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complemented lattice

related elements in lattice, complement, complemented element
perspective elements, complemented
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Mathematics Subject Classification

06C15 no label found06B05 no label found


A small question:

Which additional properties are necessary/sufficient to make a lattice $(L, <)$ isomorphic to a power set (\Pot(M), \subsetneq)?

Ok, the following properties are necessary:
- complete (every subset has an infimum and a supremum)
- complemented
- distributive

But what set of properties is sufficient?

A complemented distributive lattice is a Boolean algebra. A complete Boolean algebra A is lattice isomorphic to the power set of some set S if A is atomic (every element of A has an atom below it). You can show that S is in fact the set of all atoms of A. Check out the book Intro to Lattices and Order by Davey and Priestley.

Ok, so this would mean, a lattice that is
- complete and
- complemented and
- distributive and
- atomic
is order-isomorphic to a power set. Any such isomorphism will identify the atoms of the lattice with the singletons of the power set.

Are all the four requirements are necessary? Don't have the book at hand...

I think so. I saw a counter-example in the book if, say, the atomic condition is dropped. I'll find out.

I found the counter-example:

Consider the power set of the reals R. Consider the Boolean subalgebra A of R generated by the following intervals

(-oo, a)
empty set

where a,b,c,d are reals.

Then A is complete and atomless, and A is not lattice isomorphic to any power set of a set.

Hi, thanks, but..
Sorry I don't get it!

Do you mean a < b < c < d ?
Do you mean A \subseteq R or A \subseteq \Pot(R) ?
With (-oo, a), [b, c), [d, oo) \in A ?
or (-oo, a), [b, c), [d, oo) \subseteq A ?
Maybe a sup-generated subset of the power set \Pot(R) ??

Regards, Schneemann.

Sorry for the sloppiness. First, a, b, c, d are arbitrary real numbers, there are no specific orderings on these numbers. Second, the Boolean subalgebra A I am refering to is the set of all finite unions of those intervals that I mentioned. You can show that A is indeed a Boolean algebra under union and intersection and complementation, etc... Furthermore, it is complete. You can derive that A is atomless by using a proof of contradiction.

At the beginning I thought you mean a, b, c, d to be fixed numbers. I'm still not sure, but now I think you mean this set:

S := { (oo,b) | b in RR }
cup { [a, b) | a, b in RR }
cup { [a,oo) | a in RR }
cup { emptyset }

Now A is generated from S by building finite unions.. right? Ok, I'll think about that. Thanks!

Regards. Schneemann

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