Partial orders, Hasse diagrams, and lattices (intro)

Let us work through a complete example: constructing the poset, drawing its Hasse diagram, and verifying the lattice property for the set of positive divisors of 30, ordered by divisibility.

Step 1 — Identify the Set and Relation

The positive divisors of 30 are S = {1, 2, 3, 5, 6, 10, 15, 30}. The relation is divisibility: a ≤ b ⟺ a | b (a divides b). We verify that divisibility is a partial order on S: it is reflexive (a | a for all a), antisymmetric (if a | b and b | a for positive integers, then a = b), and transitive (a | b and b | c implies a | c).

Poset (S, |) confirmed with |S| = 8 elements.

Step 2 — Determine the Covering Relations

We need to find all pairs (a, b) where a | b and there is no c ∈ S with a | c | b (where a ≠ c ≠ b). Systematically: 1 ≺ 2, 1 ≺ 3, 1 ≺ 5 (the primes are directly above 1). Next, 2 ≺ 6, 2 ≺ 10, 3 ≺ 6, 3 ≺ 15, 5 ≺ 10, 5 ≺ 15. Finally, 6 ≺ 30, 10 ≺ 30, 15 ≺ 30. Note that 1 does not cover 6 because 2 is between them (1 | 2 | 6).

12 covering relations identified.

Step 3 — Draw the Hasse Diagram

Place 1 at the bottom (level 0). At level 1, place the primes 2, 3, 5. At level 2, place 6 = 2 × 3, 10 = 2 × 5, and 15 = 3 × 5. At level 3, place 30. Draw edges only for covering relations. The diagram has a symmetric diamond-like shape characteristic of the divisor lattice of a squarefree number.

Hasse diagram has 4 levels: {1} → {2,3,5} → {6,10,15} → {30}.

Step 4 — Verify the Lattice Property

Check that every pair has a join and a meet. For the pair (2, 3): their common multiples in S are 6 and 30, and the least is 6, so 2 ∨ 3 = lcm(2,3) = 6. Their common divisors in S are just {1}, so 2 ∧ 3 = gcd(2,3) = 1. For the pair (6, 10): common multiples are {30}, so 6 ∨ 10 = 30. Common divisors are {1, 2}, and the greatest is 2, so 6 ∧ 10 = 2. Repeating for all C(8,2) = 28 pairs confirms every join and meet exists.

(S, |) is a lattice with join = lcm and meet = gcd. Moreover, 1 is the bottom element ⊥ and 30 is the top element ⊤.

Step 5 — Identify Special Properties

This lattice is bounded (has ⊤ = 30 and ⊥ = 1), distributive (a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) for all elements, which holds because gcd distributes over lcm), and even complemented since 30 is squarefree. For example, the complement of 6 is 5, because gcd(6,5) = 1 = ⊥ and lcm(6,5) = 30 = ⊤.

The lattice is a bounded, distributive, complemented lattice — a Boolean algebra isomorphic to B₃ = 2³.