Richness of arithmetic progression in commutative semigroup

TL;DR

This paper extends the understanding of arithmetic progressions in piecewise syndetic sets within commutative semigroups, building on algebraic and combinatorial methods to generalize previous results.

Contribution

It generalizes Beiglboeck's elementary proof to broader settings of commutative semigroups, enhancing the applicability of combinatorial techniques.

Findings

Extended the elementary proof to new types of commutative semigroups

Demonstrated the presence of arithmetic progressions in broader algebraic structures

Provided a unified combinatorial framework for these results

Abstract

Furstenberg and Glasner proved that for an arbitrary k in N, any piecewise syndetic set contains k term arithmetic progressions 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. Beiglboeck 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 a recent work the second author of this paper and S.Jana provided an affirmative answer to Beiglboeck's question for countable commutative semigroup. In this work we will extend the result of Beiglboeck in different type of settings.

Equations

$$\{(s,t)\in S\times S:\{s,s+t,s+2t,\dots ,s+dt\}\subseteq A\}$$

$$\{(s+at+c,t):(s,t)\in M\}$$

$$\{(s,t)\in S\times S:\{s,s+dt,\dots ,s+ldt\}\subseteq A\}$$

$$Q=\{(s,t)\in B:\{s+at,s+at+dt,\dots ,s+at+ldt\}\subseteq -c_j+A\text{ for some }j\in [1,r]\}$$

$$\{(s,dt):(s,t)\in M\}$$

$$\{(s,t)\in S\times S:\{s,s+t,s+2t,\dots ,s+dt\}\subseteq A\}$$

$$A\supseteq A_1\supseteq A_2\supseteq \dots \supseteq A_n\supseteq \dots$$

$$\phi (A)\supseteq \phi (A_1)\supseteq \phi (A_2)\supseteq \dots \supseteq \phi (A_n)\supseteq \dots$$

$$1.M\supseteq A_1\supseteq A_2\supseteq \dots \supseteq A_n\supseteq \dots$$

$$2.B\supseteq B_1\supseteq B_2\supseteq \dots \supseteq B_n\supseteq \dots$$

$$A_N\subseteq \stackrel{[}{i}=0]l\bigcap (-(a+ib)+A_n).$$

$$\{a_1,a_1+b_1,a_1+2b_1,\dots ,a_1+l{b}_1\}\subseteq A_N\subseteq \stackrel{[}{i}=0]l\bigcap (-(a+ib)+A_n)$$

Full text

The second author of the paper is supported by UGC-JRF fellowship.

1. Introduction

A subset $S$ of $\mathbb{Z}$ is called syndetic if there exists $r\in\mathbb{N}$ such that $\bigcup_{i=1}^{r}(S-i)=\mathbb{Z}$ and it is called thick if it contains arbitrary long intervals in it. Sets which can be expressed as intersection of thick and syndetic sets are called piecewise syndetic sets. All these notions have natural generalization for arbitrary semigroups.

One of the famous Ramsey theoretic result is so called Van der Waerden’s Theorem [vdw] which states that atleast one cell of any partition $\{C_{1},C_{2},\ldots,C_{r}\}$ of $\mathbb{N}$ contains arithmetic progressions of arbitrary length. Since arithmetic progressions are invariant under shifts, it follows that every piecewise syndetic set contains arbitrarily long arithmetic progressions. The following theorem is due to Van der Waerden [vdw]

Theorem 1**.**

Given any $r,l\in\mathbb{N}$, there exists $N(r,l)\in\mathbb{N}$, such that for any $r-$partition of $[1,N]$, atleast one of the partition contains an $l-$length arithmetic progression.

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

Theorem 2**.**

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

In the recent work [GJ, Theorem 6], authors have extended the technique of Beigelboeck in general commutative semigroup and proved the following:

Theorem 3**.**

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

The above theorem involves more general Gallai type progression. But parallely the following problem comes from theorem 2

Problem 4**.**

Let $S$ be a countable commutative semigroup and $A$ be any piecewise syndetic subset of $S$. Then for any $l\in\mathbb{N}$, is it possible that

[TABLE]

is piecewise syndetic in $S\times S$.

In this moment we are unable to give complete answer to this question but we have given proof of a weak version of the theorem for countable commutative semigroup. We will also give an answer of 4 for some special kind of semigroups including divisible semigroups.

2. Proof of our results

The following lemma was proved in [BG, Lemma 4.6( I’)] for general semigroup by using algebraic structure of Stone-Čech compactification of arbitrary semigroup and in [GJ, lemma 8] for commutative semigroup by combinatorially.

Lemma 5**.**

Let $(S,+)$ and $(T,+)$ be commutative semigroups, $\varphi:S\rightarrow T$ be a homomorphism and $A\subseteq S$. Then if $A$ is piecewise syndetic in $S$ and $\varphi(S)$ is piecewise syndetic in $T$, implies that $\varphi(A)$ is piecewise syndetic in $T.$

Now we need the following useful lemma,

Lemma 6**.**

If $M\subseteq S\times S$ is piecewise syndetic, then for any $c\in S$ and $a\in\mathbb{N}$,

[TABLE]

is piecewise syndetic in $S\times S$.

Proof.

Let $c\in S$ and consider the following homomorphism $\psi_{c}:S\times S\longrightarrow S\times S$ by $\psi_{c}(s,t)=(s+c,t)$. This map preserves piecewise syndeticity.

As $A\subseteq S\times S$ is piecewise syndetic set, there exists a finite subset say $E_{1}=\{(a_{1},b_{1}),(a_{2},b_{2}),\ldots,(a_{r},b_{r})\}$ of $S\times S$ such that $\stackrel{{\scriptstyle[}}{{i}}=1]{r}{\cup}(-(a_{i},b_{i})+A)$ is thick and since $\stackrel{{\scriptstyle[}}{{i}}=1]{r}{\cup}(-(a_{i},b_{i})+A)\subseteq\stackrel{{\scriptstyle[}}{{i}}=1]{r}{\cup}(-(a_{i}+c,b_{i})+\psi_{c}(A))$, the set $\stackrel{{\scriptstyle[}}{{i}}=1]{r}{\cup}(-(a_{i}+c,b_{i})+\psi_{c}(A))$ is thick. So we have $\psi_{c}(A)$ is piecewise syndetic.

Now, for any $a\in\mathbb{N}$, the semigroup homomorphism defined by $\varphi_{a}:S\times S\longrightarrow S\times S$ by $\varphi_{a}(s,t)=(s+at,t)$, $\varphi_{a}(S\times S)$ is thick in $S\times S$ and hence piecewise syndetic. So from 5, this map preserves piecewise syndeticity.

∎

The following is a weaker version of problem 4.

Theorem 7**.**

Let $S$ be any countable commutative semigroup and $A$ be piecewise syndetic in $S$. Then for $l\in\mathbb{N}$, there exists $d\in\mathbb{N}$ such that

[TABLE]

is piecewise syndetic in $S\times S$.

Proof.

Since $A$ is piecewise syndetic in $S$, then there exists a finite subset $E$ of $S$, such that $\cup_{t\in E}-t+A$ is thick in $S$.

Let $|E|=r$ and say $E=\{c_{1},c_{2},\ldots,c_{r}\}$ and let $M(r,l)=N$ be the Van der Waerden number.

The set of all possible $l-$length arithmetic progressions in $[1,N]$ is finite as $[1,N]$ is finite. et $H=\{h_{1},h_{2},\ldots,h_{n}\}$ be the set of such progressions with $|H|=n$ (say).

Then, for any $(s_{1},t_{1})\in S\times S$, if the set $\{s_{1}+t_{1},s_{1}+2t_{1},\ldots,s_{1}+Nt_{1}\}$ will be partitioned into $r$ cells, one of the partition will contain a $l$-length arithmetic progression.

Consider the set $B=\{(s,t)\in S\times S:\>s+[1,N]t\subseteq\cup_{t\in E}-t+A\}.$ It is easy to verify that $B$ is thick in $S\times S$.

Of course for any finite set $K=\{(s_{1},t_{1}),(s_{2},t_{2}),\ldots,(s_{m},t_{m})\}$ and $t\in S$ a translation of the set $\{s_{i}+[1,N](t_{i}+t)\}_{i=1}^{m}$ by an element $a\in S$ (say) will be contained in $\cup_{t\in E}-t+A$. This gives the required translation of $K$ by $(a,t)\in S\times S$.

Now the set $B$ can be $|H\times E|-$colored in a way that we will give an element $(s,t)$ of $B$ the color $(i,j)\in[1,n]\times[1,r]$ if for the least $i$, the set $\{s+h_{i}t\}\subseteq-c_{j}+A$ with the least $j\in[1,r]$.

Then, as we have partitioned the thick set $B$, one of them will be piecewise syndetic. Let the set

[TABLE]

is piecewise syndetic in $S\times S.$

Now, the set $\tilde{Q}=\{(s+at+c_{j},t):(s,t)\in Q\}$ is piecewise syndetic by lemma 6 and this proves the theorem.

∎

Since for commutative semigroup $G$ it is not necessary that for any $g\in G$, $gG$ is a piecewise syndetic in $G$,e.g. take any $n\in\mathbb{N}$, $n\mathbb{Z}[x]$ isn’t piecewise syndetic in $\mathbb{Z}[x]$.

Now we are taking $\mathscr{A}$ as the collection of all those countable commutative semigroups $(S,+)$ for which $dS=\{dx:\>x\in S\}\subseteq S$ is piecewise syndetic in $S$. Clearly $\mathscr{A}$ includes all the divisible semigroups such as $(\mathbb{Q},+),(\mathbb{Q^{+}},+),(\mathbb{Q}/\mathbb{Z},+)$ etc. and others like $\mathbb{Z}$,$\mathbb{N}$,$\mathbb{Z}[i]$ etc. We will say a semigroup $(S,+)$ is a semigroup of class $\mathscr{A}$ if $S\in\mathscr{A}$.

Lemma 8**.**

Let $S$ be a countable commutative semigroup of class $\mathscr{A}$ and $M\subseteq S\times S$ is piecewise syndetic then for any $d\in\mathbb{N}$,

[TABLE]

is piecewise syndetic in $S\times S$.

Proof.

Let $d\in\mathbb{N}$ and define $\chi_{d}:S\times S\longrightarrow S\times S$ as $\chi_{d}(s,t)=(s,dt)$. Then $\chi_{d}$ preserves piecewise syndeticity as from 5 and the fact that $\chi_{d}(S\times S)$ is piecewise syndetic in $S\times S$.

∎

So we have the following result:

Proposition 9**.**

Let $S$ be a countable commutative semigroup of class $\mathscr{A}$ and $A$ be piecewise syndetic in $S$. Then for $l\in\mathbb{N}$, then

[TABLE]

is piecewise syndetic in $S\times S$.

In this moment we are unable to derive the above proposition for general commutative semigroup which will give an affirmtive answer of problem 4 and leave the question open.

3. Applications

The set $AP^{l+1}=\{(a,a+b,a+2b,\ldots,a+lb):a,b\in S\}$ is a commutative subsemigroup of $S^{l+1}$. Using a result deduced in [BH, Theorem 3.7 (a)] it is easy to see that for any piecewise syndetic set $A\subseteq S,\,A^{l+1}\cap AP^{l+1}$ is piecewise syndetic in $AP^{l+1}$. Now as a consequence of proposition 9 we will derive this result not for all but for a large class of semigroups in the following.

Corollary 10**.**

Let $(S,+)\in\mathscr{A}$ be a commutative semigroup then for any piecewise syndetic set $M\subseteq S$, $M^{l+1}\cap AP^{l+1}$ is piecewise syndetic in $AP^{l+1}$.

Proof.

Let us take a surjective homomorphism $\varphi:S\times S\rightarrow AP^{l+1}$ by, $\varphi(a,b)=(a,a+b,a+2b,\ldots,a+lb)$.

Then from lemma 5 the map $\varphi$ preserves the piecewise syndeticity.

Let $B=\{(s,t)\in S\times S:\{s,s+t,s+2t,\ldots,s+dt\}\subseteq A\}$ and from proposition 9 $\varphi(B)$ is piecewise syndetic in $AP^{l+1}$.

Now clearly, $\varphi(B)\subseteq M^{l}\cap AP^{l+1}$ and from lemma 9 we get our required result.

This proves the claim.

∎

Now we will give a combinatorial proof of proposition 9 replacing the condition of piecewise syndeticity by Quasi-central set which is another notion of largeness and is very close to the famous central set.

A quasi-central set is genarally defined in terms of algebraic structure of $\beta\mathbb{N}$. But it has an combinatorial characterisation which will be needed for our purpose, stated below.

Theorem 11**.**

[HMS*, Theorem 3.7]*For a countable semigroup $(S,.)$, $A\subseteq S$ is said to be Quasi-central iff there is a decreasing sequence $\langle C_{n}\rangle_{n=1}^{\infty}$ of subsets of $A$ such that,

$(1)$* for each $n\in\mathbb{N}$ and each $x\in C_{n}$, there exists $m\in\mathbb{N}$ with $C_{m}\subseteq x^{-1}C_{n}$ and*

$(2)$* $C_{n}$ is piecewise syndetic $\forall n\in\mathbb{N}$.*

The following lemma is essential for our result:

Lemma 12**.**

The notion of quasi-central is preserved under surjective semigroup homorphism

Proof.

Let $\varphi:S_{1}\longrightarrow S_{2}$ be a surjective semigroup homomorphism. Let $A$ be quasi-central in $S_{1}$ and then the following holds as in property 1 in theorem 11.

[TABLE]

Now in $S_{2}$ consider the following sequence,

[TABLE]

and due to surjectivity of $\varphi$, $\varphi(A)$ and $\varphi(A_{i})\text{ for }i\in\mathbb{N}$ are piecewise syndetic.

Choose $y\in\varphi(A_{m})$ for some $m\in\mathbb{N}$ and then there exists some $x\in A_{m}$ such that $\varphi(x)=y$ and consider the set $-y+\varphi(A_{m})$ . Now as $-x+A_{m}\supseteq A_{n}$ for some $n$, we have for any $z\in A_{n}$, $x+z\in A_{m}$ and then $y+\varphi(z)\in\varphi(A_{m})$ and so $\varphi(z)\in-y+\varphi(A_{m})$.

Hence $-y+\varphi(A_{m})\supseteq\varphi(A_{n})$ and as all $y,m,n$ are chosen arbitrarily, we have the required proof.

∎

Now we will deduce proposition 9 for quasi-central sets:

Theorem 13**.**

Let $(S,+)$ be a countable commutative semigroup of class $\mathscr{A}$. Then for any quasi-central $M\subseteq S$ the collection $\{(a,b):\,\{a,a+b,a+2b,\ldots,a+lb\}\subset M\}$ is quasi-central in $(S\times S,+)$.

Proof.

As, $M$ is quasi-central, theorem 11 guarantees that there exists a decreasing sequence $\{A_{n}\}_{n\in\mathbb{N}}$ of piecewise syndetic subsets of $S$, such that property 1 of theorem 11 is satiesfied.

As $A_{n}$ is piecewise syndetic $\forall n\in\mathbb{N}$ in the following sequence,

[TABLE]

The set $B=\{(a,b):\,\{a,a+b,a+2b,\ldots,a+lb\}\subset M\}$ is piecewise syndetic in $S\times S$ from proposition 9.

And for $i\in\mathbb{N},$ $B_{i}=\{(a,b)\in S\times S:\,\{a,a+b,a+2b,\ldots,a+lb\}\subset A_{i}\}\neq\phi$ is piecewise syndetic $\forall i\in\mathbb{N},$ proposition 9.

Consider,

[TABLE]

Now choose $n\in\mathbb{N}$ and $(a,b)\in B_{n}$, then $\{a,a+b,a+2b,\ldots,a+lb\}\subset A_{n}$. Then by property 1 we have

[TABLE]

As for any $(a_{1},b_{1})\in B_{N}$ we have

[TABLE]

and $(a_{1}+a)+i(b_{1}+b)\in A_{n}\forall i\in\{0,1,2,\ldots,l\}$. Therefore $(a_{1},b_{1})\in-(a,b)+B_{n}$. Which implies $B_{N}\subseteq-(a,b)+B_{n}$, showing the property 1 of theorem 11.

This proves the theorem.

∎

The following is an extension of corollary 10.

Corollary 14**.**

Let $(S,+)$ be a commutative semigroup of class $\mathscr{A}.$ Then for any quasi-central set $M\subseteq S$, $M^{l+1}\cap AP^{l+1}$ is quasi-central in $AP^{l+1}$.

Proof.

Let us take a surjective homomorphism $\varphi:S\times S\rightarrow AP^{l+1}$ by, $\varphi(a,b)=(a,a+b,a+2b,\ldots,a+lb)$.

As $M$ is quasi central, from property 1 of theorem 11, it satiesfies equation 3.

Now from equation 3,

Where the set $B$ and $B_{i}\ (f\text{ or }i\in\mathbb{N})$ are from previous theorem and it was shown that $B$ is quasi central.

Now clearly, $\varphi(B_{i})\subseteq M^{l}\cap AP^{l+1}$ for each $i\in\mathbb{N}$ and from lemma 12 we get our required result.

This proves the claim.

∎

However there are other different type of notion of largeness such as $IP-sets,Central\,sets,J,sets,C\,sets,D\,sets$ all of those have combinatorial characterizations described in [HS] but we don’t know if it is possible to give an affirmative answer of the problem 4.