Combinatorics (Fall 2010)/Extremal set theory: Difference between revisions

From TCS Wiki
Jump to navigation Jump to search
imported>WikiSysop
imported>WikiSysop
Line 50: Line 50:
==Intersecting Families ==
==Intersecting Families ==
=== Sunflowers ===
=== Sunflowers ===
{{Theorem|Definition (sunflower)|
: A set family <math>\mathcal{F}\subseteq 2^X</math> is a sunflower of size <math>r</math> with a core <math>C\subseteq X</math> if
::<math>\forall S,T\in\mathcal{F}</math> that <math>S\neq T</math>, <math>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>\emptyset</math>).
{{Theorem|Sunflower Lemma|
{{Theorem|Sunflower Lemma|
:Let <math>\mathcal{F}\subseteq {X\choose k}</math>. If <math>|\mathcal{F}|>k!(r-1)^k</math>, then <math>\mathcal{F}</math> contains a sunflower of size  <math>r</math>.
:Let <math>\mathcal{F}\subseteq {X\choose k}</math>. If <math>|\mathcal{F}|>k!(r-1)^k</math>, then <math>\mathcal{F}</math> contains a sunflower of size  <math>r</math>.
}}
{{Proof|
We proceed by induction on <math>k</math>. For <math>k=1</math>, <math>\mathcal{F}\subseteq{X\choose 1}</math>, thus all sets in <math>\mathcal{F}</math> are disjoint. And since <math>|\mathcal{F}|>r-1</math>, we can choose <math>r</math> of these sets and form a sunflower.
Now let <math>k\ge 2</math> and assume the lemma holds for all smaller <math>k</math>. Take a maximal family <math>\mathcal{G}\subseteq \mathcal{F}</math> whose members are disjoint, i.e. for any <math>S,T\in \mathcal{G}</math> that <math>S\neq T</math>, <math>S\cap T=\emptyset</math>.
If <math>|\mathcal{G}|\ge r</math>, then <math>\mathcal{G}</math> is a sunflower of size at least <math>r</math> and we are done.
Assume that <math>|\mathcal{G}|\le r-1</math>, and let <math>Y=\bigcup_{S\in\mathcal{G}}S</math>. Then <math>|Y|=k|\mathcal{G}|\le k(r-1)</math> (since all members of <math>\mathcal{G}</math>) are disjoint). We claim that <math>Y</math> intersets all members of <math>\mathcal{F}</math>, since if otherwise, there exists an <math>S\in\mathcal{F}</math> such that <math>S\cap Y=\emptyset</math>, then we can enlarge <math>\mathcal{G}</math> by adding <math>S</math> into <math>\mathcal{G}</math> and still have all members of <math>\mathcal{G}</math> disjoint, which contradicts the assumption that <math>\mathcal{G}</math> is the maximum of such families.
By the pigeonhole principle, some elements <math>y\in Y</math> must contained in at least
:<math>\frac{|\mathcal{F}|}{|Y|}>\frac{k!(r-1)^k}{k(r-1)}=(k-1)!(r-1)^{k-1}</math>
members of <math>\mathcal{F}</math>. We delete this <math>y</math> from these sets and consider the family
:<math>\mathcal{H}=\{S\setminus\{y\}\mid S\in\mathcal{F}\wedge y\in S\}</math>.
We have <math>\mathcal{H}\subseteq {X\choose k-1}</math> and <math>|\mathcal{H}|>(k-1)!(r-1)^{k-1}</math>, thus by the induction hypothesis, <math>\mathcal{H}</math>contains a sunflower of size <math>r</math>. Adding <math>y</math> to the members of this sunflower, we get the desired sunflower in the original family <math>\mathcal{F}</math>.
}}
}}



Revision as of 03:13, 28 October 2010

Sperner system

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)

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].
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 of Sperner's theorem
(original proof of Sperner)
[math]\displaystyle{ \square }[/math]

Second proof (double counting)

Proof of Sperner's theorem
(Lubell 1966)
[math]\displaystyle{ \square }[/math]

The LYM inequality

Theorem (Lubell, Yamamoto 1954; Meschalkin 1963)
Let [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] where [math]\displaystyle{ |X|=n }[/math], and let [math]\displaystyle{ f_k=|\{S\in\mathcal{F}\mid |S|=k\}| }[/math], for [math]\displaystyle{ k=0,1,\ldots,n }[/math].
If [math]\displaystyle{ \mathcal{F} }[/math] is an antichain, then
[math]\displaystyle{ \sum_{S\in\mathcal{F}}\frac{1}{{n\choose |S|}}=\sum_{k=0}^n\frac{f_k}{{n\choose k}}\le 1 }[/math].
Third proof (the probabilistic method)
(Due to Alon.)
[math]\displaystyle{ \square }[/math]
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].

Intersecting Families

Sunflowers

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]).

Sunflower Lemma
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]

Erdős–Ko–Rado theorem

Erdős–Ko–Rado theorem
Let [math]\displaystyle{ \mathcal{F}\subseteq {X\choose k} }[/math] where [math]\displaystyle{ |X|=n }[/math]. If for any [math]\displaystyle{ S,T\in\mathcal{F} }[/math], [math]\displaystyle{ S\cap T\neq\emptyset }[/math], then
[math]\displaystyle{ |\mathcal{F}|\le{n-1\choose k-1} }[/math].

Katona's proof

Erdős' shifting technique

Definition (shifting)
For [math]\displaystyle{ |X|=n }[/math], [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] and [math]\displaystyle{ 1\le i\lt j\le n }[/math], define the [math]\displaystyle{ (i,j) }[/math]-shift [math]\displaystyle{ S_{ij} }[/math] by [math]\displaystyle{ S_{ij}(\mathcal{F})=\{S_{ij}(T)\mid T\in \mathcal{F}\} }[/math], where
[math]\displaystyle{ S_{ij}(T)= \begin{cases} (T\setminus\{j\})\cup\{i\} & \mbox{if }j\in T, i\not\in T, \mbox{ and }(T\setminus\{j\})\cup\{i\} \not\in\mathcal{F},\\ T & \mbox{otherwise.} \end{cases} }[/math]
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].

Sauer's lemma and VC-dimension

Definition
Let [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] be set family and let [math]\displaystyle{ R\subseteq X }[/math] be a subset. The restriction of [math]\displaystyle{ \mathcal{F} }[/math] on [math]\displaystyle{ R }[/math], denoted [math]\displaystyle{ \mathcal{F}|_R }[/math] is defined as
[math]\displaystyle{ \mathcal{F}|_R=\{S\cap R\mid S\in\mathcal{F}\} }[/math].
We say that [math]\displaystyle{ \mathcal{F} }[/math] shatters [math]\displaystyle{ R }[/math] if [math]\displaystyle{ \mathcal{F}|_R=2^R }[/math], i.e. for all [math]\displaystyle{ T\subseteq R }[/math], there exists an [math]\displaystyle{ S\in\mathcal{F} }[/math] such that [math]\displaystyle{ T=S\cap R }[/math].
Sauer's Lemma (Sauer; Shelah-Perles; Vapnik-Chervonenkis)
Let [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math] where [math]\displaystyle{ |X|=n }[/math]. If [math]\displaystyle{ |\mathcal{F}|\gt \sum_{1\le i\lt k}{n\choose i} }[/math], then there exists an [math]\displaystyle{ R\in{X\choose k} }[/math] such that [math]\displaystyle{ \mathcal{F} }[/math] shatters [math]\displaystyle{ R }[/math].
Definition (VC-dimension)
The Vapnik–Chervonenkis dimension (VC-dimension) of a set family [math]\displaystyle{ \mathcal{F}\subseteq 2^X }[/math], denoted [math]\displaystyle{ \text{VC-dim}(\mathcal{F}) }[/math], is the size of the largest [math]\displaystyle{ R\subseteq X }[/math] shattered by [math]\displaystyle{ \mathcal{F} }[/math].