Let $(L,\leq,0,1)$ be a lattice, and let's denote by $JI(L)$ the set of its join-irreducibles (i.e. elements that are not the lowest grater bound of two other elements). We suppose that $\sup JI(L)=1$. and that $\mathbb P$ is a probability on $L$ such that ideals and filters are measurable and singleton have probability 0. Let's call such a lattice a "reach-irreducible lattice" (RIB). If the cardinality of $L$ is that of the continuum" we can say "continuum reach-irreducible lattice" (CRIL)

Does there exist in any $L$ that is a continuum-reach-irreducible lattice, $g\in JI(L)$ such that $\mathbb P(\left\{x\in L,\, g\leq x\right\})<1$


Related discussions on the MSE are https://math.stackexchange.com/questions/2737618/lattice-generated-by-join-irreducible-elements-such-that-any-principal-filter-c and https://math.stackexchange.com/questions/2742357/probability-martin-axiom-and-the-weak-frankl-conjecture

The first link is an imperfect attempt to fit with the motivation of this question, that is explained in the second link. If the answer to this question is yes, then a weak version of the Frankl Conjecture ("WFC")- that is also an open problem - is true (see MSE links).

Note that in both links, the lattices that I built assuming that WFC is wrong are not exactly CRIL, but up to small details, one can easily get CRIL from them.

Note also that the cardinality condition is just given to fit with WFC, but it would be nice indeed if the answer of the question is yes for any RIL. And if it is not, it might also give some useful informations, so I would also be happy to read answers and comments about general RIL.

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    $\begingroup$ Yes I realized I was wrong, that's why I deleted the comment. However - take $(0,1)\times(0,1)$ with area measure, and with extra top (and bottom if you need). Then if I am not mistaken this top will be the only join-irreducible, no? $\endgroup$ Apr 19 '18 at 5:20
  • $\begingroup$ Yes, but it doesn't seem to me as a counterexample because then the mesure of the set of its grater bounds is zero $\endgroup$
    – jcdornano
    Apr 19 '18 at 5:27
  • $\begingroup$ But if $\mathbb P({\downarrow}g)=1$ then necessarily $\mathbb P({\uparrow}g)=0$ because ${\uparrow}g\cap{\downarrow}g=\{g\}$ has measure zero $\endgroup$ Apr 19 '18 at 7:27
  • $\begingroup$ So it is a CRIL but I don't understand how this answer the question. Do you mean it's "yes" according to you? Then you have to prove if for all CRIL. $\endgroup$
    – jcdornano
    Apr 19 '18 at 8:16
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    $\begingroup$ I thought I understood correctly. Don't you ask if there is $L$ such that $\mathbb P(\leqslant g)=1$ for all join-irreducible $g$? Because it is so in my example: there is only one join-irreducible, which is top. $\endgroup$ Apr 19 '18 at 9:04

I think the answer is no.

Let $(p_i)_{n\in \mathbb N}$ be a sequence of real number such that $\Pi p_i>0$ and let's make the supposition that we can get for any $i\in \mathbb N$ some finite lattice $(L_i,\leq,0_i,1_i)$ that the set of non zero minimal elements is $M_i$, such that $\sup M_i=1_i$ and such that for any $m\in M_i$ we have $\mathbb P_i(m\vee L_i)=:|m\vee L_i|/|L_i|>p_i$. We will see later how to get the conditions of the supposition.

If $L= L_1\times L_2\times...$ is equiped with the structure of lattice product , and $\mathbb P$ is the probability product of the $\mathbb P_i$.

We now define a pre-order on $L$ : for any $(a_1,a_2,...)\in L$ and for any $b\in L$ :

$a\leq^* b$ iff $a=(0,0,..., a_k,a_{k+1},...)\leq b$ for some $k\in \mathbb N$.

For each integer $i$, Let's denote by $L_i^*$ the non zero elements of $L_i$ and by $=^*$ the equivalence relation associate to $\leq^*$, where $a^*$ will be the equivalent class of $a\in L$ and $\sup^*$ the lowest upper bound relatively to $\leq^*$

$L^*=:(L_1^*\times L_2^*...\cup\left\{0\right\})/=^*$ is a lattice and for any $A\subset L^*$ the equality $\mathbb P^*(A)=\mathbb P(\bigcup A)$ defines a probability on $L^*$ .

Note that if $m^*\in JI(L^*)$, then $m=(m_1,m_2...)\in L_1\times L_2...\times M_k\times M_{k+1}...$ for some integer $k$.

Indeed, for any sub-sequence $(m_{i_n})$ of $(m_i)$ such that $m_{i_n}\notin M_{i_n}$ one can chose respectively in $M_{i_{2n}}$ and $M_{i_{2n+1}}$, $m'_{i_{2n}}\leq m_{i_{2n}}$ and $m''_{i_{2n+1}}\leq m_{i_{2n+1}}$ and define $m'_j=m''_j=m_j$ for any $j\in \mathbb N\setminus\left\{i_n,\,n\in \mathbb N\right\}$ is not equal to $i_n$ for some $n$. Then $\sup^*\left\{m'^*,m''^*\right\}=m^*\notin JI(L^*)$. We now use the Lebesgue dominated convergence theorem to state that $\mathbb P(\left\{x\in L,m\leq^*x\right\})=\sup_{k\in \mathbb N}\Pi_{i>k}p_i=1$

We also note that $\sup M_i=1_i$ implies that $\sup^* M_1\times M_2...=1^*$.

We now just have to find for each integer $i$, $L_i$ such that we get the supposition

It is sufficient to have $p_i\to 1$ because if we don't have the non zero convergence of their infinite product we still can take a fine sub-sequence of the $p_i$ that has the property, and take the corresponding sub-sequence of finite lattices .

Let's défine :

$M_n:=\left\{[(k-1)2^{2^n}\,,\,k2^{2^n}]\subset \mathbb N, 1\leq k\leq n\right\}$

$P_n:=\left\{[1\,,\,n2^{2^n}]\setminus\left\{k\right\}\subset \mathbb N, 1\leq k\leq n2^{2^n}\right\}$

We define $L^*_n$ to be the union closure of $M_n\cup P_n$, and $L_n:=L_n\cup\left\{\emptyset\right\}$ is a lattice for inclusion. Note that $|L_n|-|M_n\cup P_n|=2^n+1$ witch is small compare to $|M_n\cup P_n|=n+n2^{2^n}$

If $\mathbb P_n$ is the uniform probability on $L_n$ we see that for any $m\in M$, $\mathbb P_n(m\cup L)=(n2^{2^n}-2^{n-1})/(n2^{2^n}+2^n+n)=:p_n\to 1$ when $n\to \infty$


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