# 数学代写|组合数学代写Combinatorial mathematics代考|МАТН418

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

The result in this section does not exactly fit the chapter title. However, it is a central result that is frequently used in combinatorics, and we’ll need it in the next section. If you’ve taken a course in graph theory, you may have seen it formulated in terms of bipartite graphs, in close connection with Kốnig’s Theorem. Here we stick to a formulation in terms of set systems.
4.4.1 Definition. Let $A_1, \ldots, A_n$ be finite sets. An $n$-tuple $\left(x_1, \ldots, x_n\right)$ is a system of distinct representatives (SDR) if

• $x_i \in A_i$ for $i \in[n]$;
• $x_i \neq x_j$ for $i, j \in[n]$ with $i \neq j$.
The question we wish to answer is: when does a set system $A_1, \ldots, A_n$ have an SDR? Clearly each $A_i$ needs to contain an element, and $A_1 \cup \cdots \cup A_n$ needs to contain $n$ elements. Write, for $J \subseteq[n], A(J):=\cup_{i \in J} A_i$. A more general necessary condition, which is equally obvious, is Hall’s Condition:
$$|A(J)| \geq|J| \quad \text { for all } J \subseteq N .$$
As it turns out, this condition is not only necessary but also sufficient:
4.4.2 THEOREM (Hall’s Marriage Theorem). The finite sets $A_1, \ldots, A_n$ have an SDR if and only if $(\mathrm{HC})$ holds.

Proof: If the sets have an SDR, then clearly (HC) holds. For the converse, suppose (HC) holds. We prove the result by induction on $n$, the case $n=1$ being obvious. Say a subset $J \subseteq[n]$ is critical if $|A(J)|=|J|$.

Case I. Suppose only $J=\emptyset$ and (possibly) $J=[n]$ are critical. Pick any $x_n \in A_n$, and let $A_i^{\prime}:=A_i \backslash\left{x_n\right}$ for $i \in[n-1]$. For $J \subseteq[n-1]$ with $J \neq \emptyset$ we have
$$\left|A^{\prime}(J)\right| \geq|A(J)|-1 \geq|J|,$$
since we removed only $x_n$ from $A(J)$, and $J$ is not critical. Hence (HC) holds for the $A_i^{\prime}$, and by induction the sets $A_1^{\prime}, \ldots, A_{n-1}^{\prime}$ have an $\operatorname{SDR}\left(x_1, \ldots, x_{n-1}\right)$. Then $\left(x_1, \ldots, x_n\right)$ is an SDR for the original problem.

## 数学代写|组合数学代写Combinatorial mathematics代考|Sperner families

Let us look at a milder restriction on our subsets:
4.6.1 Definition. A family $\mathscr{F}$ of sets is a Sperner family (or antichain, or clutter) if, for all $A, B \in \mathscr{F}$ with $A \neq B$ we have $A \not \subset B$ and $B \not \subset A$.

This condition is easy to satisfy: take $\mathscr{F}$ to be the collection of all size- $k$ subsets of [n]. This gives a family of size $\left(\begin{array}{c}n \ k\end{array}\right)$, and the size of the family is maximal for $k=\lfloor n / 2\rfloor$. Sperner proved that we cannot, in fact, do better:
4.6.2 THEOREM (Sperner). If $\mathscr{F}$ is a Sperner family of subsets of $[n]$, then $|\mathscr{F}| \leq\left(\begin{array}{c}n \ \lfloor n / 2\rfloor\end{array}\right)$.
Sperner’s Theorem is an easy consequence of the following result, known as the LYM inequality, named after Lubell, Meshalkin, and Yamamoto who each independently discovered it.
4.6.3 THEOREM (LYM inequality). If $\mathscr{F}$ is a Sperner family of subsets of $[n]$, then
$$\sum_{A \in F}\left(\begin{array}{c} n \ |A| \end{array}\right)^{-1} \leq 1$$
Proof of Sperner’s Theorem using the LYM inequality: As seen above, $\left(\begin{array}{l}n \ k\end{array}\right)$ is maximal when $k=\lfloor n / 2\rfloor$. Hence
$$1 \geq \sum_{A \in \mathscr{F}}\left(\begin{array}{c} n \ |A| \end{array}\right)^{-1} \geq \sum_{A \in \mathscr{F}}\left(\begin{array}{c} n \ \lfloor n / 2\rfloor \end{array}\right)^{-1}=|\mathscr{F}|\left(\begin{array}{c} n \ \lfloor n / 2\rfloor \end{array}\right)^{-1}$$

# 组合数学代考

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

4.4.1 定义。让 $A_1, \ldots, A_n$ 是有限集。一个 $n$-元组 $\left(x_1, \ldots, x_n\right)$ 是不同代表系统 (SDR)，如果

• $x_i \in A_i$ 为了 $i \in[n] ;$
• $x_i \neq x_j$ 为了 $i, j \in[n]$ 和 $i \neq j$.
我们要回答的问题是: 什么时候设置系统 $A_1, \ldots, A_n$ 有特别提款权? 明明每一个 $A_i$ 需要包含一个元 溸，并且 $A_1 \cup \cdots \cup A_n$ 需要包含 $n$ 元傃。写，为了J $J[n], A(J):=\cup_{i \in J} A_i$. 一个更一般的、同样 明显的必要条件是霍尔条件:
$$|A(J)| \geq|J| \quad \text { for all } J \subseteq N \text {. }$$
事实证明，这个条件不仅必要而且充分:
4.4.2 定理（霍尔姖姻定理）。有限集 $A_1, \ldots, A_n$ 拥有特别提款权当且仅当 $(\mathrm{HC})$ 持有。
证明：如果集合有 $\mathrm{SDR}$ ，则显然 $(\mathrm{HC})$ 成立。相反，假设 $(\mathrm{HC})$ 成立。我们通过归纳法证明结果 $n$ ，案子 $n=1$ 很明显。说一个子集 $J \subseteq[n]$ 如果 $|A(J)|=|J|$.
情况 I. 仅假设 $J=\emptyset$ 和 (可能) $J=[n]$ 很关键。选择任何 $x_n \in A_n$ ，然后让
$$\left|A^{\prime}(J)\right| \geq|A(J)|-1 \geq|J|,$$
因为我们只删除了 $x_n$ 从 $A(J)$ ，和 $J$ 并不重要。因此 $(\mathrm{HC})$ 成立 $A_i^{\prime}$ ，并通过归纳集合 $A_1^{\prime}, \ldots, A_{n-1}^{\prime}$ 有一 个 $\operatorname{SDR}\left(x_1, \ldots, x_{n-1}\right)$. 然后 $\left(x_1, \ldots, x_n\right)$ 是原始问题的 $\operatorname{SDR}$ 。

## 数学代写|组合数学代写Combinatorial mathematics代考|Sperner families

4.6.1 定义。一个家庭庅的集合是一个 Sperner 族（或反链，或杂乱）如果，对于所有 $A, B \in \mathscr{F}$ 和 $A \neq B$ 我们有 $A \not \subset B$ 和 $B \not \subset A$.

4.6.2 定理 (Sperner) 。如果 $\mathscr{F}$ 是一个 Sperner 族的子集 $[n]$ ， 然后 $|\mathscr{F}| \leq(n\lfloor n / 2\rfloor)$.
Sperner 定理是以下结果的简单结果，称为 LYM 不等式，以 Lubell、Meshalkin 和 Yamamoto 的名字命 名，他们各自独立地发现了它。
$4.6 .3$ 定理 (LYM 不等式) 。如果 $\mathscr{F}$ 是一个 Sperner 族的子集 $[n]$ ，然后
$$\sum_{A \in F}(n|A|)^{-1} \leq 1$$

$$1 \geq \sum_{A \in \mathscr{F}}(n|A|)^{-1} \geq \sum_{A \in \mathscr{F}}(n\lfloor n / 2\rfloor)^{-1}=|\mathscr{F}|(n\lfloor n / 2\rfloor)^{-1}$$

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