I am trying to show that a distributive lattice of finite length is itself finite.
Clearly if a lattice is finite it is of finite length. How can I show the other implication?
Note that we also have if $L$ is a distributive lattice with finite length, there are finitely many join-irreducible elements. I think we must use this fact somehow.
For any $x$ in a distributive lattice, the upper set of $\{x\}$, that is, $\uparrow x$, is a distributive sublattice.
A distributive lattice of finite length has a bottom $0$, and the set of covers of $0$, the join-irreducible elements, is finite.
Let $L(n)$ be: every distributive lattice of length $\leq n$, is finite.
Clearly $L(0)$ and $L(1)$. Now assume $L(n)$ and let $L$ be a distributive lattice of length $n + 1$.
Let $K = \{ a_1, \ldots, a_k \}$ be the set of bottom covers.
$L$ is the union of $\{0\}$ and $\uparrow a_1, \ldots, \uparrow a_k$.
Each $\uparrow a_j$ is a distributive lattice of length $\leq n$, thus finite.
Whereupon $L$, a finite union of finite sets, is finite.
Thence $L(n+1)$ and therefore every distributive lattice of finite length can be induced to be finite.
Easy exercise. Show distributive is necessary with
an example of an infinite lattice with finite length.