Mathematical-Reasoning · Unit 15 · Video 2 · Interactive Practice

Venn Diagrams for Two Sets: Four Regions and De Morgan's Law

IKey Formulas

Set	Regions	Description
$A - B$ , $A \cap B$ , $B - A$ , $A^c \cap B^c$	$1$ , $2$ , $3$ , $4$	The four regions of a two-set diagram
$A^c = \{x \in U \mid x \notin A\}$	$3$ , $4$	Everything outside circle $A$
$A^c \cup B = \{x \in U \mid x \notin A \text{ or } x \in B\}$	$2$ , $3$ , $4$	Regions of $A^c$ together with regions of $B$
$(A \cap B)^c = A^c \cup B^c$	$1$ , $3$ , $4$	De Morgan's law for sets, seen on a diagram

Key Insight: Every set built from $A$ and $B$ with $\cup$ , $\cap$ , $-$ and complement is a union of some of the four regions. Union keeps the regions in either list, intersection the regions in both, difference removes the second list from the first, and complement keeps the regions missing from the list.

IICombine the Region Lists

A compound set is shaded by finding the regions of each piece, then combining the two lists.

Set Step 1

IIIShade It Yourself

Each set named below is exactly a union of some of regions $1$ , $2$ , $3$ and $4$ .

IVOne Picture, Every Pair of Sets

Two sets that shade the same regions have the same elements, wherever the elements of $U = \{0, 1, \ldots, 10\}$ sit.

Compare

💡 Shading the same regions is a check on the diagram, not yet a written proof; a later video proves De Morgan's laws for sets from the set builder definitions.

VQuiz Questions

Problem 1 · Regions of a Compound Set

Given: subsets $A$ and $B$ of $U$ on a two-set Venn diagram with the regions numbered as in the video. Which regions represent $A^c \cap B$ ?

✅ Correct!

A^c

is regions

3

4

and

B

is regions

2

3

; the only region in both lists is

3

, the part of

B

outside

A

❌ That is the union. Regions

2

3

4

are in either list, which is

A^c \cup B

. Intersection keeps only the regions in both lists.

❌ The roles of $A$ and $B$ are swapped. Region

1

is in

A

and not in

B

, which is

A \cap B^c

. Here the element must be outside

A

and inside

B

❌ Not quite. Region

4

is outside both circles, so it is not in

B

. That region is

A^c \cap B^c

Show solution

Find the regions of each piece, then keep the regions in both lists:

A^c:\ \text{regions } 3, 4 \qquad B:\ \text{regions } 2, 3 \qquad A^c \cap B:\ \text{region } 3.

In set builder notation, $A^c \cap B = \{x \in U \mid x \notin A \text{ and } x \in B\}$ , the elements in $B$ and not in $A$ . That is region $3$ , the set $B - A$ .

Problem 2 · Which Set Is This?

Given: on a two-set Venn diagram, regions $1$ , $3$ and $4$ are shaded and only the centre overlap, region $2$ , is unshaded. Which set is shaded?

✅ Correct!

A \cap B

is region

2

, so its complement is everything else: regions

1

3

4

. The set

A^c \cup B^c

shades the same three regions.

❌ This is the De Morgan pitfall.

A^c

is regions

3

4

and

B^c

is regions

1

4

; the regions in both lists are just region

4

. Complementing each piece turns the

\cap

into a

\cup

A^c \cup B^c

is regions

1

3

4

❌ Close, but region $4$ is missing.

(A \cup B) - (A \cap B)

removes region

2

from regions

1

2

3

, leaving only regions

1

3

. Region

4

lies outside

A \cup B

❌ Not quite.

A^c \cup B

combines regions

3

4

with regions

2

3

, giving regions

2

3

4

. Its unshaded region is

1

, not

2

Show solution

Compute the regions of each option:

$A^c \cap B^c$ : regions in both $\{3, 4\}$ and $\{1, 4\}$ , so region $4$ .
$(A \cup B) - (A \cap B)$ : regions $1$ , $2$ , $3$ with region $2$ removed, so regions $1$ , $3$ .
$A^c \cup B$ : regions in either $\{3, 4\}$ or $\{2, 3\}$ , so regions $2$ , $3$ , $4$ .
$(A \cap B)^c$ : the regions missing from region $2$ , so regions $1$ , $3$ , $4$ .

Only $(A \cap B)^c$ matches. In set builder notation it is $\{x \in U \mid x \notin A \cap B\}$ , and $x \notin A \cap B$ holds exactly when $x \notin A$ or $x \notin B$ .

Problem 3 · From Regions to a Roster

Given: $U = \{0, 1, 2, \ldots, 10\}$ , $A = \{0, 1, 2, 3, 9\}$ and $B = \{2, 3, 4, 5, 6\}$ , the sets from the last video. Find the regions of $A^c \cup B$ , then its roster.

Which regions represent $A^c \cup B$ ?

What is $A^c \cup B$ in roster form?

✅ Correct! Regions

2

3

4

hold

\{2, 3\}

\{4, 5, 6\}

and

\{7, 8, 10\}

; only region

1

, the elements

0

1

9

, is left out.

❌ That is $A^c \cup B^c$ . Here the second piece is

B

, regions

2

3

, not

B^c

, regions

1

4

❌ Only half the union. Regions

3

4

are

A^c

alone; the union must also include the regions of

B

, which are

2

3

❌ That is the intersection. Region

3

is in both lists, so it is

A^c \cap B

. Union keeps every region in either list.

❌ Check the regions.

A^c

is regions

3

4

and

B

is regions

2

3

❌ That is $A^c \cup B^c$ . It contains

0

1

9

from region

1

, and those elements are in

A

but not in

B

, so they are not in

A^c \cup B

❌ That is $A^c \cap B$ . The elements

4

5

6

are region

3

only; the union also needs

2

3

from region

2

and

7

8

10

from region

4

❌ That is $A \cup B$ . It contains

0

1

9

, which are in

A

and not in

B

, and it misses

7

8

10

, which are in

A^c

❌ Check the roster. List the elements of regions

2

3

and

4

Show solution

Regions. $A^c$ is regions $3$ , $4$ and $B$ is regions $2$ , $3$ . Union keeps every region in either list, counting region $3$ once: regions $2$ , $3$ , $4$ .

Elements of each region.

\text{Region 1: } A - B = \{0, 1, 9\} \qquad \text{Region 2: } A \cap B = \{2, 3\}

\text{Region 3: } B - A = \{4, 5, 6\} \qquad \text{Region 4: } A^c \cap B^c = \{7, 8, 10\}

Roster. Combine regions $2$ , $3$ , $4$ :

A^c \cup B = \{2, 3, 4, 5, 6, 7, 8, 10\}.

Check with the definition: $A^c = \{4, 5, 6, 7, 8, 10\}$ , and adding the elements $2$ , $3$ of $B$ gives the same set.

Problem 4 · The Companion Law on a Diagram

Given: subsets $A$ and $B$ of $U$ . Which set shades exactly the same regions as $(A \cup B)^c$ ?

✅ Correct!

A \cup B

is regions

1

2

3

, so

(A \cup B)^c

is region

4

, and

A^c \cap B^c

is region

4

too. This is the companion De Morgan law, which a later video proves from the definitions.

❌ Not quite.

A^c \cup B^c

is regions

1

3

4

, the same as

(A \cap B)^c

. Complementing a union turns it into an intersection, not a union.

❌ Not quite.

(A \cup B) - (A \cap B)

is regions

1

3

, the elements in exactly one of the sets.

(A \cup B)^c

holds the elements in neither set.

❌ Too many regions.

B^c

is regions

1

4

, and region

1

lies inside circle

A

, so it is inside

A \cup B

Show solution

$A \cup B$ is regions $1$ , $2$ , $3$ , so its complement is the one region missing from that list: region $4$ , outside both circles.

$A^c \cup B^c$ : regions in either $\{3, 4\}$ or $\{1, 4\}$ , so regions $1$ , $3$ , $4$ .
$A^c \cap B^c$ : regions in both $\{3, 4\}$ and $\{1, 4\}$ , so region $4$ .
$(A \cup B) - (A \cap B)$ : regions $1$ , $3$ .
$B^c$ : regions $1$ , $4$ .

So the diagram suggests $(A \cup B)^c = A^c \cap B^c$ . In words, $x \notin A \cup B$ holds exactly when $x \notin A$ and $x \notin B$ , mirroring $\neg(P \vee Q) \equiv \neg P \wedge \neg Q$ .

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