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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 $|F_l|<n/2+1$ ?


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... $3n>|M'|/2=(4n+n)/2$.

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    $\begingroup$ The very first notation is very confusing. $\endgroup$ – Fedor Petrov Jan 27 at 23:18
  • $\begingroup$ 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) $\endgroup$ – jcdornano Jan 30 at 11:54
  • $\begingroup$ 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) $\endgroup$ – jcdornano Jan 30 at 12:01
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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 $h$.

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  • $\begingroup$ thank you, I made a mistake that I'm going to correct,the inequality is supposed to be large (like in Frankl conjecture). Sorry again. $\endgroup$ – jcdornano Jan 30 at 11:44
  • $\begingroup$ I added 1 to the right member wich is at least weaker than the large inequality (but in fact equivalent) $\endgroup$ – jcdornano Jan 30 at 11:46

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