组合数学 (Spring 2015)/Extremal set theory

Sunflowers

An set system is a sunflower if all its member sets intersect at the same set of elements.

Definition (sunflower)

A set family [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] is a sunflower of size [math]\displaystyle{ r }[/math] with a core [math]\displaystyle{ C\subseteq X }[/math] if [math]\displaystyle{ \forall S,T\in\mathcal{F} }[/math] that [math]\displaystyle{ S\neq T }[/math], [math]\displaystyle{ S\cap T=C }[/math].

Note that we do not require the core to be nonempty, thus a family of disjoint sets is also a sunflower (with the core [math]\displaystyle{ \emptyset }[/math]).

The next result due to Erdős and Rado, called the sunflower lemma, is a famous result in extremal set theory, and has some important applications in Boolean circuit complexity.

Sunflower Lemma (Erdős-Rado)

Let [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math]. If [math]\displaystyle{ |\mathcal{F}|\gt k!(r-1)^k }[/math], then [math]\displaystyle{ \mathcal{F} }[/math] contains a sunflower of size [math]\displaystyle{ r }[/math].

Proof. We proceed by induction on [math]\displaystyle{ k }[/math]. For [math]\displaystyle{ k=1 }[/math], [math]\displaystyle{ \mathcal{F}\subseteq{X\choose 1} }[/math], thus all sets in [math]\displaystyle{ \mathcal{F} }[/math] are disjoint. And since [math]\displaystyle{ |\mathcal{F}|\gt r-1 }[/math], we can choose [math]\displaystyle{ r }[/math] of these sets and form a sunflower. Now let [math]\displaystyle{ k\ge 2 }[/math] and assume the lemma holds for all smaller [math]\displaystyle{ k }[/math]. Take a maximal family [math]\displaystyle{ \mathcal{G}\subseteq \mathcal{F} }[/math] whose members are disjoint, i.e. for any [math]\displaystyle{ S,T\in \mathcal{G} }[/math] that [math]\displaystyle{ S\neq T }[/math], [math]\displaystyle{ S\cap T=\emptyset }[/math]. If [math]\displaystyle{ |\mathcal{G}|\ge r }[/math], then [math]\displaystyle{ \mathcal{G} }[/math] is a sunflower of size at least [math]\displaystyle{ r }[/math] and we are done. Assume that [math]\displaystyle{ |\mathcal{G}|\le r-1 }[/math], and let [math]\displaystyle{ Y=\bigcup_{S\in\mathcal{G}}S }[/math]. Then [math]\displaystyle{ |Y|=k|\mathcal{G}|\le k(r-1) }[/math] (since all members of [math]\displaystyle{ \mathcal{G} }[/math]) are disjoint). We claim that [math]\displaystyle{ Y }[/math] intersets all members of [math]\displaystyle{ \mathcal{F} }[/math], since if otherwise, there exists an [math]\displaystyle{ S\in\mathcal{F} }[/math] such that [math]\displaystyle{ S\cap Y=\emptyset }[/math], then we can enlarge [math]\displaystyle{ \mathcal{G} }[/math] by adding [math]\displaystyle{ S }[/math] into [math]\displaystyle{ \mathcal{G} }[/math] and still have all members of [math]\displaystyle{ \mathcal{G} }[/math] disjoint, which contradicts the assumption that [math]\displaystyle{ \mathcal{G} }[/math] is the maximum of such families. By the pigeonhole principle, some elements [math]\displaystyle{ y\in Y }[/math] must contained in at least [math]\displaystyle{ \frac{|\mathcal{F}|}{|Y|}\gt \frac{k!(r-1)^k}{k(r-1)}=(k-1)!(r-1)^{k-1} }[/math] members of [math]\displaystyle{ \mathcal{F} }[/math]. We delete this [math]\displaystyle{ y }[/math] from these sets and consider the family [math]\displaystyle{ \mathcal{H}=\{S\setminus\{y\}\mid S\in\mathcal{F}\wedge y\in S\} }[/math]. We have [math]\displaystyle{ \mathcal{H}\subseteq {X\choose k-1} }[/math] and [math]\displaystyle{ |\mathcal{H}|\gt (k-1)!(r-1)^{k-1} }[/math], thus by the induction hypothesis, [math]\displaystyle{ \mathcal{H} }[/math]contains a sunflower of size [math]\displaystyle{ r }[/math]. Adding [math]\displaystyle{ y }[/math] to the members of this sunflower, we get the desired sunflower in the original family [math]\displaystyle{ \mathcal{F} }[/math]. [math]\displaystyle{ \square }[/math]

The Erdős–Ko–Rado Theorem

A set family [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] is called intersecting, if for any [math]\displaystyle{ S,T\in\mathcal{F} }[/math], [math]\displaystyle{ S\cap T\neq\emptyset }[/math]. A natural question of extremal favor is: "how large can an intersecting family be?"

Assume [math]\displaystyle{ |X|=n }[/math]. When [math]\displaystyle{ n\lt 2k }[/math], every pair of [math]\displaystyle{ k }[/math]-subsets of [math]\displaystyle{ X }[/math] intersects. So the non-trivial case is when [math]\displaystyle{ n\ge 2k }[/math]. The famous Erdős–Ko–Rado theorem gives the largest possible cardinality of a nontrivially intersecting family.

According to Erdős, the theorem itself was proved in 1938, but was not published until 23 years later.

Erdős–Ko–Rado theorem (proved in 1938, published in 1961)

Let [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math] where [math]\displaystyle{ |X|=n }[/math] and [math]\displaystyle{ n\ge 2k }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then [math]\displaystyle{ |\mathcal{F}|\le{n-1\choose k-1} }[/math].

Katona's proof

We first introduce a proof discovered by Katona in 1972. The proof uses double counting.

Let [math]\displaystyle{ \pi }[/math] be a cyclic permutation of [math]\displaystyle{ X }[/math], that is, we think of assigning [math]\displaystyle{ X }[/math] in a circle and ignore the rotations of the circle. It is easy to see that there are [math]\displaystyle{ (n-1)! }[/math] cyclic permutations of an [math]\displaystyle{ n }[/math]-set (each cyclic permutation corresponds to [math]\displaystyle{ n }[/math] permutations). Let

[math]\displaystyle{ \mathcal{G}_\pi=\{\{\pi_{(i+j)\bmod n}\mid j\in[k]\}\mid i\in [n]\} }[/math].

The next lemma states the following observation: in a circle of [math]\displaystyle{ n }[/math] points, supposed [math]\displaystyle{ n\ge 2k }[/math], there can be at most [math]\displaystyle{ k }[/math] arcs, each consisting of [math]\displaystyle{ k }[/math] points, such that every pair of arcs share at least one point.

Lemma

Let [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math] where [math]\displaystyle{ |X|=n }[/math] and [math]\displaystyle{ n\ge 2k }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then for any cyclic permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ X }[/math], it holds that [math]\displaystyle{ |\mathcal{G}_\pi\cap\mathcal{F}|\le k }[/math].

Proof. Fix a cyclic permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ X }[/math]. Let [math]\displaystyle{ A_i=\{\pi_{(i+j+n)\bmod n}\mid j\in[k]\} }[/math]. Then [math]\displaystyle{ \mathcal{G}_\pi }[/math] can be written as [math]\displaystyle{ \mathcal{G}_\pi=\{A_i\mid i\in [n]\} }[/math]. Suppose that [math]\displaystyle{ A_t\in\mathcal{F} }[/math]. Since [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, the only sets [math]\displaystyle{ A_i }[/math] that can be in [math]\displaystyle{ \mathcal{F} }[/math] other than [math]\displaystyle{ A_t }[/math] itself are the [math]\displaystyle{ 2k-2 }[/math] sets [math]\displaystyle{ A_i }[/math] with [math]\displaystyle{ t-(k-1)\le i\le t+k-1, i\neq t }[/math]. We partition these sets into [math]\displaystyle{ k-1 }[/math] pairs [math]\displaystyle{ \{A_i,A_{i+k}\} }[/math], where [math]\displaystyle{ t-(k-1)\le i\le t-1 }[/math]. Note that for [math]\displaystyle{ n\ge 2k }[/math], it holds that [math]\displaystyle{ A_i\cap C_{i+k}=\emptyset }[/math]. Since [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, [math]\displaystyle{ \mathcal{F} }[/math] can contain at most one set of each such pair. The lemma follows. [math]\displaystyle{ \square }[/math]

The Katona's proof of Erdős–Ko–Rado theorem is done by counting in two ways the pairs of member [math]\displaystyle{ S }[/math] of [math]\displaystyle{ \mathcal{F} }[/math] and cyclic permutation [math]\displaystyle{ \pi }[/math] which contain [math]\displaystyle{ S }[/math] as a continuous path on the circle (i.e., an arc).

Katona's proof of Erdős–Ko–Rado theorem (double counting) Let [math]\displaystyle{ \mathcal{R}=\{(S,\pi)\mid \pi \text{ is a cyclic permutation of }X, \text{and }S\in\mathcal{F}\cap\mathcal{G}_\pi\} }[/math]. We count [math]\displaystyle{ \mathcal{R} }[/math] in two ways. First, due to the lemma, [math]\displaystyle{ |\mathcal{F}\cap\mathcal{G}_\pi|\le k }[/math] for any cyclic permutation [math]\displaystyle{ \pi }[/math]. There are [math]\displaystyle{ (n-1)! }[/math] cyclic permutations in total. Thus, [math]\displaystyle{ |\mathcal{R}|=\sum_{\text{cyclic }\pi}|\mathcal{F}\cap\mathcal{G}_\pi|\le k(n-1)! }[/math]. Next, for each [math]\displaystyle{ S\in\mathcal{F} }[/math], the number of cyclic permutations [math]\displaystyle{ \pi }[/math] in which [math]\displaystyle{ S }[/math] is continuous is [math]\displaystyle{ |S|!(n-|S|)!=k!(n-k)! }[/math]. Thus, [math]\displaystyle{ |\mathcal{R}|=\sum_{S\in\mathcal{F}}k!(n-k)!=|\mathcal{F}|k!(n-k)! }[/math]. Altogether, we have [math]\displaystyle{ |\mathcal{F}|\le\frac{k(n-1)!}{k!(n-k)!}=\frac{(n-1)!}{(k-1)!(n-k)!}={n-1\choose k-1} }[/math]. [math]\displaystyle{ \square }[/math]

Erdős' shifting technique

We now introduce the original proof of the Erdős–Ko–Rado theorem, which uses a technique called shifting (originally called compression).

Without loss of generality, we assume [math]\displaystyle{ X=[n] }[/math], and restate the Erdős–Ko–Rado theorem as follows.

Erdős–Ko–Rado theorem

Let [math]\displaystyle{ \mathcal{F}\subseteq {[n]\choose k} }[/math] and [math]\displaystyle{ n\ge 2k }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then [math]\displaystyle{ |\mathcal{F}|\le{n-1\choose k-1} }[/math].

We define a shift operator for the set family.

Definition (shift operator)

Assume [math]\displaystyle{ \mathcal{F}\subseteq 2^{[n]} }[/math], and [math]\displaystyle{ 0\le i\lt j\le n-1 }[/math]. Define the [math]\displaystyle{ (i,j) }[/math]-shift [math]\displaystyle{ S_{ij} }[/math] as an operator on [math]\displaystyle{ \mathcal{F} }[/math] as follows: for each [math]\displaystyle{ T\in\mathcal{F} }[/math], write [math]\displaystyle{ T_{ij}=(T\setminus\{j\})\cup\{i\} }[/math], and let [math]\displaystyle{ S_{ij}(T)= \begin{cases} T_{ij} & \mbox{if }j\in T, i\not\in T, \mbox{ and }T_{ij} \not\in\mathcal{F},\\ T & \mbox{otherwise;} \end{cases} }[/math] let [math]\displaystyle{ S_{ij}(\mathcal{F})=\{S_{ij}(T)\mid T\in \mathcal{F}\} }[/math].

It is easy to verify the following propositions of shifts.

Proposition

[math]\displaystyle{ |S_{ij}(T)|=|T|\, }[/math] and [math]\displaystyle{ |S_{ij}(\mathcal{F})|=\mathcal{F} }[/math]; if [math]\displaystyle{ \mathcal{F} }[/math] is intersecting, then so is [math]\displaystyle{ S_{ij}(\mathcal{F}) }[/math].

Proof. (1) is quite obvious. Now we prove (2). Consider any [math]\displaystyle{ A,B\in\mathcal{F} }[/math]. If [math]\displaystyle{ A\cap B }[/math] includes any element other than [math]\displaystyle{ j }[/math] then [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] are still intersecting after [math]\displaystyle{ (i,j) }[/math]-shift. Thus without loss of generality, we may consider the only unsafe case where [math]\displaystyle{ A\cap B=\{j\} }[/math], and [math]\displaystyle{ B }[/math] is successfully shifted to [math]\displaystyle{ B_{ij}=(B\setminus\{j\})\cup\{i\} }[/math] but [math]\displaystyle{ A }[/math] fails to shift to [math]\displaystyle{ A_{ij}=(A\setminus\{j\})\cup\{i\} }[/math]. Since [math]\displaystyle{ B }[/math] is successfully shifted to [math]\displaystyle{ B_{ij} }[/math], we know that it must hold that [math]\displaystyle{ i\not\in B }[/math] and [math]\displaystyle{ B_{ij}\not\in\mathcal{F} }[/math]. And the only two reasons for which [math]\displaystyle{ A }[/math] may fail to shift to [math]\displaystyle{ A_{ij} }[/math] are: (1) [math]\displaystyle{ i\in A }[/math] and (2) [math]\displaystyle{ i\not\in A }[/math] but [math]\displaystyle{ A_{ij}\in\mathcal{F} }[/math]. Case.1: [math]\displaystyle{ i\in A }[/math]. In this case, since [math]\displaystyle{ i\in B_{ij} }[/math], we have [math]\displaystyle{ S_{ij}(A)\cap S_{ij}(B)=A\cap B_{ij}=\{i\} }[/math], i.e. [math]\displaystyle{ B }[/math] is still intersecting with [math]\displaystyle{ A }[/math] after shifting. Case.2: [math]\displaystyle{ i\not\in A }[/math] but [math]\displaystyle{ A_{ij}\in\mathcal{F} }[/math]. In this case, it is easy to verify that [math]\displaystyle{ A_{ij}\cap B=(A\cap B)\setminus\{j\}=\emptyset }[/math]. Recall that we assume [math]\displaystyle{ A_{ij}\in\mathcal{F} }[/math]. This contradicts to that [math]\displaystyle{ \mathcal{F} }[/math] is intersecting. In conclusion, in all cases, [math]\displaystyle{ S_{ij}(\mathcal{F}) }[/math] remains intersecting. [math]\displaystyle{ \square }[/math]

Repeatedly applying [math]\displaystyle{ S_{ij}(\mathcal{F}) }[/math] for any [math]\displaystyle{ 0\le i\lt j\le n-1 }[/math], since we only replace elements by smaller elements, eventually [math]\displaystyle{ \mathcal{F} }[/math] will stop changing, that is, [math]\displaystyle{ S_{ij}(\mathcal{F})=\mathcal{F} }[/math] for all [math]\displaystyle{ 0\le i\lt j\le n-1 }[/math]. We call such an [math]\displaystyle{ \mathcal{F} }[/math] shifted.

The idea behind the shifting technique is very natural: by applying shifting, all intersecting families are transformed to some special forms, and we only need to prove the theorem for these special form of intersecting families.

Proof of Erdős-Ko-Rado theorem (The original proof of Erdős-Ko-Rado by shifting) By the above lemma, it is sufficient to prove the Erdős-Ko-Rado theorem holds for shifted [math]\displaystyle{ \mathcal{F} }[/math]. We assume that [math]\displaystyle{ \mathcal{F} }[/math] is shifted. First, it is trivial to see that the theorem holds for [math]\displaystyle{ k=1 }[/math] (no matter whether shifted). Next, we show that the theorem holds when [math]\displaystyle{ n=2k }[/math] (no matter whether shifted). For any [math]\displaystyle{ S\in{X\choose k} }[/math], both [math]\displaystyle{ S }[/math] and [math]\displaystyle{ X\setminus S }[/math] are in [math]\displaystyle{ {X\choose k} }[/math], but at most one of them can be in [math]\displaystyle{ \mathcal{F} }[/math]. Thus, [math]\displaystyle{ |\mathcal{F}|\le\frac{1}{2}{n\choose k}=\frac{n!}{2k!(n-k)!}=\frac{(n-1)!}{(k-1)!(n-k)!}={n-1\choose k-1} }[/math]. We then apply the induction on [math]\displaystyle{ n }[/math]. For [math]\displaystyle{ n\gt 2k }[/math], the induction hypothesis is stated as: the Erdős-Ko-Rado theorem holds for any smaller [math]\displaystyle{ n }[/math]. Define [math]\displaystyle{ \mathcal{F}_0=\{S\in\mathcal{F}\mid n\not\in S\} }[/math] and [math]\displaystyle{ \mathcal{F}_1=\{S\in\mathcal{F}\mid n\in S\} }[/math]. Clearly, [math]\displaystyle{ \mathcal{F}_0\subseteq{[n-1]\choose k} }[/math] and [math]\displaystyle{ \mathcal{F}_0 }[/math] is intersecting. Due to the induction hypothesis, [math]\displaystyle{ |\mathcal{F}_0|\le{n-2\choose k-1} }[/math]. In order to apply the induction, we let [math]\displaystyle{ \mathcal{F}_1'=\{S\setminus\{n\}\mid S\in\mathcal{F}_1\} }[/math]. Clearly, [math]\displaystyle{ \mathcal{F}_1'\subseteq{[n-1]\choose k-1} }[/math]. If only it is also intersecting, we can apply the induction hypothesis, and indeed it is. To see this, by contradiction we assume that [math]\displaystyle{ \mathcal{F}_1' }[/math] is not intersecting. Then there must exist [math]\displaystyle{ A,B\in\mathcal{F} }[/math] such that [math]\displaystyle{ A\cap B=\{n\} }[/math], which means that [math]\displaystyle{ |A\cup B|\le 2k-1\lt n-1 }[/math]. Thus, there is some [math]\displaystyle{ 0\le i\le n-1 }[/math] such that [math]\displaystyle{ i\not\in A\cup B }[/math]. Since [math]\displaystyle{ \mathcal{F} }[/math] is shifted, [math]\displaystyle{ A_{in}=A\setminus\{n\}\cup\{i\}\in\mathcal{F} }[/math]. On the other hand it can be verified that [math]\displaystyle{ A_{in}\cap B=\emptyset }[/math], which contradicts that [math]\displaystyle{ \mathcal{F} }[/math] is intersecting. Thus, [math]\displaystyle{ \mathcal{F}_1'\subseteq{[n-1]\choose k-1} }[/math] and [math]\displaystyle{ \mathcal{F}_1' }[/math] is intersecting. Due to the induction hypothesis, [math]\displaystyle{ |\mathcal{F}_1'|\le{n-2\choose k-2} }[/math]. Combining these together, [math]\displaystyle{ |\mathcal{F}|=|\mathcal{F}_0|+|\mathcal{F}_1|=|\mathcal{F}_0|+|\mathcal{F}_1'|\le {n-2\choose k-1}+{n-2\choose k-2}={n-1\choose k-1} }[/math]. [math]\displaystyle{ \square }[/math]

Sperner system

A set family [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] with the relation [math]\displaystyle{ \subseteq }[/math] define a poset. Thus, a chain is a sequence [math]\displaystyle{ S_1\subseteq S_2\subseteq\cdots\subseteq S_k }[/math].

A set family [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] is an antichain (also called a Sperner system) if for all [math]\displaystyle{ S,T\in\mathcal{F} }[/math] that [math]\displaystyle{ S\neq T }[/math], we have [math]\displaystyle{ S\not\subseteq T }[/math].

The [math]\displaystyle{ k }[/math]-uniform [math]\displaystyle{ {X\choose k} }[/math] is an antichain. Let [math]\displaystyle{ n=|X| }[/math]. The size of [math]\displaystyle{ {X\choose k} }[/math] is maximized when [math]\displaystyle{ k=\lfloor n/2\rfloor }[/math]. We wonder whether this is also the largest possible size of any antichain [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math].

In 1928, Emanuel Sperner proved a theorem saying that it is indeed the largest possible antichain. This result, called Sperner's theorem today, initiated the studies of extremal set theory.

Theorem (Sperner 1928)

Let [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] where [math]\displaystyle{ |X|=n }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is an antichain, then [math]\displaystyle{ |\mathcal{F}|\le{n\choose \lfloor n/2\rfloor} }[/math].

First proof (shadows)

We first introduce the original proof by Sperner, which uses concepts called shadows and shades of set systems.

Definition

Let [math]\displaystyle{ |X|=n\, }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math], [math]\displaystyle{ k\lt n\, }[/math]. The shade of [math]\displaystyle{ \mathcal{F} }[/math] is defined to be [math]\displaystyle{ \nabla\mathcal{F}=\left\{T\in {X\choose k+1}\,\,\bigg|\,\, \exists S\in\mathcal{F}\mbox{ such that } S\subset T\right\} }[/math]. Thus the shade [math]\displaystyle{ \nabla\mathcal{F} }[/math] of [math]\displaystyle{ \mathcal{F} }[/math] consists of all subsets of [math]\displaystyle{ X }[/math] which can be obtained by adding an element to a set in [math]\displaystyle{ \mathcal{F} }[/math]. Similarly, the shadow of [math]\displaystyle{ \mathcal{F} }[/math] is defined to be [math]\displaystyle{ \Delta\mathcal{F}=\left\{T\in {X\choose k-1}\,\,\bigg|\,\, \exists S\in\mathcal{F}\mbox{ such that } T\subset S\right\} }[/math]. Thus the shadow [math]\displaystyle{ \Delta\mathcal{F} }[/math] of [math]\displaystyle{ \mathcal{F} }[/math] consists of all subsets of [math]\displaystyle{ X }[/math] which can be obtained by removing an element from a set in [math]\displaystyle{ \mathcal{F} }[/math].

Next lemma bounds the effects of shadows and shades on the sizes of set systems.

Lemma (Sperner)

Let [math]\displaystyle{ |X|=n\, }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math]. Then [math]\displaystyle{ \begin{align} &|\nabla\mathcal{F}|\ge\frac{n-k}{k+1}|\mathcal{F}| &\text{ if } k\lt n\\ &|\Delta\mathcal{F}|\ge\frac{k}{n-k+1}|\mathcal{F}| &\text{ if } k\gt 0. \end{align} }[/math]

Proof. The lemma is proved by double counting. We prove the inequality of [math]\displaystyle{ |\nabla\mathcal{F}| }[/math]. Assume that [math]\displaystyle{ 0\le k\lt n }[/math]. Define [math]\displaystyle{ \mathcal{R}=\{(S,T)\mid S\in\mathcal{F}, T\in\nabla\mathcal{F}, S\subset T\} }[/math]. We estimate [math]\displaystyle{ |\mathcal{R}| }[/math] in two ways. For each [math]\displaystyle{ S\in\mathcal{F} }[/math], there are [math]\displaystyle{ n-k }[/math] different [math]\displaystyle{ T\in\nabla\mathcal{F} }[/math] that [math]\displaystyle{ S\subset T }[/math]. [math]\displaystyle{ |\mathcal{R}|=(n-k)|\mathcal{F}| }[/math]. For each [math]\displaystyle{ T\in\nabla\mathcal{F} }[/math], there are [math]\displaystyle{ k+1 }[/math] ways to choose an [math]\displaystyle{ S\subset T }[/math] with [math]\displaystyle{ |S|=k }[/math], some of which may not be in [math]\displaystyle{ \mathcal{F} }[/math]. [math]\displaystyle{ |\mathcal{R}|\le (k+1)|\nabla\mathcal{F}| }[/math]. Altogether, we show that [math]\displaystyle{ |\nabla\mathcal{F}|\ge\frac{n-k}{k+1}|\mathcal{F}| }[/math]. The inequality of [math]\displaystyle{ |\Delta\mathcal{F}| }[/math] can be proved in the same way. [math]\displaystyle{ \square }[/math]

An immediate corollary of the previous lemma is as follows.

Proposition 1

If [math]\displaystyle{ k\le \frac{n-1}{2} }[/math], then [math]\displaystyle{ |\nabla\mathcal{F}|\ge|\mathcal{F}| }[/math]. If [math]\displaystyle{ k\ge \frac{n-1}{2} }[/math], then [math]\displaystyle{ |\Delta\mathcal{F}|\ge|\mathcal{F}| }[/math].

The idea of Sperner's proof is pretty clear:

• we "push up" all the sets in [math]\displaystyle{ \mathcal{F} }[/math] of size [math]\displaystyle{ \lt \frac{n-1}{2} }[/math] replacing them by their shades;
• and also "push down" all the sets in [math]\displaystyle{ \mathcal{F} }[/math] of size [math]\displaystyle{ \ge\frac{n+1}{2} }[/math] replacing them by their shadows.

Repeat this process we end up with a set system [math]\displaystyle{ \mathcal{F}\subseteq{X\choose \lfloor n/2\rfloor} }[/math]. We need to show that this process does not decrease the size of [math]\displaystyle{ \mathcal{F} }[/math].

Proposition 2

Suppose that [math]\displaystyle{ \mathcal{F}\subseteq2^X }[/math] where [math]\displaystyle{ |X|=n }[/math]. Let [math]\displaystyle{ \mathcal{F}_k=\mathcal{F}\cap{X\choose k} }[/math]. Let [math]\displaystyle{ k_\min }[/math] be the smallest [math]\displaystyle{ k }[/math] that [math]\displaystyle{ |\mathcal{F}_k|\gt 0 }[/math], and let [math]\displaystyle{ \mathcal{F}'=\begin{cases} \mathcal{F}\setminus\mathcal{F}_{k_\min}\cup \nabla\mathcal{F}_{k_\min} & \mbox{if }k_\min\lt \frac{n-1}{2},\\ \mathcal{F} & \mbox{otherwise.} \end{cases} }[/math] Similarly, let [math]\displaystyle{ k_\max }[/math] be the largest [math]\displaystyle{ k }[/math] that [math]\displaystyle{ |\mathcal{F}_k|\gt 0 }[/math], and let [math]\displaystyle{ \mathcal{F}''=\begin{cases} \mathcal{F}\setminus\mathcal{F}_{k_\max}\cup \Delta\mathcal{F}_{k_\max} & \mbox{if }k_\max\ge\frac{n+1}{2},\\ \mathcal{F} & \mbox{otherwise.} \end{cases} }[/math] If [math]\displaystyle{ \mathcal{F} }[/math] is an antichain, [math]\displaystyle{ \mathcal{F}' }[/math] and [math]\displaystyle{ \mathcal{F}'' }[/math] are antichains, and we have [math]\displaystyle{ |\mathcal{F}'|\ge|\mathcal{F}| }[/math] and [math]\displaystyle{ |\mathcal{F}''|\ge|\mathcal{F}| }[/math].

Proof. We show that [math]\displaystyle{ \mathcal{F}' }[/math] is an antichain and [math]\displaystyle{ |\mathcal{F}'|\ge|\mathcal{F}| }[/math]. First, observe that [math]\displaystyle{ \nabla\mathcal{F}_k\cap\mathcal{F}=\emptyset }[/math], otherwise [math]\displaystyle{ \mathcal{F} }[/math] cannot be an antichain, and due to Proposition 1, [math]\displaystyle{ |\nabla\mathcal{F}_k|\ge|\mathcal{F}_k| }[/math] when [math]\displaystyle{ k\le \frac{n-1}{2} }[/math], so [math]\displaystyle{ |\mathcal{F}'|=|\mathcal{F}|-|\mathcal{F}_k|+|\nabla\mathcal{F}_k|\ge |\mathcal{F}| }[/math]. Now we prove that [math]\displaystyle{ \mathcal{F}' }[/math] is an antichain . By contradiction, assume that there are [math]\displaystyle{ S, T\in \mathcal{F}' }[/math], such that [math]\displaystyle{ S\subset T }[/math]. One of the [math]\displaystyle{ S,T }[/math] must be in [math]\displaystyle{ \nabla\mathcal{F}_{k_\min} }[/math], or otherwise [math]\displaystyle{ \mathcal{F} }[/math] cannot be an antichain. Recall that [math]\displaystyle{ k_\min }[/math] is the smallest [math]\displaystyle{ k }[/math] that [math]\displaystyle{ |\mathcal{F}_k|\gt 0 }[/math], thus it must be [math]\displaystyle{ S\in \nabla\mathcal{F}_{k_\min} }[/math], and [math]\displaystyle{ T\in\mathcal{F} }[/math]. This implies that there is an [math]\displaystyle{ R\in \mathcal{F}_{k_\min}\subseteq \mathcal{F} }[/math] such that [math]\displaystyle{ R\subset S\subset T }[/math], which contradicts that [math]\displaystyle{ \mathcal{F} }[/math] is an antichain. The statement for [math]\displaystyle{ \mathcal{F}'' }[/math] can be proved in the same way. [math]\displaystyle{ \square }[/math]

Applying the above process, we prove the Sperner's theorem.

Proof of Sperner's theorem (original proof of Sperner) Let [math]\displaystyle{ \mathcal{F}_k=\{S\in\mathcal{F}\mid |S|=k\} }[/math], where [math]\displaystyle{ 0\le k\le n }[/math]. We change [math]\displaystyle{ \mathcal{F} }[/math] as follows: for the smallest [math]\displaystyle{ k }[/math] that [math]\displaystyle{ |\mathcal{F}_k|\gt 0 }[/math], if [math]\displaystyle{ k\lt \frac{n-1}{2} }[/math], replace [math]\displaystyle{ \mathcal{F}_k }[/math] by [math]\displaystyle{ \nabla\mathcal{F}_k }[/math]. Due to Proposition 2, this procedure preserves [math]\displaystyle{ \mathcal{F} }[/math] as an antichain and does not decrease [math]\displaystyle{ |\mathcal{F}| }[/math]. Repeat this procedure, until [math]\displaystyle{ |\mathcal{F}_k|=0 }[/math] for all [math]\displaystyle{ k\lt \frac{n-1}{2} }[/math], that is, there is no member set of [math]\displaystyle{ \mathcal{F} }[/math] has size less than [math]\displaystyle{ \frac{n-1}{2} }[/math]. We then define another symmetric procedure: for the largest [math]\displaystyle{ k }[/math] that [math]\displaystyle{ |\mathcal{F}_k|\gt 0 }[/math], if [math]\displaystyle{ k\ge\frac{n+1}{2} }[/math], replace [math]\displaystyle{ \mathcal{F}_k }[/math] by [math]\displaystyle{ \Delta\mathcal{F}_k }[/math]. Also due to Proposition 2, this procedure preserves [math]\displaystyle{ \mathcal{F} }[/math] as an antichain and does not decrease [math]\displaystyle{ |\mathcal{F}| }[/math]. After repeatedly applying this procedure, [math]\displaystyle{ |\mathcal{F}_k|=0 }[/math] for all [math]\displaystyle{ k\ge\frac{n+1}{2} }[/math]. The resulting [math]\displaystyle{ \mathcal{F} }[/math] has [math]\displaystyle{ \mathcal{F}\subseteq{X\choose \lfloor n/2\rfloor} }[/math], and since [math]\displaystyle{ |\mathcal{F}| }[/math] is never decreased, for the original [math]\displaystyle{ \mathcal{F} }[/math], we have [math]\displaystyle{ |\mathcal{F}|\le {n\choose \lfloor n/2\rfloor} }[/math]. [math]\displaystyle{ \square }[/math]

Second proof (counting)

We now introduce an elegant proof due to Lubell. The proof uses a counting argument, and tells more information than just the size of the set system.

Proof of Sperner's theorem (Lubell 1966) Let [math]\displaystyle{ \pi }[/math] be a permutation of [math]\displaystyle{ X }[/math]. We say that an [math]\displaystyle{ S\subseteq X }[/math] prefixes [math]\displaystyle{ \pi }[/math], if [math]\displaystyle{ S=\{\pi_1,\pi_2,\ldots, \pi_{|S|}\} }[/math], that is, [math]\displaystyle{ S }[/math] is precisely the set of the first [math]\displaystyle{ |S| }[/math] elements in the permutation [math]\displaystyle{ \pi }[/math]. Fix an [math]\displaystyle{ S\subseteq X }[/math]. It is easy to see that the number of permutations [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ X }[/math] prefixed by [math]\displaystyle{ S }[/math] is [math]\displaystyle{ |S|!(n-|S|)! }[/math]. Also, since [math]\displaystyle{ \mathcal{F} }[/math] is an antichain, no permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ X }[/math] can be prefixed by more than one members of [math]\displaystyle{ \mathcal{F} }[/math], otherwise one of the member sets must contain the other, which contradicts that [math]\displaystyle{ \mathcal{F} }[/math] is an antichain. Thus, the number of permutations [math]\displaystyle{ \pi }[/math] prefixed by some [math]\displaystyle{ S\in\mathcal{F} }[/math] is [math]\displaystyle{ \sum_{S\in\mathcal{F}}|S|!(n-|S|)! }[/math], which cannot be larger than the total number of permutations, [math]\displaystyle{ n! }[/math], therefore, [math]\displaystyle{ \sum_{S\in\mathcal{F}}|S|!(n-|S|)!\le n! }[/math]. Dividing both sides by [math]\displaystyle{ n! }[/math], we have [math]\displaystyle{ \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}=\sum_{S\in\mathcal{F}}\frac{|S|!(n-|S|)!}{n!}\le 1 }[/math], where [math]\displaystyle{ {n\choose |S|}\le {n\choose \lfloor n/2\rfloor} }[/math], so [math]\displaystyle{ \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\ge \frac{|\mathcal{F}|}{{n\choose \lfloor n/2\rfloor}} }[/math]. Combining this with the above inequality, we prove the Sperner's theorem. [math]\displaystyle{ \square }[/math]

The LYM inequality

Lubell's proof proves the following inequality:

[math]\displaystyle{ \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\le 1 }[/math]

which is actually stronger than Sperner's original statement that [math]\displaystyle{ |\mathcal{F}|\le{n\choose \lfloor n/2\rfloor} }[/math].

This inequality is independently discovered by Lubell-Yamamoto, Meschalkin, and Bollobás, and is called the LYM inequality today.

Theorem (Lubell, Yamamoto 1954; Meschalkin 1963)

Let [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] where [math]\displaystyle{ |X|=n }[/math]. If [math]\displaystyle{ \mathcal{F} }[/math] is an antichain, then [math]\displaystyle{ \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\le 1 }[/math].

In Lubell's counting argument proves the LYM inequality, which implies the Sperner's theorem. Here we give another proof of the LYM inequality by the probabilistic method, due to Noga Alon.

Third proof (the probabilistic method) (Due to Alon.) Let [math]\displaystyle{ \pi }[/math] be a uniformly random permutation of [math]\displaystyle{ X }[/math]. Define a random maximal chain by [math]\displaystyle{ \mathcal{C}_\pi=\{\{\pi_i\mid 1\le i\le k\}\mid 0\le k\le n\} }[/math]. For any [math]\displaystyle{ S\in\mathcal{F} }[/math], let [math]\displaystyle{ X_S }[/math] be the 0-1 random variable which indicates whether [math]\displaystyle{ S\in\mathcal{C}_\pi }[/math], that is [math]\displaystyle{ X_S=\begin{cases} 1 & \mbox{if }S\in\mathcal{C}_\pi,\\ 0 & \mbox{otherwise.} \end{cases} }[/math] Note that for a uniformly random [math]\displaystyle{ \pi }[/math], [math]\displaystyle{ \mathcal{C}_\pi }[/math] has exact one member set of size [math]\displaystyle{ |S| }[/math], uniformly distributed over [math]\displaystyle{ {X\choose |S|} }[/math], thus [math]\displaystyle{ \mathbf{E}[X_S]=\Pr[S\in\mathcal{C}_\pi]=\frac{1}{{n\choose |S|}} }[/math]. Let [math]\displaystyle{ X=\sum_{S\in\mathcal{F}}X_S }[/math]. Note that [math]\displaystyle{ X=|\mathcal{F}\cap\mathcal{C}_\pi| }[/math]. By the linearity of expectation, [math]\displaystyle{ \mathbf{E}[X]=\sum_{S\in\mathcal{F}}\mathbf{E}[X_S]=\sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}} }[/math]. On the other hand, since [math]\displaystyle{ \mathcal{F} }[/math] is an antichain, it can never intersect a chain at more than one elements, thus we always have [math]\displaystyle{ X=|\mathcal{F}\cap\mathcal{C}_\pi|\le 1 }[/math]. Therefore, [math]\displaystyle{ \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\le \mathbf{E}[X] \le 1 }[/math]. [math]\displaystyle{ \square }[/math]

The Sperner's theorem is an immediate consequence of the LYM inequality.

Proposition

[math]\displaystyle{ \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\le 1 }[/math] implies that [math]\displaystyle{ |\mathcal{F}|\le{n\choose \lfloor n/2\rfloor} }[/math].

Proof. It holds that [math]\displaystyle{ {n\choose k}\le {n\choose \lfloor n/2\rfloor} }[/math] for any [math]\displaystyle{ k }[/math]. Thus, [math]\displaystyle{ 1\ge \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\ge \frac{|\mathcal{F}|}{{n\choose \lfloor n/2\rfloor}} }[/math], which implies that [math]\displaystyle{ |\mathcal{F}|\le {n\choose \lfloor n/2\rfloor} }[/math]. [math]\displaystyle{ \square }[/math]