I'm working out a problem from Gratzer '71. We are to show that any finite lattice $L$ $\simeq$ $I(L)$.
My attempt was as follow: prove the contrapositive. So assume $L$ is not isomorphic to $I(L)$. We have by an earlier corollary that $L$ embeds into $I(L)$ (say, under $f$), and we note that $I(L)$ is always a lattice. So let $I(L)$ be infinite. Then given $f$ we can always find a sublattice $L$ in $I(L)$ such that $L$ is infinite. END PROOF.
Problem for me is that this is very hand-wavy. And it requires me to explicitly assume that I(L) is infinite (which is a little more than is in the statement of the theorem) to get the desired result.
I'm thinking the other way to do this is to proceed directly and assume $L$ is finite. Then L is complete. . . not sure where to go from there, though.
Any suggestions are welcome. Thanks!
Lema: Every ideal $I$ of a given a finite lattice $L$ is principal and $\mathord{\downarrow}\bigvee I=I$. (Note that $\bigvee I$ makes sense because any finite lattice is complete).
Proof: Take $x\in \mathord{\downarrow}\bigvee I$. One has $x\leq \bigvee I$ so if $\bigvee I\in I$, by definition of ideal one gets $x\in I$. It is in fact the case that $\bigvee I\in I$. (To prove this recall $I=\{i_1, \ldots ,i_k\}$ for some $k\in \mathbb N$ and $i_1, \ldots ,i_k\in I$ and take joins iteratively until you find $\bigvee I$, this can rigorously proved by induction. This very same idea is what proves that $\bigvee I$ exists in $L$, so I'd say this bit can be skipped). Since any ideal is join closed, it follows that $\bigvee I\in I$). For the other inclusion let $x\in I$, then of course $x\leq \bigvee I$ and therefore $x\in \mathord{\downarrow} \bigvee I$.$\square$
Hint: Consider the map $x\mapsto \mathord{\downarrow} x$ defined on $L$.