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


1 Answer 1


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.

  • $\begingroup$ 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" $\endgroup$
    – jcdornano
    Apr 17, 2018 at 22:56
  • $\begingroup$ ...I mean "theorical hope" because I'm not "practically" expecting to prove WFC.. but maybe learn from kind of "not working arguments", about the difficulty of the problem, or maybe give some ideas to stronger people... $\endgroup$
    – jcdornano
    Apr 17, 2018 at 23:00
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    $\begingroup$ 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? $\endgroup$ Apr 17, 2018 at 23:03
  • $\begingroup$ 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...) $\endgroup$
    – jcdornano
    Apr 17, 2018 at 23:17
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    $\begingroup$ 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). $\endgroup$ Apr 18, 2018 at 0:19

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