### 数学代写|离散数学作业代写discrete mathematics代考|Finite Sums

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The result named in this subsection is a mouthful. This stems from the independent, and distinct, proofs used to prove the result that all occurred around the same time. From historical research done by Soifer [191] we find that Arnautov in [9], Folkman via personal communication, and Sanders in [178], all proved the same result, which is a generalization of Schur’s Theorem. As noted in [84], the result also follows directly from Rado’s Theorem (hence the inclusion of Rado’s name). We will offer two different proofs in this subsection: one based on van der Waerden’s Theorem (due to Folkman) and one based on Rado’s Theorem. A third proof will be given in Section 3.3.2 as it appeals to a result presented in the next chapter. Other (different) proofs can be found in $[152,199]$.

Theorem 2.55 (Arnautov-Folkman-Rado-Sanders’ Theorem). Let $k, r \in \mathbb{Z}^{+}$. There exists a minimal positive integer $n=n(k ; r)$ such that every $r$-coloring of $[1, n]$ admits $S \subseteq[1, n]$ with $|S|=k$ such that $F S(S)$ is monochromatic.
Proof based on van der Waerden’s Theorem. We start by defining the auxiliary function $a=a(k ; r)$ as the minimal positive integer such that every $r$-coloring of $[1, a]$ admits $B=\left{b_{1}<b_{2}<\cdots<b_{k}\right}$ with $F S(B) \subseteq[1, a]$ such that
$$\chi\left(b_{i_{1}}+b_{i_{2}}+\cdots+b_{i_{j}}\right)=\chi\left(b_{i_{j}}\right)$$
for any $j \in[1, k]$ with $i_{1}<i_{2}<\cdots<i_{j}$.
Assume for the moment that $a(k ; r)$ exists for all $k, r \in \mathbb{Z}^{+}$. Then, by considering $a(r(k-1)+1 ; r)$, we have $B=\left{b_{1}<b_{2}<\cdots<b_{r(k-1)+1}\right}$. By the pigeonhole principle, there must be at least $k$ elements of $B$ of the same color. Say $S=\left{b_{i_{1}}, b_{i_{2}}, \cdots, b_{i_{k}}\right}$ is monochromatic. Since $F S(S) \subseteq F S(B)$ we see that every element of $F S(S)$ is colored by the largest element of the sum. Since this largest element is necessarily in $S$, and since $S$ is monochromatic, we see that $F S(S)$ is monochromatic.

So, we will be done once we establish the existence of $a(k ; r)$. We proceed by induction on $k$, with $k=1$ being trivial. Hence, let $\ell=a(k-1 ; r)$ and let

$m=w(\ell+1 ; r)$. We will show that $a(k ; r) \leq 2 m$. To this end, consider any $r$-coloring of $[1,2 m]$. By van der Waerden’s Theorem, there exist
$$c, c+d, \ldots, c+\ell d \in[m+1,2 m]$$
all the same color. Note that $c>m$.
Since sumsets are closed under dilation, we can apply the inductive assumption to $d[1, \ell]$ to obtain
$$B=\left{d b_{1}m. Let b_{k}=c and notice that d b_{k-1}<b_{k}. As every element of F S(B) is a multiple of d, with largest possible multiple d \ell (the sum of all elements of B must remain in d[1, \ell] by definition of a(k-1 ; r); otherwise elements larger than a(k-1 ; r) would not be assigned a color ). Hence, if s \in F S(B) we see that s+b_{k}=c+j d for some j \in[1, \ell]. Since all members of c+d, \ldots, c+\ell d are the same color as c=b_{k}, we have a(k ; r) \leq 2 m. Proof based on Rado’s Theorem. Consider the following system of linear homogeneous equations:$$
\left{\sum_{i \in I} x_{i}=y_{I}: 0 \neq I \subseteq[1, k]\right},
$$where we index the y ‘s by the subset over which we are summing. By showing a monochromatic solution to this system exists we will have values of the variables for which F S\left(x_{1}, x_{2}, \ldots, x_{k}\right) is monochromatic. By Rado’s Theorem we will be done once we show that the associated coefficient matrix satisfies the columns condition. ## 数学代写|离散数学作业代写discrete mathematics代考|Hindman’s Theorem We are still investigating finite sums F S(S); however, we will now be considering the situation when S is infinite. In particular, does the conclusion of Theorem 2.55 hold when S is infinite? As you may suspect if you read the title of this subsection, an answer was achieved by Hindman [109]. Theorem 2.56 (Hindman’s Theorem). Let r \in \mathbb{Z}^{+}. Every r-coloring of \mathbb{Z}^{+} admits an infinite set A \subseteq \mathbb{Z}^{+}such that F S(A) is monochromatic. The proof of this must be fundamentally different than the proof when S is finite. We can use neither van der Waerden’s Theorem nor Rado’s Theorem as these will only give us arbitrarily large – not infinite – sets, since both are results about finite structures. The proof of Hindman’s Theorem is quite involved, so we will break it up into several lemmas before putting the pieces together. We will follow the proof given by Baumgartner [11]. But first, a definition is in order. Definition 2.57 (Disjoint sumset of S ). Let S, T \subseteq \mathbb{Z}^{+}be infinite sets. We say T \subseteq F S(S) is a disjoint sumset of S, and write T \in \mathcal{D S}(S), if every element of S is contained in at most one sum/element in T, where \mathcal{D} S(S) is the class of all disjoint sumsets of S. The benefit of this class of finite sums is that if t_{1}, t_{2} \in T \in \mathcal{D S}(S) then t_{1}+t_{2} \in F S(S) since the elements from S used in t_{1} are distinct from those used in t_{2}. Using this idea, we immediately have the following lemma, the proof of which is left to the reader as Exercise 2.17. Lemma 2.58. Let S, T, U \subseteq \mathbb{Z}^{+}be infinite sets with T \in \mathcal{D S}(S) and U \in \mathcal{D} S(T). Then (i) F S(T) \subseteq F S(S); and (ii) U \in \mathcal{D S}(S). Before the next lemma, we require another definition. Definition 2.59 (Intersective for S ). Let S \subseteq \mathbb{Z}^{+}be infinite. We say that a set X is intersective for S if for all T \in \mathcal{D S}(S) we have F S(T) \cap X \neq \emptyset. A crucial observation here is that any intersective set must be infinite. This can be confirmed by part (ii) of the next lemma. Lemma 2.60. Let X, S be subsets of \mathbb{Z}^{+}. The following hold: (i) Let n \in \mathbb{Z}^{+}and assume X=\bigsqcup_{i=1}^{n} X_{i} . If X is intersective for S, then there exists T \in \mathcal{D S}(S) and i \in{1,2, \ldots, n} such that X_{i} is intersective for T. (ii) If F is a finite subset of \mathbb{Z}^{+}and X is intersective for S, then X \backslash F is intersective for S. ## 数学代写|离散数学作业代写discrete mathematics代考|Density Results We end this chapter by presenting some analytic approaches to integer Ramsey theory. We will mainly be considering arithmetic progressions. The standard notation used for density results concerning van der Waerden’s Theorem is given next. Notation. For k, n \in \mathbb{Z}^{+}denote by r_{k}(n) the maximal size of a subset of [1, n] with no k-term arithmetic progression. We start with a result due to Behrend [14] that seems to have been overlooked. This was communicated to the author by Tom Brown. Behrend’s result appeared just a year after Erdös and Turán [71] defined r_{3}(n) and showed that$$
\frac{r_{3}(n)}{n}<\frac{3}{8}+o(1)
$$Behrend’s result is more sweeping. Theorem 2.67. For each k \in Z^{+}we have that$$
L_{k}=\lim {n \rightarrow \infty} \frac{r{k}(n)}{n}
$$exists and, more importantly,$$
L=\lim {k \rightarrow \infty} L{k} \in{0,1}
$$Proof. The fact that L_{k} exists is standard and left to the reader as Exercise 2.22. The fact that L exists follows from 0 \leq L_{1} \leq L_{2} \leq \cdots \leq L_{k} \leq \cdots \leq 1, which is a monotonic infinite sequence on a closed interval. Thus, we need only show that L is either 0 or 1 . We will first show that, for every n \in \mathbb{Z}^{+}, we have$$
\frac{r_{k}(n)}{n}>L_{k}
$$Assume, for a contradiction, that there exists m \in \mathbb{Z}^{+}such that Inequality (2.3) is false for m. We may assume that for all n \in \mathbb{Z}^{+}we have$$
\frac{r_{k}(m)}{m} \leq \frac{r_{k}(n)}{n}
$$We next note that r_{k}(m n) \leq n r_{k}(m) since if S \subseteq[1, m n] avoids k-term arithmetic progressions, then it is necessary (but not sufficient) that each of the intervals [1, m],[m+1,2 m], \ldots,[(n-1) m+1, n m] contains at most r_{k}(m) integers from S. From Inequality (2.4) and this observation, we have$$
\frac{r_{k}(m)}{m} \leq \frac{r_{k}(m n)}{m n} \leq \frac{n r_{k}(m)}{m n}=\frac{r_{k}(m)}{m}
$$and we conclude that r_{k}(m n)=n r_{k}(m) for all n \in \mathbb{Z}+. Define the intervals$$
A_{i}=[(i-1) m+1, i m], \quad i=1,2, \ldots, w\left(k ; 2^{m}\right)
$$where w\left(k ; 2^{m}\right) is the van der Waerden number. Consider$$
n=w\left(k ; 2^{m}\right)
$$## 离散数学代写 ## 数学代写|离散数学作业代写discrete mathematics代考|Arnautov-Folkman-Rado-Sanders’ Theorem 本小节中命名的结果是满口的。这源于用于证明所有结果几乎同时发生的独立且独特的证据。从 Soifer [191] 所做的历史研究中，我们发现 [9] 中的 Arnautov、通过个人交流的 Folkman 和 [178] 中的 Sanders 都证明了相同的结果，这是 Schur 定理的推广。如 [84] 中所述，结果也直接来自 Rado 定理（因此包含 Rado 的名称）。我们将在本小节中提供两种不同的证明：一种基于 van der Waerden 定理（由于 Folkman），另一种基于 Rado 定理。第 3.3.2 节将给出第三个证明，因为它与下一章中提出的结果相呼应。其他（不同的）证明可以在[152,199]. 定理 2.55（Arnautov-Folkman-Rado-Sanders 定理）。让ķ,r∈从+. 存在一个最小正整数n=n(ķ;r)这样每一个r- 着色[1,n]承认小号⊆[1,n]和|小号|=ķ这样F小号(小号)是单色的。 基于范德瓦尔登定理的证明。我们从定义辅助函数开始一种=一种(ķ;r)作为最小正整数，使得每个r- 着色[1,一种]承认B=\left{b_{1}<b_{2}<\cdots<b_{k}\right}B=\left{b_{1}<b_{2}<\cdots<b_{k}\right}和F小号(乙)⊆[1,一种]这样 χ(b一世1+b一世2+⋯+b一世j)=χ(b一世j) 对于任何j∈[1,ķ]和一世1<一世2<⋯<一世j. 暂时假设一种(ķ;r)为所有人而存在ķ,r∈从+. 然后，通过考虑一种(r(ķ−1)+1;r)， 我们有B=\left{b_{1}<b_{2}<\cdots<b_{r(k-1)+1}\right}B=\left{b_{1}<b_{2}<\cdots<b_{r(k-1)+1}\right}. 根据鸽巢原理，至少有ķ要点乙相同的颜色。说S=\left{b_{i_{1}}, b_{i_{2}}, \cdots, b_{i_{k}}\right}S=\left{b_{i_{1}}, b_{i_{2}}, \cdots, b_{i_{k}}\right}是单色的。自从F小号(小号)⊆F小号(乙)我们看到，每一个元素F小号(小号)由总和的最大元素着色。因为这个最大的元素必然在小号，并且由于小号是单色的，我们看到F小号(小号)是单色的。 所以，一旦我们确定存在一种(ķ;r). 我们通过归纳继续ķ， 和ķ=1是微不足道的。因此，让ℓ=一种(ķ−1;r)然后让 米=在(ℓ+1;r). 我们将证明一种(ķ;r)≤2米. 为此，考虑任何r- 着色[1,2米]. 根据范德瓦尔登定理，存在 C,C+d,…,C+ℓd∈[米+1,2米] 都是一样的颜色。注意C>米. 由于和集在膨胀下是闭合的，我们可以将归纳假设应用于d[1,ℓ]得到$$
B=\left{db_{1}m.大号和吨b_{k}=c一种ndn这吨一世C和吨H一种吨d b_{k-1}<b_{k}.一种s和在和r是和l和米和n吨这F前卫(乙)一世s一种米在l吨一世pl和这Fd,在一世吨Hl一种rG和s吨p这ss一世bl和米在l吨一世pl和d \ ell(吨H和s在米这F一种ll和l和米和n吨s这F乙米在s吨r和米一种一世n一世nd [1, \ell]b是d和F一世n一世吨一世这n这Fa(k-1 ; r);这吨H和r在一世s和和l和米和n吨sl一种rG和r吨H一种na(k-1 ; r)在这在ldn这吨b和一种ss一世Gn和d一种C这l这r).H和nC和,一世Fs \in FS(B)在和s和和吨H一种吨s+b_{k}=c+jdF这rs这米和j \in[1, \ell].小号一世nC和一种ll米和米b和rs这Fc+d, \ldots, c+\ell d一种r和吨H和s一种米和C这l这r一种sc=b_{k},在和H一种在和a(k ; r) \leq 2 m\$。

\left{\sum_{i \in I} x_{i}=y_{I}: 0 \neq I \subseteq[1, k]\right},\left{\sum_{i \in I} x_{i}=y_{I}: 0 \neq I \subseteq[1, k]\right},

## 数学代写|离散数学作业代写discrete mathematics代考|Hindman’s Theorem

Hindman 定理的证明非常复杂，所以我们将把它分解成几个引理，然后再把它们放在一起。我们将遵循 Baumgartner [11] 给出的证明。但首先，定义是有序的。

（一）F小号(吨)⊆F小号(小号); (
ii)在∈D小号(小号).

(i) 让n∈从+并假设X=⨆一世=1nX一世.如果X是相交的小号, 那么存在吨∈D小号(小号)和一世∈1,2,…,n这样X一世是相交的吨.
(ii) 如果F是的有限子集从+和X是相交的小号， 然后X∖F是相交的小号.

## 数学代写|离散数学作业代写discrete mathematics代考|Density Results

r3(n)n<38+这(1)

rķ(n)n>大号ķ

rķ(米)米≤rķ(n)n

rķ(米)米≤rķ(米n)米n≤nrķ(米)米n=rķ(米)米

n=在(ķ;2米)

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