Bounded, Complemented And Distributive Lattices

Theorem: Let L be a bounded distributive lattice. If a complement of any element exists, it is unique.
Proof: Suppose on the contrary that b and c are complements of the element a ∈ L. Then
a ∨ b = I a ∨ c = I
a ∧ b = 0 a ∧ c = 0
Using distributive law, we have
b = b ∨ 0
= b ∨ (a ∧ c)
= (b ∨ a) ∧ (b ∨ c)
= (a Ú b) ∧ (b Ú c)
= I ∧ (b ∨ c)
= b ∨ c
Similarly,
c = c ∨ 0
= c ∨ (a ∧ b)
= (c ∨ a) ∧ (c ∨ b)
= (a ∨ c) ∧ (c ∨ b)
= I ∧ (c ∨ b)
= I ∧ (b ∨ c)
= b ∨ c
Hence b = c.

Definition: Let (L, ∧ , ∨ ) be a lattice. An element a ∈ L is said to be joinirreducible if it cannot be expressed as the join of two distinct elements of L. In other words, a ∈ L is join-irreducible if for any b, c ∈ L
a = b ∨ c ->
a = b or a = c.
For example, prime number under multiplication have this property. In fact if p is a prime number, then p = a b
p a or p = b. Clearly 0 is join – irreducible.
Further, if a has at least two immediate predecessors, say b and c as in the diagram below:

Then a = b ∨ c and so a is not join – irreducible. On the other hand if a has a unique immediate predecessor c, then a ≠ sup(b1, b2) = b1 ∨ b2 for any other elements b1 and b2 because c would lie between b1, b2 and a.

In other words, a ≠ 0 is join irreducible if and only if a has a unique predecessor.
Definition: Those elements, which immediately succeed 0, are called atoms. From the above discussion, it follows that the atoms are join-irreducible.

However, lattices can have other join-irreducible elements. For example, the element c in five-element lattice is not an atom, even then it is join irreducible because it has only one immediate predecessor, namely a.