Set TheoryTopic #10 of 40

Set Identities

Commutative, associative, distributive laws, De Morgan's laws for sets, and proving set equality.

Complete List of Set Identities

Identity Laws

A=AA \cup \emptyset = A AU=AA \cap U = A

Domination Laws

AU=UA \cup U = U A=A \cap \emptyset = \emptyset

Idempotent Laws

AA=AA \cup A = A AA=AA \cap A = A

Complementation Law

A=A\overline{\overline{A}} = A

Commutative Laws

AB=BAA \cup B = B \cup A AB=BAA \cap B = B \cap A

Associative Laws

(AB)C=A(BC)(A \cup B) \cup C = A \cup (B \cup C) (AB)C=A(BC)(A \cap B) \cap C = A \cap (B \cap C)

Distributive Laws

A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C) A(BC)=(AB)(AC)A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

De Morgan's Laws

AB=AB\overline{A \cup B} = \overline{A} \cap \overline{B} AB=AB\overline{A \cap B} = \overline{A} \cup \overline{B}

Absorption Laws

A(AB)=AA \cup (A \cap B) = A A(AB)=AA \cap (A \cup B) = A

Complement Laws

AA=UA \cup \overline{A} = U AA=A \cap \overline{A} = \emptyset U=\overline{U} = \emptyset =U\overline{\emptyset} = U

Additional Useful Identities

Set Difference

AB=ABA - B = A \cap \overline{B} A(AB)=ABA - (A - B) = A \cap B A(BC)=(AB)(AC)A - (B \cup C) = (A - B) \cap (A - C) A(BC)=(AB)(AC)A - (B \cap C) = (A - B) \cup (A - C)

Symmetric Difference

AB=(AB)(AB)A \oplus B = (A \cup B) - (A \cap B) AB=(AB)(BA)A \oplus B = (A - B) \cup (B - A) AB=(AB)(AB)A \oplus B = (A \cap \overline{B}) \cup (\overline{A} \cap B) AA=A \oplus A = \emptyset A=AA \oplus \emptyset = A AB=BAA \oplus B = B \oplus A (AB)C=A(BC)(A \oplus B) \oplus C = A \oplus (B \oplus C)

Subset Properties

AB    AB=AA \subseteq B \iff A \cap B = A AB    AB=BA \subseteq B \iff A \cup B = B AB    AB=A \subseteq B \iff A - B = \emptyset AB    BAA \subseteq B \iff \overline{B} \subseteq \overline{A}

Proving Set Identities

Method 1: Membership Table

Similar to truth tables, use 1 for membership and 0 for non-membership.

Example: Prove A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

AABBCCBCB \cup CA(BC)A \cap (B \cup C)ABA \cap BACA \cap C(AB)(AC)(A \cap B) \cup (A \cap C)
00000000
00110000
01010000
01110000
10000000
10111011
11011101
11111111

Columns 5 and 8 are identical ✓

Method 2: Element Argument

Show both sets contain exactly the same elements.

Example: Prove AB=AB\overline{A \cap B} = \overline{A} \cup \overline{B}

Proof: xABx \in \overline{A \cap B}     x(AB)\iff x \notin (A \cap B)     ¬(xAxB)\iff \neg(x \in A \land x \in B)     xAxB (De Morgan’s for logic)\iff x \notin A \lor x \notin B \text{ (De Morgan's for logic)}     xAxB\iff x \in \overline{A} \lor x \in \overline{B}     xAB\iff x \in \overline{A} \cup \overline{B} \square

Method 3: Chain of Identities

Transform one side to the other using known identities.

Example: Prove A(AB)=AA \cup (A \cap B) = A

Proof: A(AB)A \cup (A \cap B) =(AU)(AB) (identity law)= (A \cap U) \cup (A \cap B) \text{ (identity law)} =A(UB) (distributive law)= A \cap (U \cup B) \text{ (distributive law)} =AU (domination law)= A \cap U \text{ (domination law)} =A (identity law)= A \text{ (identity law)} \square

Duality Principle

Every set identity has a dual obtained by:

  • Swapping \cup and \cap
  • Swapping \emptyset and UU

If an identity is true, its dual is also true.

Examples of Duals

IdentityDual
A=AA \cup \emptyset = AAU=AA \cap U = A
AU=UA \cup U = UA=A \cap \emptyset = \emptyset
A(BC)=(AB)(AC)A \cup (B \cap C) = (A \cup B) \cap (A \cup C)A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
AB=AB\overline{A \cup B} = \overline{A} \cap \overline{B}AB=AB\overline{A \cap B} = \overline{A} \cup \overline{B}

Common Proof Patterns

Showing ABA \subseteq B

Let xAx \in A be arbitrary. Show xBx \in B.

Showing A=BA = B

  1. Show ABA \subseteq B
  2. Show BAB \subseteq A

Showing A=A = \emptyset

Assume xAx \in A and derive a contradiction.

Showing ABA \neq B

Find an element in one set but not the other.