Sets with arithmetic progressions are abundant

TL;DR

This paper proves that various large sets, including those with unsettled properties like J-sets and C-sets, are rich in arithmetic progressions, demonstrating their abundance of such progressions.

Contribution

The paper provides an elementary proof that sets of A.P. rich contain arbitrarily long arithmetic progressions, extending known results to broader classes of sets.

Findings

Sets of A.P. rich contain arbitrarily long arithmetic progressions.

Elementary proof established for the abundance of progressions in these sets.

Includes sets like J-sets, C-sets, D-sets, previously with unsettled properties.

Abstract

Furstenberg, Glasscock, Bergelson, Beiglboeck have been studied abundance in arithmatic progression on various large sets like piecewise syndetic, central, thick, etc. but also there are so many sets in which abundance in progression is still unsettled like J-sets, C-sets, D-sets etc. But all of these sets have a common property that they contains arbitrary length of arithmatic progressions. These type of sets are called sets of A.P. rich, we have given an elementary proof of abundance of those sets.

Equations

$$B=\{(a,b):\left\{a,a+b,a+2b,....,a+lb\right\}\subset A\}$$

$$\left\{(c,d),(c,d)+(d,d),(c,d)+2(d,d),.....,(c,d)+l(d,d)\right\}.$$

$$A{P}_{l+1=}\left\{\left(a,a+b,a+2b,.....,a+lb\right):a,b\in S\right\}.$$

$$\left\{\left(a,b\right),\left(a,b\right)+\left(e,f\right),\left(a,b\right)+2\left(e,f\right),......,\left(a,b\right)+l\left(e,f\right)\right\}$$

$$\phi \left\{\left(a,b\right)+i\left(e,f\right)\right\}=\phi \left(a,b\right)+i\phi \left(e,f\right)\in A^{l+1}\cap A{P}_{l+1}\text{ for all 1≤i≤l}\text{.}$$

Full text

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

1. introduction

One of the famous Ramsey theoretic results is so called Van der Waerden’s Theorem which guarantees 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.

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 progressions.

In various times, mathematicians studied abundance in progression in different types of large sets Like **syndetic sets, central sets, thick sets, piecewise syndetic sets **etc. we have seen many abundance results from [FG], [GJ], [HS], [BH] etc. All of this results shows that if A\subseteq\mathbb{N}\text{ or S}, (Where S is any countable commutative semigroup) be large in some sense then some special configuration contained in those sets are also large in some sense. However, there also remains types of large sets where abundance are yet to be explicate, like **C-sets, D-sets, J-sets **etc.

All of these aforementioned sets have a common property: They all contain arbitrary length of arithmatic progressions, this type of sets are called sets of A.P. rich. Here we have given easiest elementary combinatorial proof of abundance for these type of sets. Also we have seen that if $A$ is a set of A.P. rich, then it is set of A.P. rich of all order.

Throughout this paper, $S$ is a countable commutative semigroup. Although sometimes countability or commutativity does not appear in the proof.

2. Main results

Theorem 2**.**

Let $S$ is any semigroup. Then If $A\subseteq S$ is a set of A.P. rich, then

[TABLE]

is a set of A.P. rich.

Proof.

Now as $A$ contains arithmatic progression of arbitrary length, fixed $l\in\mathbb{N}$ and for any $1\leq r\leq l$, it must contains arithmatic progression of length $r+(r+1)l$, which implies that $(c,d)+r(d,d)\in B$, which shows that $B$ contains

[TABLE]

∎

This proves the theorem.

Suppose, for any $l\in\mathbb{N}$, $AP_{l+1}$ be the subsemigroup of $S^{l+1}$ defined by:

[TABLE]

Now we derive a result which is one of the main application of [BH] for some large sets.

Corollary 3**.**

Let $S$ be a cancellative semigroup and $A\subseteq S$ is a set of A.P. rich, then $A^{l+1}\cap AP_{l+1}$ is also a set of A.P. rich.

Proof.

Take the semigroup epimorphism, $\varphi:\>S\times S\longrightarrow AP_{l+1}$ defined by, $\varphi\left(a,b\right)=\>\left(a,\>a+b,\>a+2b,......,\>a+lb\right)$. Then for any $A\subseteq S$ is A.P. rich, from 2 $B$ is also a set of A.P. rich. So, for any $l-$length arithmatic progression

[TABLE]

in $B$, we have

[TABLE]

Which concludes the result. ∎