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 $[a,b]$)?

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 $0$.
