Mathematical-Reasoning · Unit 15 · Video 1 · Interactive Practice

Intersection, Union, Difference, Complement: And, Or, Not for Sets

IKey Formulas

Operation	Definition	Logic word
Intersection	$A \cap B = \{x \in U \mid x \in A \text{ and } x \in B\}$	and
Union	$A \cup B = \{x \in U \mid x \in A \text{ or } x \in B\}$	or (inclusive)
Set difference	$A - B = \{x \in U \mid x \in A \text{ and } x \notin B\}$	and not
Complement	$A^c = \{x \in U \mid x \notin A\}$	not

Key Insight: each definition is a logical operator written inside the braces. Order matters for the difference ( $A - B \ne B - A$ in general), and the complement always needs a stated universal set $U$ .

IIOne Chart, Four Operations

An element of $U$ belongs to the new set exactly when the logic word holds for its membership in $A$ and $B$ .

Operation

IIIShading Compound Sets

A compound set is built inside out: form the inner set first, then apply the outer operation.

Set

💡 $(A \cap B)^c$ and $A^c \cup B^c$ shade the same regions and list the same elements; the next video shows why this equality holds for every pair of sets.

IVThe Complement Depends on U

The same set $A = \{0, 1, 2, 3, 9\}$ has a different complement in each universal set.

Universal set

VQuiz Questions

Problem 1 · Intersection by Listing

Given: $U = \{1, 2, \ldots, 10\}$ , $A = \{1, 2, 4, 7, 8\}$ and $B = \{2, 3, 4, 8, 9\}$ — find $A \cap B$ .

✅ Correct!

2

4

and

8

are the only elements that appear in both lists.

❌ That is the union.

A \cup B

collects elements in

A

or in

B

; the intersection keeps only those in

A

and in

B

❌ That is $A - B$ .

1

and

7

are in

A

but not in

B

; the intersection wants elements in both sets.

❌ That is $B - A$ .

3

and

9

are in

B

but not in

A

, so they are not in

A \cap B

Show solution

Go through $A$ one element at a time and keep those that are also in $B$ :

$1 \notin B$ , $2 \in B$ , $4 \in B$ , $7 \notin B$ , $8 \in B$ .

A \cap B = \{x \in U \mid x \in A \text{ and } x \in B\} = \{2, 4, 8\}

Problem 2 · The Order of a Difference

Given: $U = \{1, 2, \ldots, 10\}$ , $A = \{1, 2, 3, 4, 5, 6\}$ and $B = \{1, 3, 5, 7, 9\}$ — find $B - A$ .

✅ Correct!

B - A

keeps the elements of

B

that are not in

A

: only

7

and

9

❌ Order reversed.

\{2, 4, 6\}

A - B

. In

B - A

the elements come from

B

, and

2

4

6

are not in

B

❌ That is $A \cap B$ .

1

3

and

5

are in

B

but they are also in

A

, so the difference removes them.

❌ That is $A^c$ . You removed

A

's elements from

U

instead of from

B

;

8

and

10

are not in

B

Show solution

$B - A = \{x \in U \mid x \in B \text{ and } x \notin A\}$ . Check each element of $B$ :

$1, 3, 5 \in A$ , so they are removed.
$7, 9 \notin A$ , so they stay.

B - A = \{7, 9\}, \qquad A - B = \{2, 4, 6\}

The two differences are different sets, which is why the order of $A$ and $B$ matters.

Problem 3 · Where the Parentheses Go

Given: $U = \{1, 2, \ldots, 8\}$ , $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$ — find $(A \cap B)^c$ .

✅ Correct! Inside out:

A \cap B = \{3, 4\}

, and its complement is everything else in

U

❌ That is $A^c \cap B^c$ .

A^c = \{5, 6, 7, 8\}

and

B^c = \{1, 2, 7, 8\}

share only

7

and

8

, but the parentheses say to form

A \cap B

first and then take its complement.

❌ You stopped at the inner set.

\{3, 4\}

A \cap B

; the outer complement still has to be applied.

❌ Missing elements.

7

and

8

are in neither

A

nor

B

, so they are not in

A \cap B

and therefore belong to its complement.

Show solution

Inner set: $A \cap B = \{3, 4\}$ .

Outer operation: remove $3$ and $4$ from $U$ :

(A \cap B)^c = \{1, 2, 5, 6, 7, 8\}

Compare $A^c \cap B^c = \{5, 6, 7, 8\} \cap \{1, 2, 7, 8\} = \{7, 8\}$ , a different set: the parentheses matter.

Problem 4 · Operations on Infinite Sets

Given: $U = \mathbb{Z}$ , $A = \{x \in \mathbb{Z} \mid x \text{ is even}\}$ and $B = \{x \in \mathbb{Z} \mid x \ge 3\}$ — find $B - A$ .

✅ Correct!

B - A = \{x \in \mathbb{Z} \mid x \ge 3 \text{ and } x \text{ is odd}\}

❌ That is $A \cap B$ . These integers are at least

3

and even; the difference removes the even ones instead of keeping them.

❌ Order reversed. These are the even integers less than

3

, which is

A - B

❌ That is $A^c$ . These are all the odd integers;

B - A

also requires

x \ge 3

, so

-1

and

1

do not belong.

Show solution

Write the definition and substitute the two properties:

B - A = \{x \in \mathbb{Z} \mid x \in B \text{ and } x \notin A\} = \{x \in \mathbb{Z} \mid x \ge 3 \text{ and } x \text{ is not even}\}

An integer that is not even is odd, so $B - A$ is the set of odd integers that are at least $3$ :

B - A = \{3, 5, 7, 9, \ldots\}

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