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question embedding sublattice into product-lattice

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  • p_ouwehand
    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
    Message 1 of 3 , Jan 29, 2008
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      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
    • Friedrich Wehrung
      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
      Message 2 of 3 , Jan 29, 2008
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        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,
        Fred

        Le 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 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


        --
        Friedrich Wehrung
        LMNO, CNRS UMR 6139
        Universit\'e de Caen, Campus 2
        D\'epartement de Math\'ematiques, BP 5186
        14032 Caen cedex
        FRANCE

        e-mail: wehrung@...
        alternate e-mail: fwehrung@...




      • Mckenzie, Ralph N
        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
        Message 3 of 3 , Jan 30, 2008
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          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@...
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