If we consider $\omega^\omega$ as a lattice with component-wise join and meet, is there a finite distributive lattice $L$ so that there is no injective lattice homomorphism $f:L\to\omega^\omega$?



Finite distributive lattices not contained in $\omega^\omega$
I believe that by $\omega^\omega$ the original poster means the poset of order-preserving maps from the chain (=totally-ordered set) $\omega$ to itself. The answer is still "no," if my proof is correct. (It is 2:09 a.m., when all conjectures are true.)


Firstly, for any $n<\omega$, $\bf{(n+1)}^\bf n$ is a sublattice of $\omega^\omega$---$\bf m$ is the chain $\{0,1,\dots,m-1\}$---via the mapping that sends $f:\bf n\to\bf{n+1}$ to a function $\bar f$ extending $f$ such that $\bar f(k)=n+1$ for $k\ge n$. So we only have to show that every finite Boolean lattice is a sublattice of a lattice of the form $\bf{(n+1)}^\bf n$.


In fact, I will show that, for $n\ge2$, $\bf 1\oplus \bf2^{n}\oplus\bf 1$ ($\oplus$ denotes ordinal sum) is a {0,1}-sublattice of $\bf{(n+1)}^\bf n$, (i.e., there is a one-to-one lattice homomorphism that preserves the top and bottom elements). Using Priestley duality in the finite case [Theorem 5.19(i) in Davey and Priestley, *Introduction to Lattices and Order*, second edition], we just need to show that $\bf{1}$ $\oplus$ $\overline{n}\oplus\bf1$ ($\bar n$ is the $n$-element antichain) is an image of $\bf n\times\bf n$ via an order-preserving map $g$.


The poset $\bf n\times\bf n$ has an $n$-element maximal antichain $\{(0,n-1),(1,n-2),\dots,(n-1,0)\}$. Define $g$ as follows: for $x\in\bf n\times\bf n$, if $x$ is in the antichain, then $g(x)=x$; if $x$ is strictly above an element of the antichain, then $g(x)=(n-1,n-1)$; otherwise, $g(x)=(0,0)$.


