# Stronger than Frankl conjecture, with no "closure hypothesis"

Let $$n=\left\{0,1,...,n-1\right\}$$, $$n>1$$, and $$\mathcal F=\left\{F_i\subset n+1,\, i\in n \right\}$$ such that, for any $$i\in n$$,

$$i\in F_i\subset i+1,\,\,\,\,\,(*)$$

main question :

Does there exists $$l\in n$$ such that for any $$i,j\in n\setminus \left\{l\right\}$$, $$F_i\cap F_j\ne F_l$$, and $$|F_l| ?

I'm hoping for a counterexample, and not for demonstration of a "yes" answer because this "yes" implies the Frankl Conjecture. Indeed, if $$(G,\leq)=(\left\{G_i,i\in n+1\right\},\leq)$$ with $$G_n=0_G$$ is a lattice and you take as $$F_i=\left\{j\in i+1,\, G_j\leq G_i\right\}$$, then $$\left\{0,F_0,...,F_{n-1}\right\}$$ is an intersection closed family (that satisfies the condition $$(*)$$ up to a fine indexation of elements in $$G$$. $$(\mathcal F,\subset)$$ is then a lattice that is isomorphic to $$G$$. If the answer to the question is "yes", then you will find some meet irréductible $$G_i$$ with at most $$n/2$$ lower elements. (The Frankl conjecture is known to be equivalent to the existence of such a $$G_i$$ in any lattice that cardinality is $$n$$.)

So the question seems much stronger than Frankl conjecture. So I expect it to "fail", never the less, there is an interesting intermediary question that can be asked if answer to main question is found to be "no".

Question 2

If $$(H,\leq)$$ is a partial order, does there exists, $$h\in H$$, such that $$h$$ is not the greater lower bound of any pair included in $$H\setminus\left\{h\right\}$$, (let's say that $$H$$ is then "quasi-irreductible") and such that $$|\left\{x\in H, x\leq h\right\}|<|H|/2$$.

If $$H$$ is a lattice, then "quasi-irreductible" and "irreductible" is the same thing.

But answer 2, in the general case, would not be "yes" , if we replace "any pair" by "any subset" :

Take any set $$M$$ of cardinality $$n$$ that elements are subset of $$4n$$ and such that each $$m\in M$$ has cardinality $$3n$$ suppose also that the smallest intersection closed set that contain $$M$$ also contain $$S$$ the set of singleton of $$4n$$. (it is not hard to build such a set if $$n$$ is big enought let's say $$n>4$$ for example). Now consider $$M'=M\cup S$$, than each member of $$M'$$ is a quasi-irreductible and the question 2 is yes in this trivial case, because any element of $$S\subset M$$' has zero member that are smaller then them, witch is less then $$|M'|/2$$. But if your definition of $$h$$ irreductible is "$$h$$ is not the greater lower bound of some subset of $$M'\setminus\left\{h\right\}$$" than the irreductible are exactly elements of $$M$$, each one of them has cardinality $$3n$$ and then contain $$3n$$ members of $$S\subset M'$$, and now... $$3n>|M'|/2=(4n+n)/2$$.

• The very first notation is very confusing. Jan 27 '19 at 23:18
• The question is : if M is a triangular matrix with only "zero" and "one" in entry and with "one" on the diagonal, is there a row that is not the "conjonction" of two other rows, and that has at most n/2 "ones" . (the conjonction of two rows, is the multiplication coordonate by coordonate, of the two rows, like a multiplication in a ring product, that the two rows can be seen in : ex : the conjonction of 11001 and 10101 is 10001) Jan 30 '19 at 11:54
• If we ask the triangular matrix to be the adjacency matrix of a lattice (up to the 0 rows and the zero column) we get the Frankl conjecture. Actually I forget that i put this post, and then i posted about a year later , i think, the "matrix version" that i just wrote in the upper comment. I will give the link of this (then redondant ) post, as soon as I get a PC (i'm now on the telephone) Jan 30 '19 at 12:01

This is a fairly sorry counterexample to the main conjecture, but I do think it is one. Take $$n=2$$, $$F_{0}=\{0\}$$, $$F_{1}=\{1\}$$. Then neither $$0$$ or $$1$$ satisfy the conditions you're looking for on $$l$$.
Similarly, for Question 2, in a two-element anti-chain with $$x\leq x$$, $$y\leq y$$, but no relation between the two, there's no such $$h$$.