Sets · step 7 of 13

Indexed families & generalised ⋃ / ⋂

An indexed family {Aᵢ : i ∈ I} assigns a set to each index. Its union ⋃ᵢ Aᵢ contains everything belonging to at least one member, and its intersection ⋂ᵢ Aᵢ everything belonging to all of them. Written as membership conditions these are an ∃ and a ∀ over the index set, which is exactly how you should attack them when I is infinite.

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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.

Convert to a quantified statementx ∈ ⋃ᵢ Aᵢ means ∃i ∈ I with x ∈ Aᵢ. x ∈ ⋂ᵢ Aᵢ means ∀i ∈ I, x ∈ Aᵢ. Everything follows from this.
Guess the answer from small indicesCompute A₁, A₂, A₃ and watch whether the sets grow, shrink, or stabilise. Nested families make the answer obvious.
Prove by double inclusionTo show ⋂ᵢ Aᵢ = S, show every element of S is in each Aᵢ, and that anything outside S misses some particular Aᵢ.
Look for the Archimedean stepOver an infinite index set, "for every n there is an index beyond it" is usually what forces an element out of the intersection.

Worked example

For n ∈ ℕ (n ≥ 1) let Aₙ = [0, 1/n]. Find ⋃ₙ Aₙ and ⋂ₙ Aₙ.

A₁ = [0,1], A₂ = [0,½], A₃ = [0,⅓] — the sets are nested downward.
Union: every Aₙ ⊆ A₁, and A₁ is itself in the family, so ⋃ₙ Aₙ = [0, 1].
Intersection: 0 ∈ Aₙ for all n, so 0 is in the intersection.
For x > 0, choose n with 1/n < x (Archimedean property); then x ∉ Aₙ, so x is excluded.

Answer. ⋃ₙ Aₙ = [0, 1] and ⋂ₙ Aₙ = {0}.

Where marks get dropped

These are the specific errors that cost credit on indexed families & generalised ⋃ / ⋂ questions — QED's rubric penalises each of them separately.

Concluding that an infinite intersection of non-empty sets is non-empty. It is often empty — for Bₙ = (0, 1/n) the intersection is ∅.
Confusing the index set with the elements. i ranges over I; x ranges over the sets Aᵢ.
Assuming the limiting set is a member of the family. The intersection {0} above is not any single Aₙ.

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Indexed families & generalised ⋃ / ⋂ — frequently asked questions

Does De Morgan hold for indexed families?

Yes: (⋃ᵢ Aᵢ)ᶜ = ⋂ᵢ Aᵢᶜ and (⋂ᵢ Aᵢ)ᶜ = ⋃ᵢ Aᵢᶜ, for any index set. The proof is the quantifier version of the two-set case.

What if the index set is empty?

The union is ∅. The intersection over an empty family is problematic — it would be everything — so it is either left undefined or taken relative to a fixed universe.

How do nested families behave?

If A₁ ⊇ A₂ ⊇ …, the union is A₁ and the intersection is the "limit" of the shrinking sets. Ascending chains behave dually.

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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