组合数学 (Spring 2015)/Extremal set theory: Difference between revisions

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== Sperner system ==
A set family <math>\mathcal{F}\subseteq 2^X</math> with the relation <math>\subseteq</math> define a poset. Thus, a '''chain''' is a sequence <math>S_1\subseteq S_2\subseteq\cdots\subseteq S_k</math>.
A set family <math>\mathcal{F}\subseteq 2^X</math> is an '''antichain''' (also called a '''Sperner system''') if for all <math>S,T\in\mathcal{F}</math> that <math>S\neq T</math>, we have <math>S\not\subseteq T</math>.
The <math>k</math>-uniform <math>{X\choose k}</math> is an antichain. Let <math>n=|X|</math>. The size of <math>{X\choose k}</math> is maximized when <math>k=\lfloor n/2\rfloor</math>. We wonder whether this is also the largest possible size of any antichain <math>\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|Theorem (Sperner 1928)|
:Let <math>\mathcal{F}\subseteq 2^X</math> where <math>|X|=n</math>. If <math>\mathcal{F}</math> is an antichain, then
::<math>|\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.
{{Theorem|Definition|
:Let <math>|X|=n\,</math> and <math>\mathcal{F}\subseteq {X\choose k}</math>, <math>k<n\,</math>.
:The '''shade''' of <math>\mathcal{F}</math> is defined to be
::<math>\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>\nabla\mathcal{F}</math> of <math>\mathcal{F}</math> consists of all subsets of <math>X</math> which can be obtained by adding an element to a set in <math>\mathcal{F}</math>.
:Similarly, the '''shadow''' of <math>\mathcal{F}</math> is defined to be
::<math>\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>\Delta\mathcal{F}</math> of <math>\mathcal{F}</math> consists of all subsets of <math>X</math> which can be obtained by removing an element from a set in <math>\mathcal{F}</math>.
}}
Next lemma bounds the effects of shadows and shades on the sizes of set systems.
{{Theorem|Lemma (Sperner)|
:Let <math>|X|=n\,</math> and <math>\mathcal{F}\subseteq {X\choose k}</math>. Then
::<math>
\begin{align}
&|\nabla\mathcal{F}|\ge\frac{n-k}{k+1}|\mathcal{F}| &\text{ if } k<n\\
&|\Delta\mathcal{F}|\ge\frac{k}{n-k+1}|\mathcal{F}| &\text{ if } k>0.
\end{align}
</math>
}}
{{Proof|
The lemma is proved by double counting. We prove the inequality of <math>|\nabla\mathcal{F}|</math>. Assume that <math>0\le k<n</math>.
Define
:<math>\mathcal{R}=\{(S,T)\mid S\in\mathcal{F}, T\in\nabla\mathcal{F}, S\subset T\}</math>.
We estimate <math>|\mathcal{R}|</math> in two ways.
For each <math>S\in\mathcal{F}</math>, there are <math>n-k</math> different <math>T\in\nabla\mathcal{F}</math> that <math>S\subset T</math>.
:<math>|\mathcal{R}|=(n-k)|\mathcal{F}|</math>.
For each <math>T\in\nabla\mathcal{F}</math>, there are <math>k+1</math> ways to choose an <math>S\subset T</math> with <math>|S|=k</math>, some of which may not be in <math>\mathcal{F}</math>.
:<math>|\mathcal{R}|\le (k+1)|\nabla\mathcal{F}|</math>.
Altogether, we show that <math>|\nabla\mathcal{F}|\ge\frac{n-k}{k+1}|\mathcal{F}|</math>.
The inequality of <math>|\Delta\mathcal{F}|</math> can be proved in the same way.
}}
An immediate corollary of the previous lemma is as follows.
{{Theorem|Proposition 1|
:If <math>k\le \frac{n-1}{2}</math>, then <math>|\nabla\mathcal{F}|\ge|\mathcal{F}|</math>.
:If <math>k\ge \frac{n-1}{2}</math>, then <math>|\Delta\mathcal{F}|\ge|\mathcal{F}|</math>.
}}
The idea of Sperner's proof is pretty clear:
* we "push up" all the sets in <math>\mathcal{F}</math> of size <math><\frac{n-1}{2}</math> replacing them by their shades;
* and also "push down" all the sets in <math>\mathcal{F}</math> of size <math>\ge\frac{n+1}{2}</math> replacing them by their shadows.
Repeat this process we end up with a set system <math>\mathcal{F}\subseteq{X\choose \lfloor n/2\rfloor}</math>. We need to show that this process does not decrease the size of <math>\mathcal{F}</math>.
{{Theorem|Proposition 2|
:Suppose that <math>\mathcal{F}\subseteq2^X</math> where <math>|X|=n</math>. Let <math>\mathcal{F}_k=\mathcal{F}\cap{X\choose k}</math>. Let <math>k_\min</math> be the smallest <math>k</math> that <math>|\mathcal{F}_k|>0</math>, and let
::<math>
\mathcal{F}'=\begin{cases}
\mathcal{F}\setminus\mathcal{F}_{k_\min}\cup \nabla\mathcal{F}_{k_\min} & \mbox{if }k_\min<\frac{n-1}{2},\\
\mathcal{F} & \mbox{otherwise.}
\end{cases}
</math>
:Similarly, let <math>k_\max</math> be the largest <math>k</math> that <math>|\mathcal{F}_k|>0</math>, and let
::<math>
\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>\mathcal{F}</math> is an antichain, <math>\mathcal{F}'</math> and <math>\mathcal{F}''</math> are antichains, and we have <math>|\mathcal{F}'|\ge|\mathcal{F}|</math> and <math>|\mathcal{F}''|\ge|\mathcal{F}|</math>.
}}
{{Proof|
We show that <math>\mathcal{F}'</math> is an antichain and <math>|\mathcal{F}'|\ge|\mathcal{F}|</math>.
First, observe that <math>\nabla\mathcal{F}_k\cap\mathcal{F}=\emptyset</math>, otherwise <math>\mathcal{F}</math> cannot be an antichain, and due to Proposition 1, <math>|\nabla\mathcal{F}_k|\ge|\mathcal{F}_k|</math> when <math>k\le \frac{n-1}{2}</math>, so <math>|\mathcal{F}'|=|\mathcal{F}|-|\mathcal{F}_k|+|\nabla\mathcal{F}_k|\ge |\mathcal{F}|</math>.
Now we prove that <math>\mathcal{F}'</math> is an antichain . By contradiction, assume that there are <math>S, T\in \mathcal{F}'</math>, such that <math>S\subset T</math>. One of the <math>S,T</math> must be in <math>\nabla\mathcal{F}_{k_\min}</math>, or otherwise <math>\mathcal{F}</math> cannot be an antichain. Recall that <math>k_\min</math> is the smallest <math>k</math> that <math>|\mathcal{F}_k|>0</math>, thus it must be <math>S\in \nabla\mathcal{F}_{k_\min}</math>, and <math>T\in\mathcal{F}</math>. This implies that there is an <math>R\in \mathcal{F}_{k_\min}\subseteq \mathcal{F}</math> such that <math>R\subset S\subset T</math>, which contradicts that <math>\mathcal{F}</math> is an antichain.
The statement for <math>\mathcal{F}''</math> can be proved in the same way.
}}
Applying the above process, we prove the Sperner's theorem.
{{Prooftitle|Proof of Sperner's theorem | (original proof of Sperner)
Let <math>\mathcal{F}_k=\{S\in\mathcal{F}\mid |S|=k\}</math>, where <math>0\le k\le n</math>.
We change <math>\mathcal{F}</math> as follows:
* for the smallest <math>k</math> that <math>|\mathcal{F}_k|>0</math>, if <math>k<\frac{n-1}{2}</math>, replace <math>\mathcal{F}_k</math> by <math>\nabla\mathcal{F}_k</math>.
Due to Proposition 2, this procedure preserves <math>\mathcal{F}</math> as an antichain and does not decrease <math>|\mathcal{F}|</math>. Repeat this procedure, until <math>|\mathcal{F}_k|=0</math> for all <math>k<\frac{n-1}{2}</math>, that is, there is no member set of <math>\mathcal{F}</math> has size less than <math>\frac{n-1}{2}</math>.
We then define another symmetric procedure:
* for the largest <math>k</math> that <math>|\mathcal{F}_k|>0</math>, if <math>k\ge\frac{n+1}{2}</math>, replace <math>\mathcal{F}_k</math> by <math>\Delta\mathcal{F}_k</math>.
Also due to Proposition 2, this procedure preserves <math>\mathcal{F}</math> as an antichain and does not decrease <math>|\mathcal{F}|</math>. After repeatedly applying this procedure, <math>|\mathcal{F}_k|=0</math> for all <math>k\ge\frac{n+1}{2}</math>.
The resulting <math>\mathcal{F}</math> has <math>\mathcal{F}\subseteq{X\choose \lfloor n/2\rfloor}</math>, and since <math>|\mathcal{F}|</math> is never decreased, for the original <math>\mathcal{F}</math>, we have
:<math>|\mathcal{F}|\le {n\choose \lfloor n/2\rfloor}</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.
{{Prooftitle|Proof of Sperner's theorem | (Lubell 1966)
Let <math>\pi</math> be a permutation of <math>X</math>. We say that an <math>S\subseteq X</math> '''prefixes''' <math>\pi</math>, if <math>S=\{\pi_1,\pi_2,\ldots, \pi_{|S|}\}</math>, that is, <math>S</math> is precisely the set of the first <math>|S|</math> elements in the permutation <math>\pi</math>.
Fix an <math>S\subseteq X</math>. It is easy to see that the number of permutations <math>\pi</math> of <math>X</math> prefixed by <math>S</math> is <math>|S|!(n-|S|)!</math>.  Also, since <math>\mathcal{F}</math> is an antichain, no permutation <math>\pi</math> of <math>X</math> can be prefixed by more than one members of <math>\mathcal{F}</math>, otherwise one of the member sets must contain the other, which contradicts that <math>\mathcal{F}</math> is an antichain. Thus, the number of permutations <math>\pi</math> prefixed by some <math>S\in\mathcal{F}</math> is
:<math>\sum_{S\in\mathcal{F}}|S|!(n-|S|)!</math>,
which cannot be larger than the total number of permutations, <math>n!</math>, therefore,
:<math>\sum_{S\in\mathcal{F}}|S|!(n-|S|)!\le n!</math>.
Dividing both sides by <math>n!</math>, we have
:<math>\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>{n\choose |S|}\le {n\choose \lfloor n/2\rfloor}</math>, so
:<math>\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.
}}
=== The LYM inequality ===
Lubell's proof proves the following inequality:
:<math>\sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\le 1</math>
which is actually stronger than Sperner's original statement that <math>|\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|Theorem (Lubell, Yamamoto 1954; Meschalkin 1963)|
:Let <math>\mathcal{F}\subseteq 2^X</math> where <math>|X|=n</math>. If <math>\mathcal{F}</math> is an antichain, then
::<math>\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.
{{Prooftitle|Third proof (the probabilistic method)| (Due to Alon.)
Let <math>\pi</math> be a uniformly random permutation of <math>X</math>. Define a random maximal chain by
:<math>\mathcal{C}_\pi=\{\{\pi_i\mid 1\le i\le k\}\mid 0\le k\le n\}</math>.
For any <math>S\in\mathcal{F}</math>, let <math>X_S</math> be the 0-1 random variable which indicates whether <math>S\in\mathcal{C}_\pi</math>, that is
:<math>
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>\pi</math>, <math>\mathcal{C}_\pi</math> has exact one member set of size <math>|S|</math>, uniformly distributed over <math>{X\choose |S|}</math>, thus
:<math>\mathbf{E}[X_S]=\Pr[S\in\mathcal{C}_\pi]=\frac{1}{{n\choose |S|}}</math>.
Let <math>X=\sum_{S\in\mathcal{F}}X_S</math>. Note that <math>X=|\mathcal{F}\cap\mathcal{C}_\pi|</math>. By the linearity of expectation,
:<math>\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>\mathcal{F}</math> is an antichain, it can never intersect a chain at more than one elements, thus we always have <math>X=|\mathcal{F}\cap\mathcal{C}_\pi|\le 1</math>. Therefore,
:<math>\sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\le \mathbf{E}[X] \le 1</math>.
}}
The Sperner's theorem is an immediate consequence of the LYM inequality.
{{Theorem|Proposition|
:<math>\sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}\le 1</math> implies that <math>|\mathcal{F}|\le{n\choose \lfloor n/2\rfloor}</math>.
}}
{{Proof|
It holds that <math>{n\choose k}\le {n\choose \lfloor n/2\rfloor}</math> for any <math>k</math>. Thus,
:<math>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>|\mathcal{F}|\le {n\choose \lfloor n/2\rfloor}</math>.
}}
== Sunflowers ==
== Sunflowers ==
An set system is a '''sunflower''' if all its member sets intersect at the same set of elements.
An set system is a '''sunflower''' if all its member sets intersect at the same set of elements.

Revision as of 06:36, 7 December 2015

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\le 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
  1. [math]\displaystyle{ |S_{ij}(T)|=|T|\, }[/math] and [math]\displaystyle{ |S_{ij}(\mathcal{F})|=\mathcal{F} }[/math];
  2. 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]
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