# Upper and lower bound in distributive lattice

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 $$[a,b]$$)?

• What is a graded lattice? Is this condition important? Aug 1, 2019 at 9:55
• en.wikipedia.org/wiki/Graded_poset In my case the lattice is graded, don't know if it helps. Aug 1, 2019 at 10:36
• 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. Aug 1, 2019 at 12:52
• @Berci It isn't distributive. The elements $\varnothing, \{1\},\{1,2\},\{3\},\{1,2,3\}$ form a pentagon. Aug 1, 2019 at 13:14
• Ohh, then it's neither modular, hence grading must fail, too.. Aug 1, 2019 at 14:20

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 $$0$$.

• Thank you very much for this beautiful proof! Aug 2, 2019 at 10:58
• @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! Aug 2, 2019 at 13:16
• 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. Aug 4, 2019 at 21:53
• @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... Aug 5, 2019 at 8:08

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.