Hasse diagrams
A Hasse diagram draws a finite poset with all redundancy removed: it shows only the covering relations, omitting loops (implied by reflexivity) and any edge implied by transitivity. Height encodes order — larger elements are drawn higher — so arrowheads are unnecessary. Reading one correctly means remembering that a ⊑ b whenever an upward path exists, not only when a direct edge does.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Find the covering relationsa is covered by b if a ⊏ b and no c sits strictly between them. Only these become edges.
- Drop loops and shortcutsNever draw (a,a), and never draw a → c when a → b → c is already shown.
- Layer by heightPut minimal elements at the bottom and work upward, so every edge goes strictly up the page.
- Read comparability as upward pathsa ⊑ b iff you can walk from a up to b along edges. Elements on different branches with no path are incomparable.
Worked example
Build the Hasse diagram for divisors of 12 under divisibility.
- Elements: 1, 2, 3, 4, 6, 12.
- Covers of 1: 2 and 3 (nothing strictly between).
- Covers of 2: 4 and 6. Cover of 3: 6 (3 ∤ 4).
- Covers of 4 and 6: both 12. Note 1 ⊑ 12 is NOT drawn — it follows by transitivity.
Answer. Bottom 1; level two 2 and 3; level three 4 and 6; top 12. Edges: 1–2, 1–3, 2–4, 2–6, 3–6, 4–12, 6–12.
Where marks get dropped
These are the specific errors that cost credit on hasse diagrams questions — QED's rubric penalises each of them separately.
- Drawing transitive edges. Including 1–12 is not merely redundant, it is wrong for a Hasse diagram and loses marks.
- Adding loops for reflexivity. They are always implied and never drawn.
- Reading "no edge" as "incomparable". 1 and 12 have no edge but are comparable via a path.
Practise this until it is automatic
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Hasse diagrams — frequently asked questions
Why are arrows unnecessary?
Because the convention is that larger elements are drawn higher, so direction is encoded by position. A diagram drawn sideways would need arrows.
How do I recover the full order?
Take the reflexive-transitive closure of the drawn edges. The Hasse diagram is the minimal edge set from which the order can be rebuilt.
Can infinite posets have Hasse diagrams?
Only when covers exist. ℚ under ≤ has no covering pairs at all — between any two rationals lies another — so it has no Hasse diagram.
The rest of Orderings & Lattices
Partial orders, Hasse diagrams, bounds, lattices. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Partial vs total orders
- 2Hasse diagrams
- 3Minimal, maximal, least & greatest elements
- 4Upper & lower bounds, supremum & infimum
- 5Lattices: divisibility & subset orders
- 6Chains, antichains & comparability
- 7Topological sorting
- 8Well-orderings & the least-element principle
- 9Distributive & complemented lattices
- 10Product & lexicographic orders
- 11Dilworth’s theorem & chain covers
- 12Order isomorphism & comparing posets
- 13Scheduling with precedence constraints
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