# Lattice generated by join-irreducible elements, such that any principal filter contains almost any random element of $L$

Let's denote by $JI(L)$ the set of join-irreducible elements of a lattice $L$, i.e elements $g\in L$ such that for any $a,b\in L$, if $g=a\vee b$ then $g\in \left\{a,b\right\}$. Let's say that $L$ is a $\alpha$-good lattice iff

1) It's equipped with a probability mesure, such that principle filters are mesurable sets, and every singleton has mesure $0$.

2) Any element of $L$ is the lower upper bound of some subset of $JI(R)$ that cardinality is strictly smaller than the continuum

Does there exist a good lattice $L$ such that for any $g\in JI(L)$, almost every element of $L$ is greater than $g$ ?

Note that if $JI(L)$ is a countable subset of $L$ the answer is no.

Motivation Let's say that a finite lattice $(L,\leq)$ is an $\epsilon$-lattice if there exists $g\in JI(L)$, such that the (uniform) probability that a random element of $L$ is grater than $g$ is lower then $1-\epsilon$. The Frankl conjecture says that any finite lattice is a $1/2$-lattice. Even the question : "does there exists $\epsilon>0$ such that any finite lattice is an $\epsilon$-lattice" is an open question for $\epsilon\leq 1/2$. (we will call it the WFC (weak Frankl conjecture). If WFC is wrong than we can find $(L_i,\leq ,0_i)_{i\in \mathbb N}$ and equip $L:=L_0\times L_1\times L_2....$, with the product mesure $\mathbb P$ of the uniform probabilty mesure on each $L_i$, in such a way that for any $g=(g_0,g_1....)\in JI(L_0)\times JI(L_1)...$, we have $sup_{j\in \mathbb N}\mathbb P(x\in g^j)=1$ (***)

where $g^j= \left\{x\in L,\, (0_0,0_1,...0_{j-1},g_j,g_{j+1}...)\leq x\right\}$. If we consider $M$ the sublattice of $L$ such that no projetion on some $L_i$ is $0_i$, and if we say that for any $a,b\in M$, $a\leq_M b$ iff $a^k\leq b$ for some $k\in \mathbb N$, then the quotient $M^*$ obtained from $M\cup\left\{0\right\}$ by identifying $a$ and $b$ iff $a\leq_M b$ and $b\leq_M a$, is a good lattice when you equip it with the $\mathbb P^*$ in such a way that $\mathbb P^*(A)=\mathbb P(\left\{x\in L,\,\exists y\in A,\,\, y\leq_M x\,\,and\,\, x\leq_M y\right\})$. And if I'm not mistaking, the (***) condition implies that for any $g\in JI(M^*)$ we have $\mathbb P^*(x>g)=1$.

[the problem is that $M^*$ is not generated by $JI(M^*)$ (it is not even complete as a lattice), so in order make the question fit better with the motivation, one should put instead of condition 2) a weaker hypothesis and just ask "new good lattice" to satisfy $\sup M^*=\sup K$ for some $K\subset JI(M^*)$, such that $|K|<|\mathbb R|$. I opened a new post, more specific to the link between the (same) motivation and a question (a bit different) : Probability , Martin Axiom, and the "weak" Frankl Conjecture

Let $L=\omega_1$ with its usual order. Equip it with the $\sigma$-algebra of countable or cocountable sets, and the probability measure that assigns measure $0$ to every countable set and $1$ to every cocountable set. Then every element of $L$ is join-irreducible, and it is easy to see it satisfies all your conditions.
• Thank you very much! I'm going as I said in the end of motivation part, to make a new post matching with the motivation. The argument in it "might" be useful anyway, and the nice) example you give is not destroying "hope" yet, because $M^*$ has a grater element , and it is the major argument from whom I "hope" to get a contradiction (with a definitely "new good lattice" necessary definition) Indeed, the same question can still be asked if $L$ is required to have a greater element. Thank you again for this simple and natural answer. (when I write "hope" it is more like an "abstract hope" – jcdornano Apr 17 '18 at 22:56
• If $L$ has a greatest element then the answer is trivially no if CH holds, since then the greatest element is the sup of some countable set of join-irreducible elements. It's not clear to me why you care about the cardinality restriction in condition (2) though--is there a natural counterexample without it? – Eric Wofsey Apr 17 '18 at 23:03
• Ok. Let's say a "fine lattice" $(L,0,1)$ is a lattice such that $\sup K=1$ for some $K\subset JI(L)$, with $|K|<|\mathbb R|$ . Then if $\mathbb P(g\vee L)=1$ for any $g\in JI(L)$, I'm hoping to use Martin Axiom to get a contradiction. If I'm not mistaking in the use of MA, and if I can get that $M^*$ is a "fine lattice" then WFC seems to be proved, but I might have made a mistake somewhere... (because I think I can use the "tower number" to get a canonical way to get that $M^*$ is "fine" and that would be too good to be true...) – jcdornano Apr 17 '18 at 23:17
• I think that would work. That statement you've been told using MA is not true for arbitrary probability spaces, though, just for Lebesgue measure (but I think your $M^*$ is sufficiently close to Lebesgue measure that it would apply, though I haven't checked the details). – Eric Wofsey Apr 18 '18 at 0:19