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Intro to Proofs Unit 5: Sets and Set Proofs

Sets are the basic objects of modern mathematics. This unit covers the set operations and the element-chasing technique used to prove that one set is contained in, or equal to, another.

Covers: Set notation and set-builder form, Subsets and set equality, Union, intersection, difference and complement, Power sets and Cartesian products, Element-chasing proofs, Double-inclusion proofs and De Morgan's laws, Indexed families of sets

Study guide

A set is a collection of distinct objects, its elements, and x ∈ A means x is an element of A. Sets are written by listing, {1, 2, 3}, or in set-builder form, {x ∈ ℤ : x² < 10}. Order and repetition do not matter, so {1, 2} = {2, 1, 1}. The empty set ∅ has no elements, and {∅} is a different set with one element. A ⊆ B means every element of A is in B; it is a quantified conditional, ∀x (x ∈ A → x ∈ B). Since ∅ has no elements, ∅ ⊆ A holds vacuously for every set A, including ∅ itself.

The basic operations correspond to logical connectives. The union A ∪ B = {x : x ∈ A or x ∈ B}; the intersection A ∩ B = {x : x ∈ A and x ∈ B}; the difference A − B = {x : x ∈ A and x ∉ B}; and the complement Aᶜ = U − A relative to a universal set U. Because of this correspondence, De Morgan's laws carry over: (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ and (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ. For finite sets, inclusion-exclusion gives |A ∪ B| = |A| + |B| − |A ∩ B|. The symmetric difference A Δ B contains elements in exactly one of the two sets.

The power set P(A) is the set of all subsets of A; if |A| = n then |P(A)| = 2ⁿ. The Cartesian product A × B is the set of ordered pairs (a, b) with a ∈ A and b ∈ B, and |A × B| = |A| · |B|. Keep ∈ and ⊆ separate: {1} ∈ P({1, 2}) because {1} is a subset of {1, 2}, while 1 ∈ {1, 2} but 1 ∉ P({1, 2}). Indexed families such as Aₙ = [0, 1/n] for n ≥ 1 have unions and intersections defined by 'for some n' and 'for every n'; here the intersection is {0}.

To prove S ⊆ T, use element chasing: 'Let x ∈ S. ... Therefore x ∈ T.' Translate each set membership into its logical meaning and reason with the connectives; an 'or' in the hypothesis usually calls for cases. To prove S = T, prove S ⊆ T and T ⊆ S (double inclusion), or chain known identities. To disprove a set identity, give a specific small counterexample, often with sets of one or two elements. Never 'prove' an identity with a Venn diagram or a single example; a diagram can suggest a proof but is not one.

Key ideas

Formulas and rules

(A ∪ B)ᶜ = Aᶜ ∩ Bᶜ and (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ
A − B = A ∩ Bᶜ
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
|A ∪ B| = |A| + |B| − |A ∩ B|
|P(A)| = 2ⁿ and |A × B| = |A| · |B|
A ⊆ B ⇔ A ∩ B = A ⇔ A ∪ B = B

Common mistakes

Before the test, make sure you can

Words to know

Set
A collection of distinct objects called elements.
Element (∈)
x ∈ A means x is one of the objects in A.
Subset (⊆)
A ⊆ B means every element of A is an element of B.
Proper subset
A ⊊ B means A ⊆ B and A ≠ B.
Empty set
The set ∅ with no elements; a subset of every set.
Union
A ∪ B: elements in A or B (or both).
Intersection
A ∩ B: elements in both A and B.
Set difference
A − B: elements in A but not in B.
Complement
Aᶜ = U − A: elements of the universal set not in A.
Disjoint sets
Sets with no elements in common, so A ∩ B = ∅.
Power set
P(A): the set of all subsets of A.
Cartesian product
A × B: the set of ordered pairs (a, b) with a ∈ A, b ∈ B.
Symmetric difference
A Δ B = (A − B) ∪ (B − A).
Double inclusion
Proving S = T by proving S ⊆ T and T ⊆ S.
Element chasing
Proving S ⊆ T by following an arbitrary element of S into T.
Indexed family
A collection of sets Aᵢ labeled by an index set, with unions and intersections over all i.

Sample practice questions

30 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

For sets A and B, A ⊆ B means:

  1. Some element of A is also an element of B
  2. Every element of A is also an element of B
  3. Every element of B is also an element of A
  4. A and B have the same number of elements
Show the answer and steps

Answer: B) Every element of A is also an element of B

  1. A ⊆ B is the statement ∀x (x ∈ A → x ∈ B).
  2. It allows A = B, and it holds vacuously when A = ∅.
  3. 'Every element of B is in A' is B ⊆ A, the reverse inclusion.
Question 2 (level 1)

The set difference A − B (also written A \ B) is:

  1. {x : x ∉ A and x ∉ B}
  2. {x : x ∈ A and x ∉ B}
  3. {x : x ∈ B and x ∉ A}
  4. {x : x ∈ A or x ∉ B}
Show the answer and steps

Answer: B) {x : x ∈ A and x ∉ B}

  1. A − B keeps the elements of A that are not in B.
  2. In logic, x ∈ A − B means x ∈ A ∧ x ∉ B.
  3. {x : x ∈ B and x ∉ A} is B − A, which is generally different.
Question 3 (level 3)

By De Morgan's laws for sets, (A ∪ B)ᶜ equals:

  1. Aᶜ ∩ B
  2. Aᶜ ∩ Bᶜ
  3. Aᶜ ∪ Bᶜ
  4. A ∩ B
Show the answer and steps

Answer: B) Aᶜ ∩ Bᶜ

  1. x ∈ (A ∪ B)ᶜ means ¬(x ∈ A ∨ x ∈ B).
  2. By logical De Morgan, that is x ∉ A ∧ x ∉ B.
  3. So (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ; the other law gives (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ.