ideal in $A_{\infty}$ and
$(R\cup R^{\prime})+\operatorname{FS}(b_{k},\allowbreak
b_{k+1},\dotsc)\subseteq R\cup R^{\prime}$. Since

$b_{m_{2}}+\dotsb+b_{m_{n}}\in R^{\prime}\setminus R,$

we have $R\subsetneq R\cup R^{\prime}$; a contradiction to the maximality of
$R$.

_Case 3_. As above, we construct a bijective sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ of the sequence $a_{1},\allowbreak
a_{2},\allowbreak\dots$ such that for each $n$, we have

$b_{n}+\operatorname{FS}(b_{n+1},b_{n+2},\dotsc)\subseteq R.$

It follows that $\operatorname{FS}_{\geq 2}(b_{1},\allowbreak
b_{2},\allowbreak\dots)\subseteq R$ is finite, as requested. ∎

######  Corollary 13.

Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a minimal bijective sequence
with no proper sumsequence in a semigroup $S$. The semigroup
$A_{\infty}:=\bigcap_{n=1}^{\infty}\operatorname{FS}(a_{n},a_{n+1}\dotsc)$ may
be embedded in a cyclic semigroup $\langle a\rangle$. There are a bijective
sumsequence $b_{1},\allowbreak b_{2},\allowbreak\dots$ and an embedding
$\theta\colon A_{\infty}\to\langle a\rangle$ such that the following
statements hold:

  * •

$\operatorname{FS}_{\geq 2}(b_{1},\allowbreak
b_{2},\allowbreak\dots)=A_{\infty}$.

  * •

The set $A_{\infty}$ is a two-sided ideal in the semigroup $\langle
b_{1},\allowbreak b_{2},\allowbreak\dots\rangle$.

  * •

For all $n<m$, we have $\theta(b_{n}+b_{m})=2\cdot a$. If $\theta(s)=k\cdot a$
for some $s\in A_{\infty}$, then $\theta(b_{n}+s)=\theta(s+b_{n})=(k+1)\cdot
a$.

###### Proof.

It follows from Proposition 12 that the sequence $a_{1},\allowbreak
a_{2},\allowbreak\dots$ has a bijective sumsequence $c_{1},\allowbreak
c_{2},\allowbreak\dots$ such that $\operatorname{FS}_{\geq
2}(c_{1},\allowbreak c_{2},\allowbreak\dots)$ is finite. By the minimality of
the sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$, we have

$\bigcap_{n=1}^{\infty}\operatorname{FS}(c_{n},\allowbreak
c_{n+1},\dotsc)=A_{\infty},$

and hence there is $n$ such that $\operatorname{FS}_{\geq 2}(c_{n},\allowbreak
c_{n+1},\dotsc)=A_{\infty}$. Since $A_{\infty}$ is finite, we have
$A_{\infty}\subseteq\operatorname{FS}(c_{n},\allowbreak\dotsc,c_{m-1})$ for
some $m$. Thus,

$A_{\infty}+\operatorname{FS}(c_{m},c_{m+1},\dotsc)\subseteq\operatorname{FS}(c_{n},\dotsc,c_{m-1})+\operatorname{FS}(c_{m},c_{m+1},\dotsc).$

It follows that

$A_{\infty}+\operatorname{FS}(c_{m},c_{m+1},\dotsc)\subseteq\operatorname{FS}_{\geq
2}(c_{n},\allowbreak c_{n+1},\dotsc)=A_{\infty}.$

For all $m\leq k$,
$A_{\infty}\subseteq\operatorname{FS}(c_{k+1},c_{k+2},\dotsc)$, and thus

$c_{k}+A_{\infty}\subseteq
c_{k}+\operatorname{FS}(c_{k+1},c_{k+2},\dotsc)\subseteq\operatorname{FS}_{\geq
2}(c_{k},c_{k+1},\dotsc)=A_{\infty}.$

Therefore $A_{\infty}$ is a two-sided ideal in $\langle
c_{m},c_{m+1},\dotsc\rangle$.

Consider a finite coloring of the complete graph with the set of vertices
$\mathbb{N}_{\geq m}$:

$\chi\colon[\mathbb{N}_{\geq m}]^{2}\to A_{\infty}$

defined by

$\chi(\\{i<j\\}):=c_{i}+c_{j}.$

By Ramsey’s Theorem, there is an infinite complete monochromatic subgraph. It
follows that there is a subsequence $c_{n_{1}},c_{n_{2}},\dotsc$ of
$c_{m},c_{m+1},\dotsc$ such that $s_{2}:=c_{n_{i}}+c_{n_{j}}\in A_{\infty}$
are equal for all $i<j$. Using the pigeonhole principle, we construct a
subsequence $b_{1},\allowbreak b_{2},\allowbreak\dots$ of
$c_{n_{1}},c_{n_{2}},\dotsc$ such that $s_{3}:=b_{n}+s_{2}\in A_{\infty}$ are
equal for all $n$ as well. It is clear now that for all $k$ and
$n_{1}<\dotsb<n_{2\cdot k+1}$, we have

(3) $b_{n_{1}}+\dotsb+b_{n_{2\cdot k}}=k\cdot s_{2},\text{
}b_{n_{1}}+\dotsb+b_{n_{2\cdot k+1}}=(k-1)\cdot s_{2}+s_{3}.$

It follows that each finite ordered sum of $b_{1},\allowbreak
b_{2},\allowbreak\dots$ is determined only by its length. Therefore, we can
define a homomorphism $\phi\colon\mathbb{N}_{\geq 2}\to A_{\infty}$ by
$\phi(n):=b_{1}+\dotsb+b_{n}$. It is indeed a homomorphism since

$\phi(n_{1})+\phi(n_{2})=(b_{1}+\dotsb+b_{n_{1}})+(b_{1}+\dotsb+b_{n_{2}})=$
$=(b_{1}+\dotsb+b_{n_{1}})+(b_{n_{1}+1}+\dotsb+b_{n_{1}+n_{2}})=\phi(n_{1}+n_{2}).$

It is clear that the function $\phi$ is surjective.

Define now an equivalence relation on $\mathbb{N}$ by $m\sim n$ if and only if
$2\leq m,n$ and $\phi(m)=\phi(n)$. We need to show that the equivalence
relation is compatible with the addition in $\mathbb{N}$. Assume $m_{1}\sim
n_{1}$ and $m_{2}\sim n_{2}$, and we prove that $m_{1}+m_{2}\sim n_{1}+n_{2}$.
When all elements are greater or equal to $2$, it is clear since $\phi$ is
homomorphism. We only need to prove it for $2\leq m_{1},n_{1}$ and
$m_{2}=n_{2}=1$. Denote $s:=\phi(m_{1})=\phi(n_{1})\in A_{\infty}$. By
Equation 3, we have

$\phi(m_{1}+1)=b_{1}+(b_{2}+\dotsb+b_{m_{1}+1})=b_{1}+s$
$\phi(m_{2}+1)=b_{1}+(b_{2}+\dotsb+b_{m_{2}+1})=b_{1}+s.$

Therefore, $\phi(m_{1}+1)=\phi(m_{2}+1)$, and it follows that $m_{1}+m_{2}\sim
n_{1}+n_{2}$, as requested.

It is clear that for each $s\in A_{\infty}$ the set $\phi^{-1}(s)$ is an
equivalence class of $\sim$; for all $s\neq s^{\prime}\in A_{\infty}$, the
equivalence classes $\phi^{-1}(s),\phi^{-1}(s^{\prime})$ are disjoint. The
function $\theta(s):=[\phi^{-1}(s)]$ defines an embedding of $A_{\infty}$ into
the cyclic semigroup $\mathbb{N}/\sim$.

It is left to prove that for each $n$ and each $s\in A_{\infty}$ such that
$\theta(s)=[k]$, we have $\theta(b_{n}+s)=\theta(s+b_{n})=[k+1]$. Fix $n$. We
may choose $n<n_{1}<\dotsb<n_{k}$ and by Equation 3, we have

$b_{n}+s=b_{n}+(b_{n_{1}}+\dotsb+b_{n_{k}})=\phi(k+1).$

Choose $m$ such that
$A_{\infty}\subseteq\operatorname{FS}(b_{1},\dotsc,b_{m-1})$. Let $s$ be an
element in $A_{\infty}$ such that $\theta(s)=[k]$. We may assume that $k<m$
and $s=b_{n_{1}}+\dotsb+b_{n_{k}}$ for an increasing index set
$n_{1}<\dotsb<n_{k}<m$. For all $m\leq n$, it follows from Equation 3 that

$s+b_{n}=b_{n_{1}}+\dotsb+b_{n_{k}}+b_{n}=\phi(k+1).$

The sumsequence $b_{m},b_{m+1},\dotsc$ satisfies the requested conditions. ∎

The proof of the Fustenberg–Galvin–Hindman Theorem (Theorem 1) actually
establishes the following result.

######  Theorem 14 (Hindman–Strauss [7, Theorem 5.14]).

For each sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ in a semigroup
$S$, and each finite coloring of $S$, there is a sumsequence
$b_{1},\allowbreak b_{2},\allowbreak\dots$ of $a_{1},\allowbreak
a_{2},\allowbreak\dots$ such that the set $\operatorname{FS}(b_{1},\allowbreak
b_{2},\allowbreak\dots)$ is monochromatic.

Following is our main result.

######  Theorem 15.

The following assertions are equivalent for a semigroup $S$:

  1. (1)

The proper IP sets are partition regular.

  2. (2)

Every bijective sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ has a
proper sumsequence.

  3. (3)

The semigroup $S$ has no subsemigroup of any of the following types:

    1. (a)

An infinite semigroup $S^{\prime}$ with $S^{\prime}+S^{\prime}$ finite.

    2. (b)

The fan semilattice $(\mathbb{N},\wedge)$ with $m\wedge n:=1$ for distinct
$m,n$.

    3. (c)

The semigroup $(\mathbb{N}\times\mathbb{N})\cup\\{0\\}$ with
$0+(m,n),(m,n)+0:=0$ and

$(m_{1},n_{1})+(m_{2},n_{2}):=\begin{cases}(m_{1},n_{1}+n_{2})&m_{1}=m_{2},\\\
0&\text{otherwise}\end{cases}.$

  4. (4)

For each bijective sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ and
each finite coloring of $S$, there is a bijective sumsequence
$b_{1},\allowbreak b_{2},\allowbreak\dots$ such that the set
$\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)$ is
monochromatic.

  5. (5)

For each bijective sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ and
each finite coloring of $S$, there is a proper sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ such that the set $\operatorname{FS}(b_{1},\allowbreak
b_{2},\allowbreak\dots)$ is monochromatic.

###### Proof.

(1 $\Rightarrow$ 3) Each of the subsemigroups listed in (3) has a bijective
sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ such that the set
$F:=\operatorname{FS}_{\geq 2}(a_{1},\allowbreak a_{2},\allowbreak\dots)$ is
finite. In such a case, the set $S:=\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)$ is a proper IP set, and the pieces in the partition
$(S\setminus F)\cup F$ are not proper IP sets.

(2 $\Rightarrow$ 5) Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a
bijective sequence. By (2), there is a proper sumsequence $c_{1},\allowbreak
c_{2},\allowbreak\dots$ of $a_{1},\allowbreak a_{2},\allowbreak\dots$. Given a
finite coloring of the semigroup, by Theorem 14, the sequence
$c_{1},\allowbreak c_{2},\allowbreak\dots$ has a sumsequence
$b_{1},\allowbreak b_{2},\allowbreak\dots$ such that the set
$\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)$ is
monochromatic. Since the sequence $c_{1},\allowbreak c_{2},\allowbreak\dots$
is proper, so is its sumsequence $b_{1},\allowbreak b_{2},\allowbreak\dots$.

(5 $\Rightarrow$ 4) Proper sequences are bijective.

(4 $\Rightarrow$ 1) Let $A$ be a proper IP set. We may assume that
$A=\operatorname{FS}(a_{1},\allowbreak a_{2},\allowbreak\dots)$, where the
sequence $a_{1},\allowbreak a_{2},\allowbreak\dots$ is bijective. For each
finite coloring of the set $A$, there is by (4) a bijective sumsequence
$b_{1},\allowbreak b_{2},\allowbreak\dots$ such that the (proper IP) set
$\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)$ is
monochromatic.

(3 $\Rightarrow$ 2) Let $S$ be a semigroup containing a bijective sequence
with no proper sumsequence. Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be
a bijective sequence as in Corollary 13. In particular, the set
$A_{\infty}:=\operatorname{FS}_{\geq 2}(a_{1},\allowbreak
a_{2},\allowbreak\dots)$ is a finite ideal in $\langle a_{1},\allowbreak
a_{2},\allowbreak\dots\rangle$ and $A_{\infty}$ may be embedded as a semigroup
in a cyclic semigroup $\langle a\rangle$. For convenience, we denote the
elements of $A_{\infty}$ by appropriate multiples of $a$. As we have seen,
$a_{n}+a_{m}=2\cdot a$ for all $n<m$. Moreover, for each element $k\cdot a\in
A_{\infty}$ we have $a_{n}+k\cdot a=k\cdot a+a_{n}=(k+1)\cdot a$.

We continue the proof addressing three possible cases.

  1. (1)

There is a natural number $n$ such that the set $\\{\,a_{m}+a_{n}:m>n\,\\}$ is
infinite. Otherwise, there must be a subsequence $a_{n_{1}},a_{n_{2}},\dotsc$
such that for each $k$, the sums $a_{n_{j}}+a_{n_{k}}$ are equal for all
$j>k$. Denote $b_{k}:=a_{n_{j}}+a_{n_{k}}$. There are two possible cases:

  2. (2)

The set $\\{\,b_{k}:k\in\mathbb{N}\,\\}$ is infinite.

  3. (3)

The set $\\{\,b_{k}:k\in\mathbb{N}\,\\}$ is finite.

_Case 1_. There is $n$ such that $\\{\,a_{m}+a_{n}:m>n\,\\}$ is infinite. For
$m_{1},m_{2}>n$, denote the sum $s:=a_{n}+a_{m_{2}}\in A_{\infty}$. The set
$A_{\infty}$ is a two-sided ideal so we have:

$(a_{m_{1}}+a_{n})+(a_{m_{2}}+a_{n})=a_{m_{1}}+(a_{n}+a_{m_{2}})+a_{n}=a_{m_{1}}+s+a_{n}\in
A_{\infty}.$

Consider the infinite subsemigroup $S^{\prime}:=\\{\,a_{m}+a_{n}:m>n\,\\}\cup
A_{\infty}\leq S$. We have seen that $S^{\prime}+S^{\prime}\subseteq
A_{\infty}$ is finite as requested.

_Case 2_. Consider the subset $S^{\prime}:=\\{\,b_{k}:k\in\mathbb{N}\,\\}\cup
A_{\infty}\subseteq S$. Consider two elements $b_{k},b_{r}$ for some $k$ and
$r$ (not necessarily distinct). Choose a number $m>k,r$. Denote the sum
$s:=a_{n_{k}}+a_{n_{m}}\in A_{\infty}$. Once again, by the assumption that
$A_{\infty}$ is a two-sided ideal we may receive the following computation:

$b_{k}+b_{r}=(a_{n_{k+1}}+a_{n_{k}})+(a_{n_{m}}+a_{n_{r}})=a_{n_{k+1}}+(a_{n_{k}}+a_{n_{m}})+a_{n_{r}}=a_{n_{k+1}}+s+a_{n_{r}}\in
A_{\infty}.$

It follows that the subset $S^{\prime}\subseteq S$ is a subsemigroup and that
$S^{\prime}+S^{\prime}\subseteq A_{\infty}$ is finite as requested.

_Case 3_. Denote $B:=\\{\,b_{n}:n\in\mathbb{N}\,\\}$. The set $B$ is finite,
and by moving to a subsequence of the original sequence we may assume, using
Ramsey theorem, that for all $n<m$, the sum $a_{m}+a_{n}\in B$ is equal.
Denote $b:=a_{m}+a_{n}$. For each $n\geq 2$, we have the following equations:

(4) $\displaystyle
b+a_{n}=(a_{n}+a_{n-1})+a_{n}=a_{n}+(a_{n-1}+a_{n})=a_{n}+2\cdot a=3\cdot a,$
$\displaystyle a_{n}+b=a_{n}+(a_{n+1}+a_{n})=(a_{n}+a_{n+1})+a_{n}=2\cdot
a+a_{n+1}=3\cdot a.$

It follows that

(5) $\text{for }k\geq 3\text{, if }n_{1},\dotsc,n_{k}\geq 2\text{ are not all
equal, }a_{n_{1}}+\dotsb+a_{n_{k}}=k\cdot a.$

Once again, we continue the proof addressing three possible cases:

  1. (1)

There is a subsequence $c_{1},\allowbreak c_{2},\allowbreak\dots$ of
$a_{2},a_{3},\dotsc$ such that $\\{\,m\cdot c_{n}:m\in\mathbb{N}\,\\}\cap
A_{\infty}\neq\emptyset$ for all $n$.

  2. (2)

There is a subsequence $c_{1},\allowbreak c_{2},\allowbreak\dots$ of
$a_{2},a_{3},\dotsc$ such that $\\{\,m\cdot c_{n}:m\in\mathbb{N}\,\\}$ is
finite and

$\\{\,m\cdot c_{n}:m\in\mathbb{N}\,\\}\cap A_{\infty}=\emptyset$

for all $n$.

  3. (3)

There is a subsequence $c_{1},\allowbreak c_{2},\allowbreak\dots$of
$a_{2},a_{3},\dotsc$ such that $\\{\,m\cdot
c_{n}:m\in\mathbb{N}\,\\}\cong\mathbb{N}$ for each $n$.

_Case 3.1_. For each $n$, let $m_{n}$ be the minimal number such that
$m_{n}\cdot c_{n}\in A_{\infty}$. Denote
$m^{\prime}_{n}:=\lceil\frac{m_{n}}{2}\rceil$ and $s_{n}:=m^{\prime}_{n}\cdot
c_{n}$.

By moving to an appropriate subsequence, we may assume that either $m_{n}\geq
4$ or $m_{n}\leq 3$ for each $n$. Assume that $m_{n}\geq 4$ for all $n$. It
follows that $s_{n}\neq s_{k}$ for all $n<k$. If $s_{n}=s_{k}$,

$(m^{\prime}_{n}+1)\cdot c_{n}=c_{n}+m^{\prime}_{k}\cdot c_{k}\in A_{\infty},$

contradicting the minimality of $m_{n}$. Hence

$S^{\prime}:=\\{\,s_{n}:n\in\mathbb{N}\,\\}\cup A_{\infty}\subseteq S$

is an infinite subset. For each $n$, we have $s_{n}+s_{n}=2\cdot
m^{\prime}_{n}\cdot c_{n}\in A_{\infty}$ by the definition of
$m^{\prime}_{n}$. For each $n\neq k$, we have

$s_{n}+s_{k}=m^{\prime}_{n}\cdot c_{n}+m^{\prime}_{k}\cdot c_{k}\in
A_{\infty}$

by Statement 5. It follows that $S^{\prime}$ is an infinite semigroup and
$S^{\prime}+S^{\prime}\subseteq A_{\infty}$ is finite as requested.

Otherwise, we may assume that $m_{n}\leq 3$ for all $n$. By moving to an
appropriate subsequence using Ramsey’s theorem, we may assume that either
$2\cdot c_{n}=2\cdot c_{k}$ for all $n<k$, or $2\cdot c_{n}\neq 2\cdot c_{k}$
for all $n<k$.

In the first case, it follows from Equation 5 that $2\cdot c_{1}+c_{n}=3\cdot
a\in A_{\infty}$ for $n\neq 1$, and $2\cdot c_{1}+c_{1}=2\cdot
c_{2}+c_{1}=3\cdot a\in A_{\infty}$. Consider the subset

$S^{\prime}:=\\{\,c_{n}:n\in\mathbb{N}\,\\}\cup\\{\,2\cdot c_{1},b\,\\}\cup
A_{\infty}\leq S.$

For each $n<m$, $c_{n}+c_{m}\in A_{\infty}$ and $c_{m}+c_{n}=b$. By Equation
4, we have $b+c_{n}=c_{n}+b\in A_{\infty}$ for all $n$. It is clear now that
$S^{\prime}$ is an infinite subsemigroup with
$S^{\prime}+S^{\prime}\subseteq\\{\,2\cdot c_{1},b\,\\}\cup A_{\infty}$
finite.

In the second case, $S^{\prime}:=\\{\,2\cdot c_{n}:n\in\mathbb{N}\,\\}\cup
A_{\infty}$ is an infinite subset. Since $m_{n}\leq 3$ for each $n$, we have
$2\cdot c_{n}+2\cdot c_{n}=4\cdot c_{n}\in A_{\infty}$. By Equation 5, $2\cdot
c_{n}+2\cdot c_{m}\in A_{\infty}$ for all $n\neq m$. It follows that
$S^{\prime}$ is an infinite subsemigroup, and $S^{\prime}+S^{\prime}\subseteq
A_{\infty}$ is finite.

_Case 3.2_. There is a subsequence $c_{1},\allowbreak c_{2},\allowbreak\dots$
such that $\\{\,m\cdot c_{n}:m\in\mathbb{N}\,\\}$ is finite and

$\\{\,m\cdot c_{n}:m\in\mathbb{N}\,\\}\cap A_{\infty}=\emptyset$

for all $n$. For each $n$, let $e_{n}:=m_{n}\cdot c_{n}$ be an idempotent.
There is an idempotent $e=k\cdot a\in A_{\infty}$, and we can assume that
$k\geq 2$. By Equation 5, we may compute for all $n\neq l$,

$e_{n}+e_{l}=k\cdot e_{n}+k\cdot e_{l}=k\cdot m_{n}\cdot c_{n}+k\cdot
m_{l}\cdot c_{l}=(m_{n}+m_{l})\cdot k\cdot a=(m_{n}+m_{l})\cdot e=e.$

We can define now an embedding of the fan lattice into $S$ given by:

$\phi(n):=\begin{cases}e&n=1,\\\ e_{n}&2\leq n\end{cases}.$

_Case 3.3_. There is a subsequence $c_{1},\allowbreak c_{2},\allowbreak\dots$
such that $\\{\,m\cdot c_{n}:m\in\mathbb{N}\,\\}\cong\mathbb{N}$ for all $n$.
First, notice that for $n<k$ and $m_{1},m_{2}\in\mathbb{N}$, $m_{1}\cdot
c_{n}\neq m_{2}\cdot c_{k}$. Otherwise, we would have a contradiction by

$(m_{1}+1)\cdot c_{n}=c_{n}+m_{1}\cdot c_{n}=c_{n}+m_{2}\cdot c_{k}\in
A_{\infty}$

since $c_{n}+c_{k}\in A_{\infty}$ and $A_{\infty}$ is an ideal.

Once again, let $e=k\cdot a\in A_{\infty}$ be the idempotent of $A_{\infty}$
(for some $k\geq 2$). By Equation 5, for all $n_{1}\neq n_{2}$ and all $l_{1}$
and $l_{2}$, we have

$l_{1}\cdot k\cdot c_{n_{1}}+l_{2}\cdot k\cdot c_{n_{2}}=(l_{1}+l_{2})\cdot
k\cdot a=(l_{1}+l_{2})\cdot e=e.$

In a similar way,

$l_{1}\cdot k\cdot c_{n_{1}}+e=e+l_{1}\cdot k\cdot c_{n_{1}}=(l_{1}+1)\cdot
k\cdot a=(l_{1}+1)\cdot e=e$

We can define now an embedding of the semigroup of type $(c)$ into $S$:

$\phi(s):=\begin{cases}m\cdot k\cdot c_{n}&s=(n,m)\\\ e&s=0\end{cases}.\qed$

##  4\. A solution of a problem of Andrews and Goldbring

An infinite semigroup $S$ is _moving_ if for each infinite $A\subseteq S$ and
each finite $F\subseteq S$, there are elements $a_{1},\dotsc,a_{k}\in A$ such
that $\\{a_{1}+s,\dotsc,a_{k}+s\\}\not\subseteq F$ for all but finitely many
$s\in S$. Golan and Tsaban [4] proved that in every moving semigroup, the
proper IP sets are partition regular. Andrews and Goldbring [1] asked whether
partition regularity of proper IP sets characterizes moving semigroups. Our
main theorem implies the following answer.

######  Corollary 16.

There is a non-moving semigroup $S$ where the proper IP sets are partition
regular.

###### Proof.

Steinberg [9] proved that the following semigroup has finite-to-one right
addition, but is not moving:

$S:=\langle\,t,x_{0},x_{1},\dotsc:x_{0}+t=x_{0},x_{i}+t=x_{i-1},i>0\,\rangle.$

Since this semigroup has finite-to-one right addition, it does not include a
subsemigroup of the types in Theorem 15. It follows that the proper IP sets in
the semigroup $S$ are partition regular. ∎

##  5\. Hindman’s Theorem with infinitely many proper IP sets

We use our results to prove that every semigroup satisfying Hindman’s Finite
Sums Theorem with the monochromatic set being proper IP, has a much stronger
property.

######  Lemma 17.

Every sequence in a semigroup has a sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ of one of the following types:

  1. (1)

For all natural numbers $n<m$, $b_{n}+b_{m}=b_{n}$.

  2. (2)

For all natural numbers $n$, we have

$\operatorname{FS}(b_{1},\dotsc,b_{n})\cap(\operatorname{FS}(b_{1},\dotsc,b_{n})+b_{n+1})=\emptyset.$

###### Proof.

Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a sequence in a semigroup
$S$. Assume that $a_{1},\allowbreak a_{2},\allowbreak\dots$ does not have a
sumsequence of the second type. We construct by induction a sumsequence
$c_{1},\allowbreak c_{2},\allowbreak\dots$ together with a sumsequence
$b^{(n)}_{1},b^{(n)}_{2},\dotsc$ for each $n$, such that the following
conditions hold for each $n$:

  1. (1)

$c_{n}\in\operatorname{FS}(b^{(n)}_{1},b^{(n)}_{2},\dotsc)$.

  2. (2)

For each $s\in\operatorname{FS}(b^{(n+1)}_{1},b^{(n+1)}_{2},\dotsc)$, we have
$c_{n}+s=c_{n}$.

Define $b^{(1)}_{1},b^{(1)}_{2},\dotsc:=a_{1},\allowbreak
a_{2},\allowbreak\dots$. Assume we have defined $c_{1},\dotsc,c_{n-1}$ and a
sumsequence $b^{(i)}_{1},b^{(i)}_{2},\dotsc$ for all $1\leq i\leq n$. By
moving to a tail of the sumsequence, we may assume there is a number $k$ such
that $c_{1},\dotsc,c_{n-1}\in\operatorname{FS}(a_{1},\dotsc,a_{k})$ and
$b^{(n)}_{1},b^{(n)}_{2},\dotsc\in\operatorname{FS}(a_{k+1},a_{k+2},\dotsc)$.

Since the sumsequence $b^{(n)}_{1},b^{(n)}_{2},\dotsc$ does not have a
sumsequence of the second type, there is a maximal _finite_ sumsequence
$s_{1},\dotsc,s_{k}$ of the second type. Let $F_{1}<\dotsb<F_{k}$ be the
corresponding index sets. For each finite index set $F_{k}<H$, we have

$\operatorname{FS}(s_{1},\dotsc,s_{k})\cap(\operatorname{FS}(s_{1},\dotsc,s_{k})+b^{(n)}_{H})\neq\emptyset.$

Thus, there are elements $x,y\in\operatorname{FS}(s_{1},\dotsc,s_{k})$ such
that $x+b^{(n)}_{H}=y$. Let $m:=\max F_{k}$. Define a finite coloring of
$\operatorname{FS}(b^{(n)}_{m+1},b^{(n)}_{m+2},\dotsc)$ by

$\chi\colon\operatorname{FS}(b^{(n)}_{m+1},b^{(n)}_{m+2},\dotsc)\to(\operatorname{FS}(s_{1},\dotsc,s_{k}))^{2}$

such that $\chi(b^{(n)}_{H})=(x,y)$ implies $x+b^{(n)}_{H}=y$. By Theorem 14,
there is a sumsequence $t_{1},\allowbreak t_{2},\allowbreak\dots$ of
$b^{(n)}_{m+1},\allowbreak b^{(n)}_{m+2},\dotsc$ such that the set
$\operatorname{FS}(t_{1},\allowbreak t_{2},\allowbreak\dots)$ is
monochromatic. Let $x,y\in\operatorname{FS}(s_{1},\dotsc,s_{k})$ be such that
$x+t_{H}=y$ for each finite sum $t_{H}\in\operatorname{FS}(t_{1},\allowbreak
t_{2},\allowbreak\dots)$. For each
$t_{H}\in\operatorname{FS}(t_{2},t_{3},\dotsc)$, we have

$y+t_{H}=x+t_{1}+t_{H}=x+t_{\\{1\\}\cup H}=y.$

Define $c_{n}:=y$ and
$b^{(n+1)}_{1},b^{(n+1)}_{2},\dotsc:=t_{2},t_{3},\dotsc$.

We thus have a sumsequence $c_{1},\allowbreak c_{2},\allowbreak\dots$ of
$a_{1},\allowbreak a_{2},\allowbreak\dots$ such that for all $n<m$, we have
$c_{m}\in\operatorname{FS}(b^{(n+1)}_{1},\allowbreak b^{(n+1)}_{2},\dotsc)$
and therefore $c_{n}+c_{m}=c_{n}$. This is a sumsequence of the first type. ∎

While Lemma 17 holds for arbitrary sequences, it is mainly interesting for
_proper_ sequences. Indeed, if a sequence has no proper sumsequence, then
there is an idempotent element such that the constant sequence $e,e,e,\dotsc$
is a sumsequence of the given sequence [10, Proposition 1.7]. This is a
sequence of the first type in the Lemma, but we are interested in bijective
sequences. Sumsequences of proper sequences are proper and, in particular,
bijective.

######  Proposition 18.

Every proper sequence in a semigroup has a sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ such that $b_{F}\neq b_{G}$ for all disjoint finite
index sets $F$ and $G$.

###### Proof.

Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a proper sequence. There is
a (necessarily, bijective) sumsequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ of $a_{1},\allowbreak a_{2},\allowbreak\dots$ of one
of the types specified in Lemma 17.

If $b_{1},\allowbreak b_{2},\allowbreak\dots$ is a sequence of the first type,
then

$b_{F}=b_{\min F}\neq b_{\min G}=b_{G}.$

Assume that $b_{1},\allowbreak b_{2},\allowbreak\dots$ is a sequence of the
second type. We may assume that $\max F<\max G$. If $G$ is a singleton, then
$F<G$ and $b_{F}\neq b_{G}$, by the properness of the sequence. And if not,
then $b_{F}\in\operatorname{FS}(b_{1},\dotsc,b_{\max G-1})$ and
$b_{G}\in\operatorname{FS}(b_{1},\dotsc,b_{\max G-1})+b_{\max G}$. Since

$\operatorname{FS}(b_{1},\dotsc,b_{\operatorname{max}G-1})\cap(\operatorname{FS}(b_{1},\dotsc,b_{\operatorname{max}G-1})+b_{\operatorname{max}G})=\emptyset,$

we have $b_{F}\neq b_{G}$. ∎

######  Corollary 19.

Let $a_{1},\allowbreak a_{2},\allowbreak\dots$ be a proper sequence in a
semigroup. The IP set $A=\operatorname{FS}(a_{1},\allowbreak
a_{2},\allowbreak\dots)$ can be partitioned into infinitely many proper IP
sets.

###### Proof.

Let $b_{1},\allowbreak b_{2},\allowbreak\dots$ be a sumsequence of
$a_{1},\allowbreak a_{2},\allowbreak\dots$ as in Proposition 18. Let
$\mathbb{N}=\bigcup_{n=1}^{\infty}I_{n}$ be a partition of the natural numbers
into infinitely many infinite sets. For each $n$, enumerate
$I_{n}=\\{\,i^{(n)}_{1},i^{(n)}_{2},\dots\,\\}$ in increasing order, and let
$A_{n}:=\operatorname{FS}(b_{i^{(n)}_{1}},\allowbreak
b_{i^{(n)}_{2}},\dotsc)$. We obtain infinitely many disjoint proper IP subsets
of the set $A$. (The elements that do not belong to any set $A_{n}$ may be
added to $A_{1}$, for example.) ∎

######  Theorem 20.

Let $S$ be a semigroup where every finite coloring has a monochromatic proper
IP set. For each finite coloring of the semigroup $S$, there are infinitely
many, pairwise disjoint, monochromatic proper IP sets.

###### Proof.

If the semigroup has a proper sequence, then the assertion follows from
Theorem 14 and Corollary 19.

Thus, assume there is no proper sequence in the semigroup. Let $c$ be a finite
coloring of $S$. By induction, we construct a sequence $A_{1},\allowbreak
A_{2},\allowbreak\dots$ of monochromatic, pairwise disjoint, proper IP sets.
We construct each set $A_{i}$ together with a sequence
$a^{(i)}_{1},a^{(i)}_{2},\dotsc$ such that
$A_{i}=\operatorname{FS}(a^{(i)}_{1},a^{(i)}_{2},\dotsc)$ and the set
$B_{i}:=\operatorname{FS}_{\geq 2}(a^{(i)}_{1},a^{(i)}_{2},\dotsc)$ is finite.

Assume we have defined the sets $A_{1},\dotsc,A_{m}$, with the attached
sequences $a^{(i)}_{1},a^{(i)}_{2},\dotsc$ and sets $B_{i}$ for $1\leq i\leq
m$. Let $A:=\bigcup_{i=1}^{m}A_{i}$ and $B:=\bigcup_{i=1}^{m}B_{i}$. Define a
new finite coloring $c^{\prime}$ of the semigroup $S$ by fixing two new colors
$\alpha$ and $\beta$, and setting

$c^{\prime}(s):=\begin{cases}c(s)&s\in S\setminus(\bigcup_{i=1}^{m}A_{i}),\\\
\alpha&s\in A\setminus B\\\ \beta&s\in B\end{cases}.$

Let $\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)$, for a
bijective sequence $b_{1},\allowbreak b_{2},\allowbreak\dots$, be
monochromatic for the new coloring.

Since the set $B$ is finite, the color is not $\beta$.

Assume that the color is $\alpha$. By moving to a subsequence, we may assume
that there is $i$ with $b_{1},\allowbreak b_{2},\allowbreak\dots\in
A_{i}\setminus B$. Since the sequence $b_{1},\allowbreak
b_{2},\allowbreak\dots$ is bijective in the set $\\{\,a_{1},\allowbreak
a_{2},\allowbreak\dots\,\\}$, there are indices $j_{1}<k_{1}$ and
$j_{2}<k_{2}$ with $b_{j_{1}}=a_{j_{2}}$ and $b_{k_{1}}=a_{k_{2}}$. It follows
that

$b_{j_{1}}+b_{k_{1}}=a_{j_{2}}+a_{k_{2}}\in B,$

and thus the color of the sum $b_{j_{1}}+b_{k_{1}}$ is $\beta$; a
contradiction.

Thus, $\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)\subseteq
S\setminus(\bigcup_{i=1}^{m}A_{i})$. Since the colorings $c$ and $c^{\prime}$
agree on the set $S\setminus(\bigcup_{i=1}^{m}A_{i})$, and the set
$\operatorname{FS}(b_{1},\allowbreak b_{2},\allowbreak\dots)$ is
$c^{\prime}$-monochromatic, it is also $c$-monochromatic.

Since there are no proper sequences in the semigroup, by Proposition 12, there
is a bijective sumsequence $s_{1},\allowbreak s_{2},\allowbreak\dots$ of
$b_{1},\allowbreak b_{2},\allowbreak\dots$ such that the set
$\operatorname{FS}_{\geq 2}(s_{1},\allowbreak s_{2},\allowbreak\dots)$ is
finite. Define $a^{(m+1)}_{1},a^{(m+1)}_{2},\dotsc:=s_{1},\allowbreak
s_{2},\allowbreak\dots$, $A_{m+1}:=\operatorname{FS}(s_{1},\allowbreak
s_{2},\allowbreak\dots)$, and $B_{m+1}:=\operatorname{FS}_{\geq
2}(s_{1},\allowbreak s_{2},\allowbreak\dots)$. Since the sequence
$s_{1},\allowbreak s_{2},\allowbreak\dots$ is a sumsequence of
$b_{1},\allowbreak b_{2},\allowbreak\dots$, the set $A_{m+1}$ is a subset of
the $c$-monochromatic set $\operatorname{FS}(b_{1},\allowbreak
b_{2},\allowbreak\dots)$, which in turn is disjoint from the sets
$A_{1},\dotsc,A_{m}$, as required. ∎

We point out that the second part of the last proof cannot be omitted: There
are semigroups with no proper sequences, where every finite coloring has a
monochromatic proper IP set.

### Acknowledgments

This paper is based on a part of Yonatan Gadot’s M.Sc. thesis, supervised by
Boaz Tsaban. Tsaban wishes to stress that the major contributions in this
paper are due to Gadot.

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