STEP-BY-STEP LESSON · §6.2

Properties of Sets

Prove set equality using equivalent membership conditions.

Before you begin

Set Theory: Definitions and the Element Method of Proof

The statement x∈A concerns one object. A⊆B says that every element of A belongs to B.

A⊆B ⇔ ∀x(x∈A ⇒ x∈B)

X=Y means X⊆Y and Y⊆X. One inclusion still allows extra elements in the second set.

X=Y ⇔ (X⊆Y ∧ Y⊆X)

More in lesson §6.1 Set Subset Intersection

Logical Form and Logical Equivalence

p and q stand for statements, each true or false. Compound formulas combine them with logical operations.

p∧q

p∧q is true only when both parts are true. p∨q is true when at least one part is true. ¬p reverses the value.

¬(p∧q)

More in lesson §2.1 Negation / NOT Conjunction / AND Disjunction / OR

Symbols

⇔: equivalent in both directions
Aᶜ: complement relative to the specified U
U: fixed universe containing all sets’ elements under discussion
p,q,r: membership statements x∈A, x∈B, x∈C for one fixed arbitrary x∈U

Definitions and notation for this topic

Universe and complement laws

Aᶜ=U∖A; (A∪B)ᶜ=Aᶜ∩Bᶜ; (A∩B)ᶜ=Aᶜ∪Bᶜ

Fix a universe U and subsets A,B⊆U. Complement means outside the set but inside U; both De Morgan identities use this same universe.

An element is outside the union exactly when it is outside both sets. Related laws are A∪Aᶜ=U, A∩Aᶜ=∅, and (Aᶜ)ᶜ=A.

Separate glossary example

If U={1,2,3,4}, A={1,2}, B={2,3}, then (A∪B)ᶜ={4}=Aᶜ∩Bᶜ.

Open glossary card

Distributive and difference laws

A∩(B∪C)=(A∩B)∪(A∩C); A∖B=A∩Bᶜ

For A,B,C⊆U, set identities permit replacing a set expression by an equal one. The dual distributive law swaps ∩ and ∪ throughout.

Each equals sign needs a valid law. Alternatively, fix an arbitrary x∈U and show that membership in the left and right sides is equivalent.

Separate glossary example

(A∩B)∪(A∩Bᶜ)=A∩(B∪Bᶜ)=A∩U=A.

Open glossary card

Step by step

Step 1 / 6

Name the membership statements

Fix an arbitrary x in a common universe U. Write p for x∈A, q for x∈B and r for x∈C. Set operations become logical operations on these three statements.

∩ ↔ ∧; ∪ ↔ ∨; complement ↔ ¬

The truth table now refers to the same object throughout the comparison.

Worked example

Prove that intersection distributes over union.

x∈A∩(B∪C) iff x∈A and (x∈B or x∈C), by the definitions.
By logical distributivity this is equivalent to (x∈A and x∈B) or (x∈A and x∈C).
Reassemble the definitions: x∈(A∩B)∪(A∩C). The sets contain exactly the same elements.

Colour cues

Build sets and follow membership

Change which elements belong to A and B. Compare the diagram, the roster and the rule for one element.

Given U = {1,2,3,4,5,6}
A={1, 2, 3} · B={3, 4, 5}

View

Operation

Inspect element x

Regions show membership, not quantity. Read the elements or the counts; a larger area does not mean more elements. All complements are taken inside U. The area outside both circles is still inside U.

Neither A nor B · U∖(A∪B)
{6}

B only · B∖A
{4, 5}

A only · A∖B
{1, 2}

Both A and B · A∩B
{3} · Included

Result: A∩B = {3}

Keep x only when it belongs to both A and B.

For x=3: 3∈A; 3∈B. Therefore 3∈A∩B.

A≠B in this example. Witness x=1: 1∈A; 1∉B. One such element is enough to refute equality.

Membership controls: select whether each element belongs to A and B.
x	x∈A	x∈B	Region
1			A only
2			A only
3			Both A and B
4			B only
5			B only
6			Neither A nor B

Try a boundary case

Optional self-check

How can x∈(A∪B)ᶜ be expressed without complementing a union?

Show answer

For x∈U: x∉A and x∉B, meaning x∈Aᶜ∩Bᶜ. Negating “or” gives “and”.

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