Are all set lattices isomorphic to a power set?

Question

A set lattice is a lattice where the meet is the intersection, the join is the union, and the partial order is $\subseteq$. The standard example is is $P(A)$ where $A$ is a set. I’ve thought about other possible examples, yet have failed to come up with any. Are there any set/distributive lattices that aren’t isomorphic to $P(A)$ for some $A$?

Answer 1

Well, not every lattice is isomorphic to $P(A)$. For example the $P(A)$ lattice has $2^{|A|}$ elements, and there are lattices for any order, e.g.

$$K_n=\big\{\{1\},\{1,2\},\ldots,\{1,2,\ldots,n-1,n\}\big\}$$

with union and intersection is a latice of exactly $n$ elements.

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So the more interesting question would be: can every lattice be embedded into $P(A)$ for some set $A$? The answer is still no. To see that we need to look at some property of (a sublattice of) $P(A)$ that other lattices don't have to have. One of those properties is the distributivity of $\vee$ over $\wedge$ (and vice versa). Union is always distributive over intersection (and vice versa), and so any subtlatice of $P(A)$ is distributive. But there are non-distributive lattices, e.g the $M_3$:

$$M_3=\{0,a,b,c,1\}$$ $$0< a<1$$ $$0< b<1$$ $$0< c<1$$

In fact this is if and only if: a lattice is isomorphic to a lattice of sets (with union and intersection), and thus embeddable into some $P(A)$, if and only if it is distributive. Note that this is a non-trivial theorem (see the wiki I linked earlier).