Combinatorics (Fall 2010)/Extremal set theory: Difference between revisions
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== Erdős–Ko–Rado theorem == | == Erdős–Ko–Rado theorem == | ||
== Kruskal–Katona theorem == | == Kruskal–Katona theorem == |
Revision as of 08:50, 23 October 2010
Sperner system
Theorem (Sperner 1928) - Let [math]\displaystyle{ |S|=n }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq 2^S }[/math] be an antichain. Then
- [math]\displaystyle{ |\mathcal{F}|\le{n\choose \lfloor n/2\rfloor} }[/math].
- Let [math]\displaystyle{ |S|=n }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq 2^S }[/math] be an antichain. Then
First proof (symmetric chain decomposition)
Proof of Sperner's theorem - [math]\displaystyle{ \square }[/math]
Second proof (shadowing)
Definition - Let [math]\displaystyle{ |S|=n\, }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq {S\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\{A\in {S\choose k+1}\,\,\bigg|\,\, \exists B\in\mathcal{F}\mbox{ such that } B\subset A\right\} }[/math].
- Thus the shade [math]\displaystyle{ \nabla\mathcal{F} }[/math] of [math]\displaystyle{ \mathcal{F} }[/math] consists of all subsets of [math]\displaystyle{ S }[/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\{A\in {S\choose k-1}\,\,\bigg|\,\, \exists B\in\mathcal{F}\mbox{ such that } A\subset B\right\} }[/math].
- Thus the shadow [math]\displaystyle{ \Delta\mathcal{F} }[/math] of [math]\displaystyle{ \mathcal{F} }[/math] consists of all subsets of [math]\displaystyle{ S }[/math] which can be obtained by removing an element from a set in [math]\displaystyle{ \mathcal{F} }[/math].
Lemma (Sperner) - Let [math]\displaystyle{ |S|=n\, }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq {S\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]
- Let [math]\displaystyle{ |S|=n\, }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq {S\choose k} }[/math]. Then
Proof of Sperner's theorem (original proof of Sperner)
- [math]\displaystyle{ \square }[/math]
Third 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{ |S|=n }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq 2^S }[/math] be an antichain. For [math]\displaystyle{ k=0,1,\ldots,n }[/math], let [math]\displaystyle{ f_k=|\{A\in\mathcal{F}\mid |A|=k\}| }[/math]. Then
- [math]\displaystyle{ \sum_{A\in\mathcal{F}}\frac{1}{{n\choose |A|}}=\sum_{k=0}^n\frac{f_k}{{n\choose k}}\le 1 }[/math].
- Let [math]\displaystyle{ |S|=n }[/math] and [math]\displaystyle{ \mathcal{F}\subseteq 2^S }[/math] be an antichain. For [math]\displaystyle{ k=0,1,\ldots,n }[/math], let [math]\displaystyle{ f_k=|\{A\in\mathcal{F}\mid |A|=k\}| }[/math]. Then
Another proof (the probabilistic method) - [math]\displaystyle{ \square }[/math]
Proposition - [math]\displaystyle{ \sum_{A\in\mathcal{F}}\frac{1}{{n\choose |A|}}\le 1 }[/math] implies that [math]\displaystyle{ |\mathcal{F}|\le{n\choose \lfloor n/2\rfloor} }[/math].