Let $a\in L$, where $L$ is a graded (if needed) distributive lattice. Let $x_1, \ldots, x_k$ - the set of elements which cover $a$ ($x$ covers $a$ if $a < x$ and there is no element $t$ such that $a < t < x$). Denote least upper bound of $x_1, \ldots, x_k$ as $b$. Consider sublattice $[a, b]$. Is it true (and if it is, how to prove it?) that greatest lower bound of elements covered by $b$ is $a$ (in sublattice $[a,b]$)?
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$\begingroup$ What is a graded lattice? Is this condition important? $\endgroup$– BerciAug 1, 2019 at 9:55
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$\begingroup$ en.wikipedia.org/wiki/Graded_poset In my case the lattice is graded, don't know if it helps. $\endgroup$– Ordev AgensAug 1, 2019 at 10:36
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$\begingroup$ I might have a counterexample: the poset $P(\{1,2,3\}) \setminus \{\{1,3\}\} $, ordered by inclusion. It's a graded lattice, however I'm not sure if it's distributive. $\endgroup$– BerciAug 1, 2019 at 12:52
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$\begingroup$ @Berci It isn't distributive. The elements $\varnothing, \{1\},\{1,2\},\{3\},\{1,2,3\}$ form a pentagon. $\endgroup$– amrsaAug 1, 2019 at 13:14
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$\begingroup$ Ohh, then it's neither modular, hence grading must fail, too.. $\endgroup$– BerciAug 1, 2019 at 14:20
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2 Answers
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Let $\rho : L \to \mathbb N$ be the rank function of $L$.
Then the interval $[a,b]$ has finite length, given by $\rho(b) - \rho(a)$.
Since $[a,b]$ is a sublattice of $L$, it is distributive.
A distributive lattice is finite iff it has finite length.
Notice that the elements which cover $a$, in $L$, are the atoms of $[a,b]$;
analogously the elements covered by $b$ which are above $a$ are co-atoms of $[a,b]$.
So we're left with the task of proving that in a finite distributive lattice, if the join of the atoms is $1$, then the meet of the co-atoms is $0$.
That just follows from the fact that if the join of the atoms is $1$, then those are the only join-irreducible elements of the lattice, which is then Boolean, and it is clear that in a Boolean lattice the meet of the co-atoms is $0$.
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$\begingroup$ Thank you very much for this beautiful proof! $\endgroup$ Aug 2, 2019 at 10:58
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$\begingroup$ @OrdevAgens I'm glad to help. I'm must also add that I'm not sure that being graded is strictly necessary, that is, I admit there is a proof of the result you want without using that hypothesis, but that one eluded me, and I had to use all the hypothesis available. Good luck! $\endgroup$– amrsaAug 2, 2019 at 13:16
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$\begingroup$ Ahh, Boolean algebra! So, working in $[a, b]$, which is distributive, $x_1,\dots, x_k$ will be the atoms and $\bigvee_i x_i=1$. Then for any $u$, $u=\bigvee_i(x_i\land u)$, thus every element is a union of atoms, and hence it is indeed going to be a Boolean algebra. $\endgroup$– BerciAug 4, 2019 at 21:53
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$\begingroup$ @Berci Yes! As as noted in the previous comment to the OP, this question eluded me for a while. I still think that being a graded lattice might not be strictly necessary, but it works as it is, so I let it be like that... $\endgroup$– amrsaAug 5, 2019 at 8:08
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As I suspected (and commented) the property of being graded is redundant here, and this alternative answer doesn't use it.
Notation. We write $x \prec y$ to denote that $x$ is covered by $y$, and $x \preceq y$ means that $x \prec y$ or $x = y$.
A lattice $L$ is said to satisfy the Upper Covering Condition if
$$x \preceq y \;\Longrightarrow\; x \vee z \preceq y \vee z$$
holds for every $x,y,z \in L$.
A modular lattice always satisfies the Upper Covering Condition (see G. Grätzer, General Lattice Theory, IV§1. Modular Lattices, Theorem 4).
Since a distributive lattice is modular, it satisfies the Upper Covering Condition too.
Now, let $a_1, \ldots, a_k$ be the atoms of a distributive lattice.
Form
$$0 \prec a_i$$
it follows that
$$x = 0 \vee x \preceq a_i \vee x,$$
for every $x \in L$.
In particular,
$$a_1 \preceq a_1 \vee a_2 \preceq \cdots \preceq a_1 \vee \cdots \vee a_k.$$
If additionally, we have that $a_1 \vee \cdots \vee a_k = 1$, then the lattice has finite length.
Now we follow the reasoning in the previous answer: if the lattice is distributive and has finite length then it is finite, and if it is a finite distributive lattice with the join of the atoms equal to $1$, then it is Boolean, whence the meet of the co-atoms is $0$.
Update. Actually we can circumvent the use of the Upper Covering Condition, and prove directly that, in a distributive lattice $L$, if $a$ is an atom and $x \in L$ is any other element, then $x \preceq a \vee x$ as plug in the remaining of the answer above.
Indeed, if $a \leq x$ then there is nothing to prove.
If $a \nleq x$ and there exists $c \in L$ such that $x < c < a \vee x$, then $a \nleq c$, for otherwise $a \vee x \leq c$, a contradiction;
thus, as $a$ is an atom, $a \wedge c =0$, and the elements
$$0,\,a,\,x,\,c,\,a\vee x$$
for a pentagon, whence the lattice is not distributive, a contradiction.