Abundance of Progression in commutative semigroup by elementary means

TL;DR

This paper provides an elementary proof that piecewise syndetic sets in commutative semigroups contain arbitrarily long arithmetic progressions, extending classical results with a more accessible combinatorial approach.

Contribution

It offers an elementary combinatorial proof for the existence of arithmetic progressions in piecewise syndetic sets within commutative semigroups, answering Beiglbock's question.

Findings

Elementary proof of arithmetic progressions in commutative semigroups

Extension of classical results to more abstract algebraic structures

Simplification of proof techniques for combinatorial properties

Abstract

Furstenberg and Glasner proved that for an arbitrary k in N, any piecewise syndetic set contains k term arithmetic progression and such collection is also piecewise syndetic in Z. They used algebraic structure of beta N. The above result was extended for arbitrary semigroups by Bergelson and Hindman, again using the structure of Stone-Cech compactification of general semigroup. Beiglbock provided an elementary proof of the above result and asked whether the combinatorial argument in his proof can be enhanced in a way which makes it applicable to a more abstract setting. In this work we provide a positive answer to Beiglbock question for commutative semigroup.

Equations

$$K+n+\varphi(m)\subseteq\bigcup_{j\in\varphi(F)+E}\big{(}-j+\varphi(A)\big{)}$$

$$\textstyle\bigcup_{(a,n)\in E}(-(a,n)+A)\subseteq\bigcup_{(a,n)\in E}\big{(}-(a+y,n)+\psi_{y}(A)\big{)}\,,$$

$$a+n\cdot (b+u\cdot F)\subseteq a+n\cdot A\cap (-t_i+M).$$

Full text

In this work we extend that technique of Beiglböck in commutative semigroups.

1 Introduction

A subset $S$ of $\mathbb{Z}$ is called syndetic if there exists $r\in\mathbb{N}$ such that $\bigcup_{i=1}^{r}(S-i)=\mathbb{Z}$. Again a subset $S$ of $\mathbb{Z}$ is called thick if it contains arbitrarily long intervals. Sets which can be expressed as the intersection of thick and syndetic sets are called piecewise syndetic.

For a general commutative semigroup $(S,+)$, a set $A\subseteq S$ is said to be syndetic in $(S,+)$, if there exist a finite nonempty set $F\subseteq S$ such that $\bigcup_{t\in F}(-t+A)=S$ where $-t+A=\{s\in S:t+s\in A\}$. A set $A\subseteq S$ is said to be thick if for every finite nonempty set $E\subseteq S$, there exists an element $x\in S$ such that $E+x\subseteq A$. A set $A\subseteq S$ is said to be piecewise syndetic set if there exist a finite nonempty set $F\subseteq S$ such that $\bigcup_{t\in F}(-t+A)$ is thick in $S$. It can be proved that a piecewise syndetic set is the intersection of a thick set and a syndetic set [HS, Theorem 4.49].

One of the famous Ramsey theoretic results is van der Waerden’s Theorem [V] which states that, given $r,k\in\mathbb{N}$ there is some $l\in\mathbb{N}$ such that one cell of any partition $\{C_{1},C_{2},\ldots,C_{r}\}$ of $\{1,2,\ldots,l\}$ contains an arithmetic progressions of length $k$. It follows from van der Waerden’s Theorem that any piecewise syndetic subset $A$ of $\mathbb{N}$ contains arbitrarily long arithmetic progressions. To see this, pick finite $F\subseteq\mathbb{N}$ such that $\bigcup_{t\in F}(-t+A)$ is thick in $\mathbb{N}$. Let $r=|F|$ and let a length $k$ be given. Pick $l$ as guaranteed for $r$ and $k$ and pick $x$ such that $\{1,2,\ldots,l\}+x\subseteq\bigcup_{t\in F}(-t+A)$. For $t\in F$, let $C_{t}=\{y\in\{1,2,\ldots,l\}:y+x\in(-t+A)\}$. Pick $a$ and $d$ in $\mathbb{N}$ and $t\in F$ such that $\{a,a+d,a+2d,\ldots,a+(k-1)d\}\subseteq C_{t}$ and let $a^{\prime}=a+x+t$. Then $\{a^{\prime},a^{\prime}+d,a^{\prime}+2d,\ldots,a^{\prime}+(k-1)d\}\subseteq A$.

A homothetic copy of a finite set $F$ in a commutative semigroup $(S,+)$, is of the form $a+n\cdot F=\{a+n\cdot x:x\in F\}$, where $a\in S$, $n\in\mathbb{N}$ and $n\cdot x$ is the sum of $x$ with itself $n$ times. Given a set $A\subseteq S$, we denote by $\hbox{\rm Homth}_{A}(F)$ the set of homothetic copies of $F$ that are contained in $A$.

Furstenberg and E. Glasner [FG] proved that if $S$ is a piecewise syndetic subset of $\mathbb{Z}$ and $l\in\mathbb{N}$ then the set of all length $l$ progressions contained in $S$ is also large.

Theorem 1**.**

Let $k\in\mathbb{N}$ and assume that $S\subseteq\mathbb{Z}$ is piecewise syndetic. Then $\{(a,d)\,:\,a,a+d,\ldots,a+kd\in S\}$ is piecewise syndetic in $\mathbb{Z}^{2}$.

Furstenberg and Glasner’s proof was algebraic in nature. Beiglböck provided an elementary proof of Theorem 1 in [B]. In [BH, Theorem 3.7], Theorem 1 was extended to more general semigroups and for various other notions of largeness using ultrafilter techniques.

Theorem 2**.**

Let $(S,\cdot)$ be a semigroup, let $l\in\mathbb{N}$, let $E$ be a subsemigroup of $S^{l}$ with $\left\{(a,a,\ldots,a):a\in S\right\}\subseteq E$, and let $I$ be an ideal of $E$. If $M$ is a piecewise syndetic subset of $S$, then $M^{l}\cap I$ is piecewise syndetic in $I$.

Proof.

[BH, Theorem 3.7(a)]. ∎

In this article we will extend that technique of Beiglböck to commutative semigroups and provide an elementary proof of a special case of Theorem 2 which is Corollary 10.

Conventionally $[t]$ denotes the set $\{1,2,\ldots,t\}$ and words of length $N$ over the alphabet $[t]$ are the elements of $[t]^{N}$. A variable word is a word over $[t]\cup\{*\}$ in which $*$ occurs at least once and $*$ denotes the variable. A combinatorial line is denoted by $L_{\tau}=\{\tau(1),\tau(2),\ldots,\tau(t)\}$ where $\tau(*)$ is a variable word and for $i\in\{1,2,\ldots,t\}$, $\tau(i)$ is obtained by replacing each occurrence of $*$ by $i$.

The following theorem is due to Hales and Jewett.

Theorem 3**.**

For all values $t,r\in\mathbb{N}$, there exists a number $HJ(r,t)$ such that, if $N\geq HJ(r,t)$ and $[t]^{N}$ is $r$ colored then there will exists a monochromatic combinatorial line.

Proof.

[HJ, Theorem 1]. ∎

The above theorem may be strengthening by requiring that the variable word include at least one constant.

Lemma 4**.**

Let $r,m\in\mathbb{N}$. There exists some $N\in\mathbb{N}$ such that whenever the set of length $N$ words over $[m]$ is r-colored, there is a variable word $w(*)$ such that $w(*)$ begins and ends with a constant, and $\{w(a):a\in[m]\}$ is monochromatic.

Proof.

This is an immediate consequence of [HS, Lemma 14.8.1]. ∎

The following lemma will be used in our proof of the main theorem.

Lemma 5**.**

Let $(S,+)$ be a commutative semigroup, let $F$ be a finite nonempty subset of $S$, and let $r\in\mathbb{N}$. There exists a finite subset $A$ of $S$ such that for any $r$-coloring of $A$, there exist a monochromatic homothetic copy of $F$.

Proof.

Let $m=|F|$ and enumerate $F$ as $\{s_{1},s_{2},\ldots,s_{m}\}$. Let $N$ be as guaranteed by Lemma 4 for $r$ and $m$. Let $A=\{\sum_{j=1}^{N}a_{j}:$ each $a_{j}\in F\}$ and let $\chi:A\to\{1,2,\ldots,r\}$ be an $r$-coloring of $A$. Define an $r$-coloring $\chi^{\prime}$ of $[m]^{N}$ by \chi^{\prime}(x)=\chi\big{(}\sum_{j=1}^{N}x(j)\big{)} for x=\big{(}x(1),x(2),\ldots,x(N)\big{)}\in[m]^{N}. Pick a variable word $w(*)=w_{1}w_{2}\cdots w_{N}$ which begins with a constant and pick $l\in\{1,2,\ldots,r\}$ such that for each $t\in\{1,2,\ldots,m\}$, \chi^{\prime}\big{(}w(t)\big{)}=l.

Let $I=\{j\in\{1,2,\ldots,N\}:w_{j}=*\}$. Let $a=\sum_{j\in[N]\setminus I}s_{w_{j}}$ and let $n=|I|$. Since $w(*)$ begins with a constant, $[N]\setminus I\neq\emptyset$ so $a\in S$. Then for $t\in\{1,2,\ldots,m\}$, l=\chi^{\prime}\big{(}w(t)\big{)}=\chi(\sum_{i\in[N]\setminus I}s_{w_{i}}+\sum_{i\in I}s_{t})=\chi(a+n\cdot s_{t}). ∎

For a commutative semigroup $(S,+)$, consider $(S\times\mathbb{N},+)$ with operation defined by $(s_{1},n_{1})+(s_{2},n_{2})=(s_{1}+s_{2},n_{1}+n_{2})$. Then $(S\times\mathbb{N},+)$ is a commutative semigroup.

Using essentially the same proof that established that piecewise syndetic subsets of $\mathbb{N}$ contain arbitrarily long arithmetic progressions, one can derive the following theorem from Lemma 5. (We will not be using this theorem, which is also a consequence of Theorem 7.)

Theorem 6**.**

Let $(S,+)$ be a commutative semigroup and let $F$ be a finite nonempty subset of $S$. Then for any piecewise syndetic set $M\subseteq S$ the collection $\{(a,n)\in S\times\mathbb{N}:\,a+n\cdot F\subseteq M\}$ is non empty.

The following is the main result of this paper.

Theorem 7**.**

Let $(S,+)$ be a commutative semigroup, let $F$ be a finite nonempty subset of $S$, and let $M$ be a piecewise syndetic subset of $S$. Then the collection $\{(a,n)\in S\times\mathbb{N}:\,a+n\cdot F\subseteq M\}$ is piecewise syndetic in $(S\times\mathbb{N},+)$.

2 Proof of Theorem 7

In this section we will prove Theorem 7, enhancing the combinatorial arguments used in [B]. The following lemma will be needed for our purposes. This lemma is [BG, Lemma 4.6($I^{\prime}$)], and was proved by using the algebraic structure of the Stone-Čech compactification of an arbitrary semigroup. Since we have promised a combinatorial proof, we will give a purely elementary proof of the above lemma for commutative semigroups. We thank the anonymous referee for simplifying the proof.

Lemma 8**.**

Let $(S,\cdot)$ and $(T,.)$ be semigroups, let $\varphi:S\to T$ be a homomorphism, and let $A\subseteq$S. If $A$ is piecewise syndetic in $S$ and $\varphi(S)$ is piecewise syndetic in $T$, then $\varphi(A)$ is piecewise syndetic in $T$.

Proof.

Assume that $A$ is piecewise syndetic in $S$ and $\varphi(S)$ is piecewise syndetic in $T$. Pick finite $F\subseteq S$ such that $\bigcup_{s\in F}(-s+A)$ is thick in $S$ and pick finite $E\subseteq T$ such that \bigcup_{t\in E}\big{(}-t+\varphi(S)\big{)} is thick in $T$. To show that $\varphi(A)$ is piecewise syndetic in $T$, it suffices to show that \bigcup_{j\in\varphi(F)+E}\big{(}-j+\varphi(A)\big{)} is thick in $T$.

To see this, let $K$ be a finite nonempty subset of $T$. Pick $n\in T$ such that K+n\subseteq\bigcup_{t\in E}\big{(}-t+\varphi(S)\big{)}. For each $k\in K$, pick $t_{k}\in E$ such that $t_{k}+k+n\in\varphi(S)$ and pick $b_{k}\in S$ such that $\varphi(b_{k})=t_{k}+k+n$. Let $B=\{b_{k}:k\in K\}$ and pick $m\in S$ such that $B+m\subseteq\bigcup_{s\in F}(-s+A)$. We claim that

[TABLE]

so let $k\in K$ be given. Pick $s\in F$ such that $s+b_{k}+m\in A$. Then $\varphi(s)+t_{k}\in\varphi(F)+E$ and $(\varphi(s)+t_{k})+k+n+\varphi(m)=\varphi(s+b_{k}+m)\in\varphi(A)$. ∎

Lemma 9**.**

If $M\subseteq S\times\mathbb{N}$ is piecewise syndetic in $S\times\mathbb{N}$, then for any $x,y\in S$ and $t\in\mathbb{N}$, $\{(a+nx+y,n\cdot t):(a,n)\in M\}$ is piecewise syndetic in $S\times\mathbb{N}$.

Proof.

For a fixed $x\in S$, let $\varphi_{x}:S\times\mathbb{N}\rightarrow S\times\mathbb{N}$ be a homomorphism defined by $\varphi_{x}(a,n)=(a+nx,n)$. We claim that $\varphi_{x}(S\times\mathbb{N})$ is thick in $S\times\mathbb{N}$ and so piecewise syndetic and hence this transformation preserves piecewise syndeticity.

To prove the claim, let us take a finite subset $F=\{(s_{1},n_{1}),(s_{2},n_{2}),\ldots,\break(s_{m},n_{m})\}$ of $S\times\mathbb{N}$ and observe that F+\big{(}(n_{1}+n_{2}+\ldots+n_{m}+1)x,1\big{)}\subseteq\varphi_{x}(S\times\mathbb{N}).

Now for a fixed $t\in\mathbb{N},$ let $\chi_{t}:S\times\mathbb{N}\rightarrow S\times\mathbb{N}$ be a homomorphism defined by $\chi_{t}(a,n)=(a,n\cdot t)$. We claim that, $\chi_{t}(S\times\mathbb{N})$ is syndetic in $S\times\mathbb{N}$, in particular piecewise syndetic. Therefore this transformation also preserves piecewise syndeticity.

To prove the claim, pick $x\in S$. We show that $S\times\mathbb{N}\subseteq\bigcup_{j=1}^{t}(-(x,j)+\chi_{t}(S\times\mathbb{N}))$. So let $(a,n)\in S\times\mathbb{N}$ be given. Pick $j\in\{1,2,\ldots,t\}$ such that $t$ divides $n+j$ and let $m=\frac{n+t}{m}$. Then $(x,j)+(a,n)=\chi_{t}(x+a,m)$.

Again for any fixed $y\in S$, we claim that $\psi_{y}:S\times\mathbb{N}\to S\times\mathbb{N}$ defined by $\psi_{y}(a,n)=(a+y,n)$ preserves piecewise syndeticity. To see this, let $A\subseteq S\times\mathbb{N}$ be piecewise syndetic in $S\times\mathbb{N}$. Pick a finite nonempty subset $E$ of $S\times\mathbb{N}$ such that $\bigcup_{(a,n)\in E}(-(a,n)+A)$ is thick. Since

[TABLE]

we have that \bigcup_{(a,n)\in E}\big{(}-(a+y,n)+\psi_{y}(A)\big{)} is thick, so $\psi_{y}(A)$ is piecewise syndetic.

As $(a+nx+y,n.t)=\psi_{y}\circ\chi_{t}\circ\varphi_{x}(a,n)$ and all the three maps preserves piecewise syndeticity, we have the desired result. ∎

Now we will prove the main theorem.

Proof of Theorem 7.

As $M$ is piecewise syndetic, pick a finite nonempty set $E\subseteq S$ such that $\bigcup_{t\in E}(-t+M)$ is thick and let $r=|E|$. Pick by Lemma 5 a finite subset $A$ of $S$ such that, whenever $A$ is $r$-colored, there exists a monochromatic homothetic copy of $F$. Then, given $a\in S$ and $n\in\mathbb{N}$, if $a+n\cdot A$ is $r$-colored, there exist $b\in S$ and $t\in\mathbb{N}$ such that $a+n\cdot(b+t\cdot F)$ is monochromatic.

Let $B=\{(a,n)\in S\times\mathbb{N}:a+n\cdot A\subseteq\bigcup_{t\in E}(-t+M)\}$. We claim that $B$ is thick in $S\times\mathbb{N}$. To see this, let $C$ be a finite nonempty subset of $S\times\mathbb{N}$. Let $G=\bigcup_{(a,n)\in C}(a+(n+1)\cdot A)$. By the thickness of $\bigcup_{t\in E}(-t+M)$, pick $z\in S$ such that $G+z\subseteq\bigcup_{t\in E}(-t+M)$. Then, given $(a,n)\in C$, $a+(n+1)\cdot A\subseteq G$ so $a+z+(n+1)\cdot A\subseteq\bigcup_{t\in E}(-t+M)$ and thus $(a,n)+(z,1)\in B$.

Let $s=|\hbox{\rm Homth}_{A}(F)|$, enumerate $\hbox{\rm Homth}_{A}(F)$ as $\{M_{1},M_{2},\ldots,M_{s}\}$, and enumerate $E$ as $\{t_{1},t_{2},\ldots,t_{r}\}$. Note that, for any $(a,n)\in B$, there are some $i\in\{1,2,\ldots,r\}$ and some $(b,u)\in S\times\mathbb{N}$ such that

[TABLE]

Define $\varphi:B\to E\times\hbox{\rm Homth}_{A}(F)$ as follows. First pick the least $i\in\{1,2,\ldots,r\}$ such that there exist $b\in S$ and $u\in\mathbb{N}$ such that $a+n\cdot(b+u\cdot F)\subseteq(a+n\cdot A)\cap(-t_{i}+M)$ and note that $b+u\cdot F\in\hbox{\rm Homth}_{A}(F)$. Now pick the least $j\in\{1,2,\ldots,s\}$ such that $a+n\cdot M_{j}\subseteq(a+n\cdot A)\cap(-t_{i}+M)$. Then define $\varphi(a,n)=(t_{i},M_{j})$.

Now as $E\times\hbox{\rm Homth}_{A}(F)$ is finite, the mapping $\varphi$ gives a finite coloring of $B$ so pick $(i,j)\in\{1,2,\ldots,r\}\times\{1,2,\ldots,s\}$ such that $Q=\{(a,n)\in B:\varphi(a,n)=(t_{i},M_{j})\}$ is piecewise syndetic. (We are using here the elementary fact that if the union of finitely many sets is piecewise syndetic, then one of them is.)

Choose some $b\in S$ and $u\in\mathbb{N}$ such that $b+u\cdot F=M_{j}$. then $(a,n)\in Q$ implies $a+n\cdot(b+u\cdot F)\subseteq(-t_{i}+M)$ and hence $a+t_{i}+n\cdot(b+u\cdot F)\subseteq M$. Now by Lemma 9, $\widetilde{Q}=\{(a+n\cdot b+t_{i},n\cdot u)\,:\,(a,n)\in Q\}$ is piecewise syndetic. For $(a_{1},n_{1})\in\widetilde{Q}$ , $a_{1}+n_{1}\cdot F\subseteq M$ and this proves the theorem. ∎

The following corollary gives an elementary proof of a special case of Theorem 2.

Here we denote the set of all homothetic copies of $F=\{s_{1},s_{2},\ldots,s_{l}\}$ by $HC_{F}$, i.e. the set $\{(a+n\cdot s_{1},a+n\cdot s_{2},\ldots,a+n\cdot s_{l})\,:\,a\in S$ and $n\in\mathbb{N}\}$. It is easy to check that $HC_{F}$ is a subsemigroup of $S^{l}$.

Corollary 10**.**

Let $(S,+)$ be a commutative semigroup and $F\subseteq S$ be a given finite nonempty set of cardinality $l$. Then for any piecewise syndetic subset $M\subseteq S$, $M^{l}\cap HC_{F}$ is piecewise syndetic in $HC_{F}$.

Proof.

Let us define a surjective homomorphism $\varphi:S\times\mathbb{N}\to HC_{F}$ by, $\varphi(a,n)=(a+n\cdot s_{1},a+n\cdot s_{2},\ldots,a+n\cdot s_{l})$.

Now let $C=\{(a,n)\in S\times\mathbb{N}:a+n\cdot F\subseteq M\}$. Then $C$ is piecewise syndetic in $S\times\mathbb{N}$ by Theorem 7. Now $\varphi(C)\subseteq M^{l}\cap HC_{F}$ and as $C$ is piecewise syndetic in $S\times\mathbb{N}$, $\varphi(C)$ is piecewise syndetic in $HC_{F}$ by Lemma 8 and so $M^{l}\cap HC_{F}$ is piecewise syndetic in $HC_{F}$. This proves the claim. ∎

Acknowledgements: The first author acknowledges the grant UGC-NET SRF fellowship with id no. 421333 of CSIR-UGC NET December 2016. The second author acknowledges the grant CSIR-SRF fellowship with file no. 09/106(0149)/2014-EMR-1. We also acknowledge the anonymous referee for several helpful comments on the paper.