Lattices

Lattices are fundamental structures in order theory, combining mathematics, computer science, and various applied fields.

A Lattice is a particular kind of partially ordered set ( POSET ) that has additional properties. In a lattice, every pair of elements has both a unique least upper bound (called the join) and a unique greatest lower bound (called the meet).

These operations make lattices useful in areas like logic, algebra, data structures (e.g., in databases or programming languages), and cryptography.

Formally: A lattice is a poset (L, ≤) in which every pair of elements a, b ∈ L has:

• A join (least upper bound): a ∨ b = lub{a, b}, the smallest element ≥ both a and b.
• A meet (greatest lower bound): a ∧ b = glb{a, b}, the largest element ≤ both a and b.

Properties of Lattices

Lattices satisfy several algebraic properties, which hold for all elements a, b, c ∈ L:

1. Idempotent Laws: Joining or meeting an element with itself gives the same element.

• a ∨ a = a
• a ∧ a = a

2. Commutative Laws: Order of operands does not matter.

• a ∨ b = b ∨ a
• a ∧ b = b ∧ a

3. Associative Laws: Grouping of operations does not matter.

• a ∨ (b ∨ c) = (a ∨ b) ∨ c
• a ∧ ( b ∧ c) = (a ∧ b) ∧ c

4. Absorption Laws: An element absorbs another through a join /meet in this pattern.

• a ∨ (a ∧ b) = a
• a ∧ (a ∨ b) = a

These properties ensure that (L, ∨) and (L, ∧) form semilattices, linked by absorption.

Types of Lattice

A lattice is a partially ordered set where every pair of elements has a meet (∧) and a join (∨). Now, depending on extra properties, lattices can be of different types:

1. Bounded Lattice

A lattice L is bounded if it has a greatest element (top, often denoted 1 or ⊤) and a least element (bottom, often denoted 0 or ⊥). The top is ≥ all elements, and the bottom is ≤ all elements.

Example: Consider the lattice with elements {a, b, c, d} where a is bottom (a ≤ b, a ≤ c, a ≤ d), and d is top (d ≥ b, d ≥ c, d ≥ a). Here, b and c are incomparable.

Note: Every Finite lattice is always bounded. (because they have a least element and a greatest element)

2. Complemented Lattice

A bounded lattice L is complemented if every element a ∈ L has at least one complement b such that:

• a ∧ b = 0 (bottom)
• a ∨ b = 1 (top)

Complements may not be unique in general complemented lattices.

Example (Complemented): Elements {a, b, c, d} with a (bottom), d (left middle), b (right middle), c (top). Here, b is the complement of d (d ∧ b = a, d ∨ b = c), d is the complement of b, c is the complement of a (c ∧ a = a = bottom, c ∨ a = c = top), and a is the complement of c.

Example (Not Complemented): A larger lattice with elements {a, b, c, d, e, f, g, h, i} where a is the bottom and i is the top. The complement of e does not exist, as no element x satisfies e ∧ x = a and e ∨ x = i without contradictions in the order.

3. Distributive Lattice

If a lattice satisfies the following two distributive properties, it is called a distributive lattice.

• x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z)
• x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)

A lattice L is distributive if every element in L has "at most one complement".

• A complemented distributive lattice is a Boolean algebra or Boolean lattice.
• A lattice is distributive if and only if none of its sublattices is isomorphic to N₅ or M₃.
• For a distributive lattice, each element can have at most one complement. This can be used as a theorem to prove that a lattice is not distributive.

Complement of d: doesn't exist.
Complement of c: doesn't exist
Complement of e is b, and b is e:

• glb(b, e) = glb(e, b) = a
• lub(b, e) = lub(e, b) = f

Similarly complement of a is f, and f is a.

Therefore, every element in the given lattice has at most one complement.

Complement of b is c and d:

• glb(b, c) = glb(b, d) = a
• lub(b, c) = lub(b, d) = e

Therefore, some element in the given lattice has more than one complement.

4. Modular Lattice

A lattice is called a modular lattice if it satisfies the following property:

• a ∨ (b ∧ c) = (a ∨ b) ∧ c whenever a ≤ c for all a, b, c ϵ L.

For (0, b, 1):

0 ∨ (b ∧ 1) = (0 ∨ b) ∧ 1
0 ∨ (b) = (b) ∧ 1
b = b

Similarly, in the given condition, it is true for every combination.

5. Complete Lattice

A lattice is complete if every subset (not just pairs) has a meet (infimum) and join (supremum).

• Every complete lattice is bounded (the join of all elements is top; the meet is bottom).
• However, the converse is false: a bounded lattice may not be complete if infinite subsets lack bounds.
• Counterexample: Rational numbers in [1, 2] under the usual order. Bounded (1 bottom, 2 top), but the subset of rational approximations to √2 < 2 has no supremum in the set.
• Finite lattices are always complete.

Solved Examples on Lattices

Example 1: Identify meet and join in a given set. Let L = {1, 2, 4, 8} be partially ordered by divisibility (a ≤ b if a divides b). Find:
1. 2 ∨ 4,
2. 2 ∧ 4

Solution:

• Join (∨) = Least Common Multiple (LCM) in divisibility order

2 ∨ 4 = lcm(2, 4) = 4

• Meet (∧) = Greatest Common Divisor (GCD) in divisibility order

2 ∧ 4 = gcd(2, 4) = 2

Example 2. Check if a given lattice is distributive. Consider L = {∅, {a}, {b}, {a, b}} ordered by set inclusion. Is L distributive?

Solution:

Meet = set intersection (∩)
Join = set union (∪)
For distributivity, check:

x ∩ (y ∪ z) = (x ∩ y) ∪ (x ∩ z) x

and

x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z)

For example, take x = {a}, y = {b}, z = {a, b}:
LHS = {a} ∩ ({b} ∪ {a,b}) = {a} ∩ {a,b} = {a}
RHS = ({a} ∩ {b}) ∪ ({a} ∩ {a,b}) = ∅ ∪ {a} = {a}

This holds for all cases → L is distributive.

Example 3: Modular law verification. Let L be the subspace lattice of {\displaystyle \mathbb {R} }^2 , where:

• A = span{(1, 0)}
• B = span{(0, 1)}
• C = {\displaystyle \mathbb {R} }^2

Check if the modular law holds:

A∨(B∧C) = (A∨B)∧C

Solution:

• B ∧ C = B ∩ C = B
• LHS = A ∨ B = span{(1,0),(0,1)} = R2
• A∨B = {\displaystyle \mathbb {R} }^2

Modular law holds.

Example 4: Identify a non-modular lattice.

The pentagon lattice N₅​ is given below (elements: 0, a, b, c, 1 with 0 < a < 1, 0 < b < c < 1, a and b incomparable). Show it is not modular.

Solution:

Take x = a, y = b, z = c where a ≤ c:

• y ∧ z = b
• LHS = a ∨ (b ∧ c) = a ∨ b = 1
• x ∨ y = a ∨ b = 1
• RHS = 1 ∧ c = c
  Since 1 ≠ c → violates modular law → not modular.

Example 5: Problem mixing meet, join, and complement

In the Boolean lattice P({1, 2, 3}) under set inclusion:

• Meet = ∩
• Join = ∪
• Complement = relative to {1, 2, 3}

Find:

1. Complement of {1, 2}
2. {1, 2} ∨ {2, 3}
3. {1, 2} ∧ {2, 3}

Solution:

1. Complement = {3}
2. Join = union → {1, 2, 3}
3. Meet = intersection → {2}