Lattice

**dwsmith** · November 30th 2010, 06:48 PM

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?

**Drexel28** · November 30th 2010, 06:50 PM

Originally Posted by dwsmith

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?

**dwsmith** · November 30th 2010, 06:58 PM

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?

**Drexel28** · November 30th 2010, 07:04 PM

Originally Posted by dwsmith

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 $f(x\vee y)=f(x)\vee f(y)$ .

**dwsmith** · November 30th 2010, 07:07 PM

I have the bijection and what you are saying is the same as the lub and glb are the same.

**Drexel28** · November 30th 2010, 07:12 PM

Originally Posted by dwsmith

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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