Sets · step 5 of 13

Proving set identities

To prove two sets are equal you have two standard routes. The double-inclusion method shows X ⊆ Y and Y ⊆ X separately, chasing an arbitrary element through the definitions. The algebraic method rewrites one side using established laws — De Morgan, distributivity, absorption — until it becomes the other. Exams usually specify which one they want, and the method mark depends on it.

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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.

Choose the method the question namesIf it says "using the laws", an element-chase earns no method marks, and vice versa.
For ⊆, start with an arbitrary elementWrite "let x ∈ X" and unfold the definition of X into a logical statement about x.
Convert the logic, then fold back upApply propositional equivalences to the membership condition, then re-read it as membership in Y. Set identities are propositional identities in disguise.
Do the second inclusionEither repeat the argument in reverse, or note that every step was an iff and say so explicitly.

Worked example

Prove (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ.

Let x be arbitrary in the universe.
x ∈ (A ∪ B)ᶜ ⟺ ¬(x ∈ A ∪ B) ⟺ ¬(x ∈ A ∨ x ∈ B).
By propositional De Morgan: ⟺ x ∉ A ∧ x ∉ B.
⟺ x ∈ Aᶜ ∧ x ∈ Bᶜ ⟺ x ∈ Aᶜ ∩ Bᶜ.

Answer. Every step is an equivalence, so the two sets have exactly the same members and (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ.

Where marks get dropped

These are the specific errors that cost credit on proving set identities questions — QED's rubric penalises each of them separately.

Proving only one inclusion and claiming equality. Unless every step is a stated iff, both directions are required.
Using a Venn diagram as the proof. A diagram is good evidence and a fine sanity check, but most schemes do not accept it as a proof of an identity.
Applying De Morgan without switching ∪ to ∩. The operation must flip, exactly as ∧ and ∨ do.

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Proving set identities — frequently asked questions

Which method is safer under exam pressure?

Double inclusion — it always works and needs no memorised law list. The algebraic method is faster when you can see the route.

Can I use iff-chains instead of two inclusions?

Yes, and it is elegant, but every step really must be reversible. If one step is only an implication the chain proves a subset relation, not equality.

Why do set identities mirror logical ones?

Because membership is a proposition: x ∈ A ∪ B is literally (x ∈ A) ∨ (x ∈ B). Set algebra and propositional algebra are two readings of the same Boolean algebra.

The rest of Sets

Set-builder notation, operations, and set identities. Each subtopic below has its own method, worked example and mark-losing traps.

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