M=13 case of FUNC

On this page, we attempt a proof of the m=13 case of Frankl's conjecture. This will probably end up being a long case analysis, with the cases being tackled out of order, so bear with me while the page is in progress. In all results below, we will try to prove that the set [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's (satisfies the conjecture), but will assume the opposite. We will use the terminology defined in the paper "The 11 element case of Frankl's Conjecture" by Ivica Bosnjak and Petar Markovic.

The smallest size of a non-empty set must be 3, 4, 5, or 6, as if there were a set of size 1 or 2 in [math]\displaystyle{ \mathcal{A} }[/math], it would be Frankl's. Also, if all the nonempty sets have size 7 or larger, then the average set size is greater than 13/2=6.5, so [math]\displaystyle{ \mathcal{A} }[/math] is Frankl's (to prove this more rigorously, use the uniform weight function over all the elements in X (the ground set).

Case 4: The smallest set has size 6

In this case we will assume that the set is 123456 (omitting commas and brackets for simplicity) use the weight function w(x)=2 for x<=6, and w(x)=1 otherwise. With K as the bottom set we consider the set of sets in the interval [K, KU{1,2,3,4,5,6}] with K removed from each set. Going through each size of K, we have:

|K|=0:

The only sets in here are the [math]\displaystyle{ C_0 }[/math] set and the [math]\displaystyle{ C_6 }[/math] set. The deficit of the [math]\displaystyle{ C_0 }[/math] set is 9, and the surplus of the [math]\displaystyle{ C_6 }[/math] set is 3 (weight 12). So the deficit for this case is 6 more than the surplus.

|K|=1, 2 or 3:

Since all nonempty sets have size at least 6, the only sets in here have at least 3 elements from the original 123456 set, and there is no deficit.