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Math Help - Lattice

  1. #1
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    Lattice

    If I am proving a lattice is non-distributive i.e. is shown it isomorphic to one of the two lattices in the theorem, is there a quick way to prove it is isomorphic without going through every vertex and showing \displaystyle a\land b \ \mbox{and} \ a\lor b for every vertex?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    If I am proving a lattice is non-distributive i.e. is shown it isomorphic to one of the two lattices in the theorem, is there a quick way to prove it is isomorphic without going through every vertex and showing \displaystyle a\land b \ \mbox{and} \ a\lor b for every vertex?
    Can you state this in different terms?
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    I have lattice I want to show is isomorphic to another lattice that is always non-distributive so can I prove the lattice is non-distributive. In order to show it is isomorphic, I must show all \displaystyle \forall a,b\in L \ glb(a,b)=glb(x,y) \ x,y\in L_2 \ \mbox{and} \ lub(a,b)=lub(x,y)

    glb=greatest lower bound
    lub=least upper bound
    L is the lattice I want to show is isomorphic to L2 since L2 is always non-distributive.

    If I check the glb and lub for ever combination of as and bs, I will have to do this 48 times. Is there a simpler way?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    I have lattice I want to show is isomorphic to another lattice that is always non-distributive so can I prove the lattice is non-distributive. In order to show it is isomorphic, I must show all \displaystyle \forall a,b\in L \ glb(a,b)=glb(x,y) \ x,y\in L_2 \ \mbox{and} \ lub(a,b)=lub(x,y)

    glb=greatest lower bound
    lub=least upper bound
    L is the lattice I want to show is isomorphic to L2 since L2 is always non-distributive.

    If I check the glb and lub for ever combination of as and bs, I will have to do this 48 times. Is there a simpler way?
    I think you're missing something here. Two lattices \mathfrak{L}_1,\mathfrak{L}_2 are isomorphic if there exists a bijection f:\mathfrak{L}_1\to\mathfrak{L}_2 such that f(x\wedge y)=f(x)\wedge f(y) and f(x\vee y)=f(x)\vee f(y).
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  5. #5
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    I have the bijection and what you are saying is the same as the lub and glb are the same.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dwsmith View Post
    I have the bijection and what you are saying is the same as the lub and glb are the same.
    Then yes, without more information it sounds like you might have to check it for all 48 elements.
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