Supersaturation for Intersecting Families Shagnik Das University of California, Los Angeles / ETH Z¨ urich April 1, 2014 Joint work with Wenying Gan and Benny Sudakov
Supersaturation for Intersecting Families Shagnik Das University of California, Los Angeles / ETH Z¨ urich April Fools’ Day, 2014 Joint work with Wenying Gan and Benny Sudakov
Background Counting Probabilistic Our results Conclusion Intersecting families Definition (Intersecting families) A family of sets F is said to be intersecting if F 1 ∩ F 2 � = ∅ for all F 1 , F 2 ∈ F .
Background Counting Probabilistic Our results Conclusion Intersecting families Definition (Intersecting families) A family of sets F is said to be intersecting if F 1 ∩ F 2 � = ∅ for all F 1 , F 2 ∈ F . Notation: [ n ] = { 1 , 2 , . . . , n } - ground set for our set families � [ n ] � = { F ⊂ [ n ] : | F | = k } k � [ n ] � = { F ⊂ [ n ] : | F | ≥ k } ≥ k
Background Counting Probabilistic Our results Conclusion Non-uniform families Observation If a family F on [ n ] is intersecting, |F| ≤ 2 n − 1 .
Background Counting Probabilistic Our results Conclusion Non-uniform families Observation If a family F on [ n ] is intersecting, |F| ≤ 2 n − 1 . Proof: F can contain at most one of F , F c for all F ⊂ [ n ].
Background Counting Probabilistic Our results Conclusion Non-uniform families Observation If a family F on [ n ] is intersecting, |F| ≤ 2 n − 1 . Proof: F can contain at most one of F , F c for all F ⊂ [ n ]. Constructions: Star : F = { F ⊂ [ n ] : 1 ∈ F }
Background Counting Probabilistic Our results Conclusion Non-uniform families Observation If a family F on [ n ] is intersecting, |F| ≤ 2 n − 1 . Proof: F can contain at most one of F , F c for all F ⊂ [ n ]. Constructions: Star : F = { F ⊂ [ n ] : 1 ∈ F } Large sets : � [ n ] � � � [ n ] � � F = ∪ F ∈ : 1 ∈ F ≥ n +1 n 2 2
Background Counting Probabilistic Our results Conclusion Non-uniform families Observation If a family F on [ n ] is intersecting, |F| ≤ 2 n − 1 . Proof: F can contain at most one of F , F c for all F ⊂ [ n ]. Constructions: Star : F = { F ⊂ [ n ] : 1 ∈ F } Large sets : � [ n ] � � � [ n ] � � F = ∪ F ∈ : 1 ∈ F ≥ n +1 n 2 2 Large sets make intersections easier
Background Counting Probabilistic Our results Conclusion Erd˝ os-Ko-Rado Theorem (Erd˝ os-Ko-Rado, 1961) � [ n ] � n − 1 � � Suppose n ≥ 2 k, and F ⊂ is intersecting. Then |F| ≤ . k k − 1
Background Counting Probabilistic Our results Conclusion Erd˝ os-Ko-Rado Theorem (Erd˝ os-Ko-Rado, 1961) � [ n ] � n − 1 � � Suppose n ≥ 2 k, and F ⊂ is intersecting. Then |F| ≤ . k k − 1 Extremal families: stars
Background Counting Probabilistic Our results Conclusion Erd˝ os-Ko-Rado Theorem (Erd˝ os-Ko-Rado, 1961) � [ n ] � n − 1 � � Suppose n ≥ 2 k, and F ⊂ is intersecting. Then |F| ≤ . k k − 1 Extremal families: stars 1 A star with centre 1
Background Counting Probabilistic Our results Conclusion Beyond the threshold Previous results answer the typical extremal problem Question How large can a structure be without containing a forbidden configuration?
Background Counting Probabilistic Our results Conclusion Beyond the threshold Previous results answer the typical extremal problem Question How large can a structure be without containing a forbidden configuration? Gives rise to natural extension Question How many forbidden configurations must appear in larger structures?
Background Counting Probabilistic Our results Conclusion One extra set Warm up : How many disjoint pairs must a family of 2 n − 1 + 1 sets contain?
Background Counting Probabilistic Our results Conclusion One extra set Warm up : How many disjoint pairs must a family of 2 n − 1 + 1 sets contain? Answer :
Background Counting Probabilistic Our results Conclusion One extra set Warm up : How many disjoint pairs must a family of 2 n − 1 + 1 sets contain? Answer : 1!
Background Counting Probabilistic Our results Conclusion One extra set Warm up : How many disjoint pairs must a family of 2 n − 1 + 1 sets contain? Answer : 1! Any maximal intersecting family can be extended with only one disjoint pair:
Background Counting Probabilistic Our results Conclusion One extra set Warm up : How many disjoint pairs must a family of 2 n − 1 + 1 sets contain? Answer : 1! Any maximal intersecting family can be extended with only one disjoint pair: Let F be an intersecting family of 2 n − 1 sets
Background Counting Probabilistic Our results Conclusion One extra set Warm up : How many disjoint pairs must a family of 2 n − 1 + 1 sets contain? Answer : 1! Any maximal intersecting family can be extended with only one disjoint pair: Let F be an intersecting family of 2 n − 1 sets For minimal F 0 ∈ F , add F c 0 = [ n ] \ F 0
Background Counting Probabilistic Our results Conclusion One extra set Warm up : How many disjoint pairs must a family of 2 n − 1 + 1 sets contain? Answer : 1! Any maximal intersecting family can be extended with only one disjoint pair: Let F be an intersecting family of 2 n − 1 sets For minimal F 0 ∈ F , add F c 0 = [ n ] \ F 0 G ∩ F c 0 = ∅ ⇒ G ⊂ F 0 ⇒ G = F 0
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2 Theorem (Frankl, 1977; Ahlswede, 1980) � �� � �� [ n ] � [ n ] � Suppose � ≤ m ≤ � . Then the minimum number of � � � � ≥ k +1 ≥ k � � disjoint pairs in a family of m sets is attained by some F with [ n ] � [ n ] � � � ⊆ F ⊆ . ≥ k +1 ≥ k
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2 Theorem (Frankl, 1977; Ahlswede, 1980) � �� � �� [ n ] � [ n ] � Suppose � ≤ m ≤ � . Then the minimum number of � � � � ≥ k +1 ≥ k � � disjoint pairs in a family of m sets is attained by some F with [ n ] � [ n ] � � � ⊆ F ⊆ . ≥ k +1 ≥ k . . . . . . n − 3 n − 1 n +1 n +3 n − 1 n 0 1 2 2 2 2
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2 Theorem (Frankl, 1977; Ahlswede, 1980) � �� � �� [ n ] � [ n ] � Suppose � ≤ m ≤ � . Then the minimum number of � � � � ≥ k +1 ≥ k � � disjoint pairs in a family of m sets is attained by some F with [ n ] � [ n ] � � � ⊆ F ⊆ . ≥ k +1 ≥ k . . . . . . n − 3 n − 1 n +1 n +3 n − 1 n 0 1 2 2 2 2
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2 Theorem (Frankl, 1977; Ahlswede, 1980) � �� � �� [ n ] � [ n ] � Suppose � ≤ m ≤ � . Then the minimum number of � � � � ≥ k +1 ≥ k � � disjoint pairs in a family of m sets is attained by some F with [ n ] � [ n ] � � � ⊆ F ⊆ . ≥ k +1 ≥ k . . . . . . n − 3 n − 1 n +1 n +3 n − 1 n 0 1 2 2 2 2
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2 Theorem (Frankl, 1977; Ahlswede, 1980) � �� � �� [ n ] � [ n ] � Suppose � ≤ m ≤ � . Then the minimum number of � � � � ≥ k +1 ≥ k � � disjoint pairs in a family of m sets is attained by some F with [ n ] � [ n ] � � � ⊆ F ⊆ . ≥ k +1 ≥ k . . . . . . n − 3 n − 1 n +1 n +3 n − 1 n 0 1 2 2 2 2
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2 Theorem (Frankl, 1977; Ahlswede, 1980) � �� � �� [ n ] � [ n ] � Suppose � ≤ m ≤ � . Then the minimum number of � � � � ≥ k +1 ≥ k � � disjoint pairs in a family of m sets is attained by some F with [ n ] � [ n ] � � � ⊆ F ⊆ . ≥ k +1 ≥ k . . . . . . n − 3 n − 1 n +1 n +3 n − 1 n 0 1 2 2 2 2
Background Counting Probabilistic Our results Conclusion Many extra sets � [ n ] � Previous argument: construction best to extend ≥ n +1 2 Theorem (Frankl, 1977; Ahlswede, 1980) � �� � �� [ n ] � [ n ] � Suppose � ≤ m ≤ � . Then the minimum number of � � � � ≥ k +1 ≥ k � � disjoint pairs in a family of m sets is attained by some F with [ n ] � [ n ] � � � ⊆ F ⊆ . ≥ k +1 ≥ k . . . . . . n − 3 n − 1 n +1 n +3 n − 1 n 0 1 2 2 2 2
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