Union, intersection & complement of relations
Since relations are sets of pairs, they can be united, intersected and complemented. The interesting question is which properties are preserved. Reflexivity and symmetry survive both union and intersection; transitivity survives intersection but NOT union; and the complement of a reflexive relation is irreflexive. Knowing which combinations break is worth more than the mechanics.
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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.
- Compute pairwise(a,b) ∈ R ∪ S iff it is in either; ∈ R ∩ S iff in both; ∈ R̄ iff in neither R nor the ambient A × A.
- Check preservation by definitionFor intersection and transitivity: if (a,b) and (b,c) are in both R and S, then (a,c) is in both, so it is in the intersection.
- Find counterexamples for the failuresTransitivity under union breaks with R = {(1,2)} and S = {(2,3)}: neither has a chain, but the union does and lacks (1,3).
- Watch the complementThe complement of a symmetric relation is symmetric, but the complement of a transitive relation is usually not.
Worked example
Give transitive R and S on {1,2,3} whose union is not transitive, and confirm R ∩ S is transitive.
- Let R = {(1,2)} and S = {(2,3)}. Each is vacuously transitive — neither contains a chain.
- R ∪ S = {(1,2),(2,3)} contains the chain 1→2→3.
- Transitivity would require (1,3), which is absent, so the union fails.
- R ∩ S = ∅, which is vacuously transitive.
Answer. Transitivity is not preserved by union but is preserved by intersection — the empty intersection here is transitive.
Where marks get dropped
These are the specific errors that cost credit on union, intersection & complement of relations questions — QED's rubric penalises each of them separately.
- Assuming all properties behave the same way under all operations. Transitivity is the one that breaks under union, and it is the one examiners ask about.
- Forgetting the ambient set when taking complements. R̄ = (A × A) \ R, so the answer depends on A.
- Claiming the union of two equivalence relations is an equivalence relation. It is generally not — the transitive closure of the union is.
Practise this until it is automatic
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Union, intersection & complement of relations — frequently asked questions
Is the intersection of two equivalence relations an equivalence relation?
Yes. Reflexivity, symmetry and transitivity are all preserved by intersection, which is exactly why closure operators exist for these properties.
What is the complement of a partial order?
Rarely anything useful: it fails reflexivity and usually transitivity. The related useful construction is the strict order, obtained by removing the diagonal rather than complementing.
Does composition distribute over union?
Over union yes: T∘(R ∪ S) = (T∘R) ∪ (T∘S). Over intersection only ⊆ holds in general, not equality.
The rest of Relations
Properties of relations, composition, and representations. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Reflexive, symmetric, antisymmetric & transitive
- 2Checking properties for a given relation
- 3Composition R∘S & inverse R⁻¹
- 4Matrix & digraph representations
- 5Reflexive & transitive closures
- 6Relations as subsets of A × B
- 7Powers Rⁿ & reachability
- 8Warshall’s transitive closure algorithm
- 9Counting relations with a given property
- 10Union, intersection & complement of relations
- 11n-ary relations & the relational data model
- 12Restricting a relation to a subset
- 13Symmetric closure vs transitive closure
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