- Hi all,

I have a question which looks easy, but I haven't been able to get an

answer.

Suppose that $L$ is a finite lattice, and that $\omega$ is the first

infinite cardinal. Is it true that for every natural number $n$ there

is a (finite) natural number $m$ such that:

Whenever $a_1,..., a_n$ belong to $L^\omega$, there is a sublattice

$A$ of $L^\omega$, which is isomorphic to $L^m$, such that

$a_1,...,a_n$ belong to $A$?

Has anyone seen a proof or counterexample, results that look similar,

or papers that might prove useful?

Thanks!

Peter Ouwehand - Dear Peter,It seems to me that this is essentially the same argument that says that the free object on $n$ generators in the variety generated by $L$ is the sublattice of $L^{L^n}$ generated by the projection maps. Hence $m=|L|^n$ works! Here is an outline of the argument for your problem. Let $a: \omega\to L^n$ be the map defined by\[a(k)=(a_1(k),\dots,a_n(k))\](for each $k\in\omega$), and denote by $S$ the range of $a$ (thus a subset of $L^n$). Put $X_s=a^{-1}\{s\}$, for each $s\in S$. Then the sets $X_s$, for $s\in S$, form a partition of $\omega$, and hence the lattice $A$ of all elements of $L^\omega$ which are constant on each $X_s$ is a sublattice of $L^\omega$, isomorphic to $L^{|S]}$. Now observe that each $a_i$ belongs to $A$.Incidentally, there is nothing special about lattices in that argument (Jonsson's Lemma is not used), it would work just as well in any variety of algebras.I hope that this is what you had in mind!Cheers,FredLe 29 janv. 08 à 11:04, p_ouwehand a écrit :Hi all,I have a question which looks easy, but I haven't been able to get ananswer.Suppose that $L$ is a finite lattice, and that $\omega$ is the firstinfinite cardinal. Is it true that for every natural number $n$ thereis a (finite) natural number $m$ such that:Whenever $a_1,..., a_n$ belong to $L^\omega$, there is a sublattice$A$ of $L^\omega$, which is isomorphic to $L^m$, such that$a_1,...,a_n$ belong to $A$?Has anyone seen a proof or counterexample, results that look similar,or papers that might prove useful?Thanks!Peter Ouwehand--Friedrich WehrungLMNO, CNRS UMR 6139Universit\'e de Caen, Campus 2D\'epartement de Math\'ematiques, BP 518614032 Caen cedexFRANCEe-mail: wehrung@...alternate e-mail: fwehrung@...
- I think this is easy. Let k = |L|. For f_1,...,f_n in L^{omega},

let p be the function defined on omega with

p(i)=(f_a(i),...,f_n(i)). There are at most k^n possible values of

p. Let M be the set

of all functions f in L^{omega} such that whenever p(i)=p(j) then

f(i)=f(j). Now M is isomorphic to L^r where r=|p(omega)| is no

greater than k^n. By dividing several of the equivalence classes

of elements of omega on which functions in M are constant, we

get a larger lattice M' with M' isomorphic to L^{r'}, r'=k^n.

Of course f_1,...,f_n all belong to M and to M'.

Ralph

--On Tuesday, January 29, 2008 10:04 AM +0000 p_ouwehand

<peter_ouwehand@...> wrote:

>

-----------------------------------------------------------------

>

>

>

> Hi all,

>

> I have a question which looks easy, but I haven't been able to

> get an answer.

>

> Suppose that $L$ is a finite lattice, and that $\omega$ is the

> first infinite cardinal. Is it true that for every natural

> number $n$ there is a (finite) natural number $m$ such that:

> Whenever $a_1,..., a_n$ belong to $L^\omega$, there is a

> sublattice $A$ of $L^\omega$, which is isomorphic to $L^m$, such

> that $a_1,...,a_n$ belong to $A$?

>

> Has anyone seen a proof or counterexample, results that look

> similar, or papers that might prove useful?

>

> Thanks!

>

> Peter Ouwehand

>

>

Mckenzie, Ralph N

Vanderbilt University

Email: ralph.n.mckenzie@...