# 数学代写|组合数学代写Combinatorial mathematics代考|MATH4575

## 数学代写|组合数学代写Combinatorial mathematics代考|Dilworth’s Theorem

The first proof of the LYM inequality used the existence of a relation between Sperner families and chains of subsets. This relation is more explicit in a result by Dilworth. Dilworth’s theorem holds for more general structures than subsets: we can prove it for arbitrary partial orders.
4.7.1 Definition. A partially ordered set (or poset) is a pair $(P, \leq)$ such that $\leq$ is a subset of $P \times P$ (we denote the fact that $(x, y)$ is in this subset by $x \leq y$ ), satisfying for all $x, y, z \in P$ :
Reflexivity: $x \leq x$
Antisymmetry: If $x \leq y$ and $y \leq x$ then $x=y$;
Transitivity: If $x \leq y$ and $y \leq z$ then $x \leq z$.
A key example of a poset is when $P$ is a collection of sets, and $\leq$ is set inclusion: $X \leq Y$ iff $X \subseteq Y$.
4.7.2 Definition. A chain in a poset is an ordered tuple $\left(x_1, \ldots, x_k\right)$ of distinct elements such that $x_1 \leq x_2 \leq \cdots \leq x_k$. An antichain is a subset $F$ such that, for all distinct $x, y \in F$ we have neither $x \leq y$ nor $y \leq x$.

We study the problems of partitioning a poset into disjoint chains, or into disjoint antichains. The following is easy to see, since a chain and an antichain intersect in at most one element:
4.7.3 LEMMA. Let $(P, \leq)$ be a poset.
(i) If $P$ has a chain of size $r$, then $P$ cannot be partitioned into fewer than $r$ antichains;

## 数学代写|组合数学代写Combinatorial mathematics代考|The clubs of Oddtown

In an effort to cut costs, the town council of Oddtown tries to limit the number of clubs the citizens can form. They hire two consulting firms, who come up with the following rules for clubs. Both firms agree on the first two rules, but have a different version of the third.
(i) No two clubs can have the same set of members;
(ii) Every two distinct clubs must have an even number of common members;
(iii) a) Each club has an even number of members;
b) Each club has an odd number of members.
If clubs have even size, a collection of clubs of size $2^{\lfloor n / 2\rfloor}$ is easily constructed: divide the citizens into pairs, and let each club be a union of pairs. If clubs have odd size, it is harder to come up with a large construction, and you’ll struggle to do better than ${1}, \ldots,{n}$. There is a reason for the struggle:
1.1 THEOREM (Oddtown). Let $\mathscr{F}$ be a family of subsets of $[n]$ such that, for all $A, B \in \mathscr{F}$ with $A \neq B$, we have $|A \cap B|$ is even and $|A|$ is odd. Then $|\mathscr{F}| \leq n$.
Proof: Let $\mathscr{F}=\left{A_1, \ldots, A_m\right}$. For each $A_i$ define a vector $a_i \in \mathbb{Z}2^n$ by $$\left(a_i\right)_j= \begin{cases}1 & \text { if } j \in A_i \ 0 & \text { otherwise }\end{cases}$$ Consider the inner product $\left\langle a_i, a_j\right\rangle$. $$\left\langle a_i, a_j\right\rangle=\sum{x \in[n]}\left(a_i\right)x\left(a_j\right)_x=\sum{x \in A_i \cap A_j} 1(\bmod 2)= \begin{cases}0 & i \neq j \ 1 & i=j .\end{cases}$$

# 组合数学代考

## 数学代写|组合数学代写Combinatorial mathematics代考|Dilworth’s Theorem

LYM 不等式的第一个证明使用了 Sperner 族和子集链之间存在的关系。这种关系在 Dilworth 的结果中更 为明确。Dilworth 定理适用于比子集更一般的结构：我们可以证明它适用于任意偏序。
$4.7 .1$ 定义。偏序集 (或偏序集) 是一对 $(P, \leq)$ 这样 $\leq$ 是一个子集 $P \times P$ (我们表示这样一个事实 $(x, y)$

4.7.2 定义。偏序集中的链是有序元组 $\left(x_1, \ldots, x_k\right)$ 不同的元嫊使得 $x_1 \leq x_2 \leq \cdots \leq x_k$. 反链是一个子 集 $F$ 这样，对于所有不同的 $x, y \in F$ 我们都没有 $x \leq y$ 也不 $y \leq x$.

4.7.3 LEMMA。让 $(P, \leq)$ 成为一个小人物。
(i) 如果 $P$ 有大小链 $r$ ，然后 $P$ 不能分割成少于 $r$ 反链;

## 数学代写|组合数学代写Combinatorial mathematics代考|The clubs of Oddtown

(i) 任何两个倶乐部都不能拥有同一组会员;
(ii) 每两个不同的倶乐部必须有偶数的共同会员；
(iii) a) 每个倶乐部的云员人数为偶数；
b) 每个倶乐部的会员人数为奇数。

1.1 THEOREM (Oddtown)。让龙是一个子集的家庭 $[n]$ 这样，对于所有 $A, B \in \mathscr{F}$ 和 $A \neq B$ ，我们有 $|A \cap B|$ 是偶数并且 $|A|$ 很奇怪。然后 $|\mathscr{F}| \leq n$.

$$\left(a_i\right)_j=\left{1 \quad \text { if } j \in A_i 0 \quad\right. \text { otherwise }$$

$$\left\langle a_i, a_j\right\rangle=\sum x \in[n]\left(a_i\right) x\left(a_j\right)_x=\sum x \in A_i \cap A_j 1(\bmod 2)={0 \quad i \neq j 1 \quad i=j .$$

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