# Probability , Martin Axiom, and the “weak” Frankl Conjecture

This is the continuation of a previous question asked here Lattice generated by join-irreducible elements, such that any principal filter contains almost any random element of $L$ - Note that I did here some small changes with conventions and notations, compare to that of the upper link.

Let $(L_i,\leq,0_i,1_i,\mathbb P_i)$ be for each $i\in \mathbb N$ a finite lattice, together with the uniform probability on $L_i$. Let's call $I_i$ the set of the joint-irréductibles including zero éléments of each $L_i$.

$L=L_0\times L_1\times...$ is then naturally equipped with the "product order" still denoted by $\leq$ that make $(L,0,1,\mathbb P)$ a (complete) lattice together with $\mathbb P$, the product probability of the $\mathbb P_i$.

But we now equip $L$ with the following pre-order defined for any $a,b\in L$ as follow :

$a=:(a_0,a_1,...)\leq^* (b_0,b_1...)=b$ iff there exists $k\in \mathbb N$ such that $a^{k\to}:=(0,0,...a_k,a_{k+1},...)<b$.

We now make two hypothesis

(1) :

for any $g\in I=I_0\times I_1\times I_2...$ we have $\mathbb P(\left\{x\in L,\, g\leq^* x\right\})=1$

(2) :

there exists $K\subset I$ such that $\left\{x\in L, \forall k\in K, k\leq x\right\}=:\sup^*K=\sup^*L:=1^*\subset L$ ,

and $|K|<|\mathbb R|$

It is clear that if CH holds, then you cannot have hypotheses (1) and hypotheses (2) at the same time. But "I've been told" that assuming MA, infinite cardinality lower than the continuum kind of act like that of $\mathbb N$... (you can look at "comments" with Eric Worsey following his smart answer in the link that stand at the begin of this post... ) Then here is :

First question :

Assuming Martin Axiom, can we still state that assuming (1) and (2) leads to a contradiction ?

second question :

Is there some model of ZFC+MA where (2) is true ?

The second question will then prove that there is a model where (1) is false. And I think that it will prove that (1) is wrong in ZFC but I'm not really sure about that, and I will be happy enough with (1) being false in some model of ZFC. And the reason for this is that (1) is equivalent to the negation of a weak form of Frankl conjecture that is also an open problem (let's call it WFC):

WFC : There exists $\alpha\in [1/2,1[$ such that for any finite lattice $L$ there exists $g\in L$ that is join-irreducible such that $|g\vee L|/|L|\leq \alpha$

If first question happens to be true, I will give an idea getting (2) true by using the "tower number", but I don't want to get this post heavy with informations that relevance depends strongly on first question in the first hand, and that I still need to check in detail, on the other hand...

Here I show why the MA argument cannot work in a very similar construction : I will take the very same construction except that I won't need to suppose that WFC is wrong. The only hypotheses I do on the $L_i$ is that elements of $I_i$, the set of join minimal-elements of $L_i$ are such that

(i) $\sup I_i=1_i$

(ii) For any $m\in M_i$, $|m\vee L_i|/|L_i|=p_i$ with $\Pi_{i\in \mathbb N}p_i>0$.

Not that $I_i$ is no longer the set of irreducible elements, I took the same letter for "minimal" elements so that the formula (1) and (2) in the question are still holding (for the new $L_i$ that I can get without assuming any open problem or axiom : such a construction really exists and is at the end of the answer I gave here (no need to read the entier post) :

Let's then assume that with the same notations and definitions (except for $I$) : We are allowed to say that :

If $(Z_{\alpha})_{\alpha<\beta}$ with $\beta<\mathbb R$ is such that for any ordinal $\alpha<\beta$, $Z_{\alpha}\subset L$ and $\mathbb P(Z_{\alpha})=0$, then $\mathbb P(\bigcup (Z_\alpha)_{\alpha<\beta})=0$

In order to get a contradiction.

Let's first notice that for any $a\in L\setminus 1^*$ we have $P(<^*a):=\mathbb P(\left\{x\in L|\, x\leq^*a\right\})=0$.

Indeed, (2) says that for any $a\in L$ there exist $m\in I$ such that $m\leq^* a$ is false. Then $a$ is in the complement of the set of grater bounds of $m$, and then by (1), we get that $P(<^*a)=0$.

Let's now build an increasing family that elements are such $P(<^*a)$, that lowest grater bound is $1^*$, and that cardinality is strictly lower than the continuum (4), assuming that $\mathfrak t$ ("the tower number") is strictly lower that the continuum (witch is true in some model independently of MA, if I'm not mistaking).

The tower number is, by definition , the cardinality of the smallest $<^*$-increasing family in $(2^{\mathbb N},1^*,0^*,\leq^*)$ that top is $1^*$ and (the common construction for $<^*$ if we take $L_i=2$)

So let $(t_{\alpha})_{\alpha\in \mathfrak t}$ be this family. Now we chose a for each integer $i$, $a_i\in L_i$, and for any $t_{\alpha}=^* (t_0,t_1,...,t_i,...)$ we define $t'_{\alpha}=^*(t'_0,t'1,...,t'_i,...)$ such that for each integer $i$, we have $t'_i=a_i$ if $t_i=0$ and $t'_i=1_i$ if $t_i=1$.

the $P(<^*t'_{\alpha})$ for $\alpha<\mathfrak t<|\mathbb R|$ are as expected in (4) and the "yellow statement" wrong for some $<^*$-lattice