EKR Sets for Large n and r

Let A⊂([n]r) be a compressed, intersecting family and let X⊂[n]. Let A(X)={A∈A:A∩X≠∅} and Sn,r=([n]r)({1}). Motivated by the Erdős-Ko-Rado theorem, Borg asked for which X⊂[2,n] do we have |A(X)|≤|Sn,r(X)| for all compressed, intersecting families A? We call X that satisfy this property EKR. Borg c...

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Bibliographic Details
Main Authors: Bond, Benjamin (Author), Bond, Benjamin R. (Contributor)
Other Authors: Massachusetts Institute of Technology. Department of Mathematics (Contributor)
Format: Article
Language:English
Published: Springer Japan, 2017-06-23T15:37:50Z.
Subjects:
Online Access:Get fulltext
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100 1 0 |a Bond, Benjamin  |e author 
100 1 0 |a Massachusetts Institute of Technology. Department of Mathematics  |e contributor 
100 1 0 |a Bond, Benjamin R.  |e contributor 
700 1 0 |a Bond, Benjamin R.  |e author 
245 0 0 |a EKR Sets for Large n and r 
260 |b Springer Japan,   |c 2017-06-23T15:37:50Z. 
856 |z Get fulltext  |u http://hdl.handle.net/1721.1/110210 
520 |a Let A⊂([n]r) be a compressed, intersecting family and let X⊂[n]. Let A(X)={A∈A:A∩X≠∅} and Sn,r=([n]r)({1}). Motivated by the Erdős-Ko-Rado theorem, Borg asked for which X⊂[2,n] do we have |A(X)|≤|Sn,r(X)| for all compressed, intersecting families A? We call X that satisfy this property EKR. Borg classified EKR sets X such that |X|≥r. Barber classified X, with |X|≤r, such that X is EKR for sufficiently large n, and asked how large n must be. We prove n is sufficiently large when n grows quadratically in r. In the case where A has a maximal element, we sharpen this bound to n>φ2r implies |A(X)|≤|Sn,r(X)|. We conclude by giving a generating function that speeds up computation of |A(X)| in comparison with the naïve methods. 
520 |a National Science Foundation (U.S.) (Grant No. 1062709) 
520 |a United States. Department of Defense (Grant No. 1062709) 
520 |a United States. National Security Agency (Grant Number H98230-11-1-0224) 
546 |a en 
655 7 |a Article 
773 |t Graphs and Combinatorics