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 $f:L\to\omega^\omega$?
$\begingroup$
$\endgroup$
0
asked Apr 6, 2015 at 18:55
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
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, $g(x)=(0,0)$.
$\begingroup$
$\endgroup$
1
By $\omega^\omega$ do you mean sequences $(a_1,a_2,\dots)$ of positive integers? The elements with $a_i=1,2$ for $1\leq i\leq n$ and $a_i=1$ for $i>n$ form a boolean sublattice $B_n$ of rank $n$, and every finite distributive lattice of height $n$ is isomorphic to a sublattice of $B_n$.
answered Apr 6, 2015 at 19:03
-
$\begingroup$ $\omega=\{0,1,2,\dots\}$ so $a_i$ are nonnegative $\endgroup$ Commented Apr 6, 2015 at 19:11
$\begingroup$
$\endgroup$
3
No, any finite distributive lattice is isomorphic to (hence can be identified with) a collection of finite sets (closed under $\cap$ and $\cup$) ordered by inclusion. Let $1_A$ be the characteristic function (indicator function) of the finite set $A\subseteq\omega$. Then $A\mapsto 1_A$ is your desired injective lattice homomorphism.
answered Apr 6, 2015 at 19:03
-
$\begingroup$ Bjørn, I think that you essentially said what Richard said just before you. Anyway, the theorem about representation of distributive lattices as lattices of sets, with their set-theoretic operations, is as classical as possible. The embedding is an instant corollary. $\endgroup$ Commented Apr 12, 2015 at 8:25
-
$\begingroup$ Agreed @WłodzimierzHolsztyński, although apparently it was in the same minute, and the seconds are not recorded. $\endgroup$ Commented Apr 12, 2015 at 8:46
-
$\begingroup$ Bjørn, that's why I said just :-) $\endgroup$ Commented Apr 12, 2015 at 18:17