Mathematical-Reasoning · Unit 15 · Video 4 · Interactive Practice

Venn Diagrams for Three Sets: Eight Regions and Subset Conditions

IKey Formulas

Idea	Rule
Regions for three sets	$2 \cdot 2 \cdot 2 = 8$ : one region for each way a point can be in or out of $A$ , $B$ and $C$
The three circles as region lists	$A$ : regions 1, 2, 4, 5; $B$ : regions 2, 3, 5, 6; $C$ : regions 4, 5, 6, 7
Operations on region lists	$X \cap Y$ : regions on both lists; $X \cup Y$ : regions on either list; $X^c$ : regions of $U$ not on the list
Subset condition $X \subseteq Y$	Every region of $X$ that is not a region of $Y$ is empty: the regions of $X - Y$

Key Insight: Shading describes a set; a subset condition says which regions contain no elements. The most general diagram for $X \subseteq Y$ empties exactly the regions of $X - Y$ and no others. The numbers 1 to 8 label regions; they are not elements of any set.

IIBuilding a Set from Region Lists

Every set built from $A$ , $B$ and $C$ is a list of regions, computed one operation at a time.

Set Step 1

IIIWhich Regions Must Be Empty?

A condition $X \subseteq Y$ forces exactly the regions of $X$ outside $Y$ to be empty, no more and no fewer.

Condition

IVRedrawing the Most General Diagram

Moving the circles until the forced regions vanish, while every other region survives, draws the most general diagram.

Condition Redraw 0%

💡 A common mistake for $B \subseteq A \cup C$ is to draw $B$ inside circle $A$ alone: that also empties region 6, which the condition never required, so that diagram is not the most general one.

VQuiz Questions

Problem 1 · Shade an Intersection

Given: the standard three-set diagram with regions numbered as in the video. Find the regions that make up $A \cap B^c$ .

✅ Correct!

A \cap B^c

is the part of circle

A

outside circle

B

: regions 1 and 4.

❌ Not quite. Regions 1, 2, 4, 5, 7, 8 make up

A \cup B^c

. A union keeps the regions on either list; an intersection keeps only the regions on both.

❌ Close, but… Region 1 is in

A

and outside

B

, but so is region 4: it is in

A

and

C

and not in

B

, so it is also on the list for

B^c

❌ Not quite. Regions 1, 4, 7, 8 make up

B^c

itself. Regions 7 and 8 are not in

A

, so the intersection with

A

drops them.

Show solution

Write each set as a list of regions:

$A$ : regions 1, 2, 4, 5 (the whole circle $A$ ).
$B^c$ : $B$ is regions 2, 3, 5, 6, so $B^c$ is every region of $U$ missing from that list: regions 1, 4, 7, 8.

An intersection keeps only the regions on both lists. Regions 1 and 4 appear on both, so $A \cap B^c$ is regions 1 and 4, the part of circle $A$ that lies outside circle $B$ .

Problem 2 · The Region Outside Every Circle

Given: the standard numbered three-set diagram. Find the regions that make up $B^c \cap C^c$ .

✅ Correct! The points in neither

B

nor

C

are the part of

A

outside both (region 1) and the points outside all three circles (region 8).

❌ Close, but… Region 1 is right, but region 8, outside all three circles yet inside

U

, is in neither

B

nor

C

, so it is on both complement lists.

❌ Not quite. Regions 1, 2, 3, 4, 7, 8 make up

B^c \cup C^c

, the regions on either list. The intersection keeps only the regions on both lists.

❌ Not quite. Regions 2 to 7 make up

B \cup C

: exactly the regions that

B^c \cap C^c

leaves out.

Show solution

$B$ is regions 2, 3, 5, 6, so $B^c$ is regions 1, 4, 7, 8.
$C$ is regions 4, 5, 6, 7, so $C^c$ is regions 1, 2, 3, 8.

The regions on both lists are 1 and 8, so $B^c \cap C^c$ is regions 1 and 8. Region 8 is easy to forget: a point outside all three circles is certainly not in $B$ and not in $C$ .

Problem 3 · Most General Diagram

Given: the condition $B \subseteq A \cap C$ . Find the regions that must be empty in the most general Venn diagram showing it.

✅ Correct! Every point of

B

must lie in both

A

and

C

, so only region 5 of circle

B

may hold elements.

❌ Not quite. Emptying region 3 alone is the answer for

B \subseteq A \cup C

. Here every element of

B

must be in

A

and in

C

, so region 2 (in

A

, not

C

) and region 6 (in

C

, not

A

) are forced empty too.

❌ Not quite. Emptying region 4 shows

A \cap C \subseteq B

, the condition with the roles reversed. Here the forced regions are the regions of

B

missing from

A \cap C

❌ Close, but… Regions 2, 3 and 6 are forced, but region 4 lies in

A \cap C

and not in

B

; the condition says nothing about it, so emptying it gives a diagram that is not the most general.

Show solution

$B$ : regions 2, 3, 5, 6.
$A \cap C$ : $A$ is regions 1, 2, 4, 5 and $C$ is regions 4, 5, 6, 7; the regions on both lists are 4 and 5.

The regions of $B$ that are not regions of $A \cap C$ are 2, 3 and 6, the regions of $B - (A \cap C)$ . The most general diagram empties regions 2, 3 and 6 and keeps the other five: circle $B$ sits inside the lens where $A$ and $C$ overlap, with region 4 (the rest of that lens) still present.

Problem 4 · From Empty Regions Back to a Condition

Given: in the most general diagram for a subset condition, regions 2 and 3 are empty and the other six regions may hold elements. Find which of these conditions it shows.

✅ Correct! Regions 2 and 3 are exactly the regions of

B

outside

C

, so emptying them says every element of

B

is in

C

❌ Not quite.

C \subseteq B

empties the regions of

C

outside

B

: regions 4 and 7.

❌ Not quite.

B \subseteq A \cup C

empties only region 3. Region 2 lies in

A

, so under that condition it may still hold elements.

❌ Not quite.

A \cap B \subseteq C

empties only region 2. Region 3 is not in

A \cap B

, so under that condition it may still hold elements.

Show solution

For each condition $X \subseteq Y$ , the forced-empty regions are the regions of $X$ that are not regions of $Y$ :

$B \subseteq C$ : $B$ is 2, 3, 5, 6 and $C$ is 4, 5, 6, 7, so regions 2 and 3 are empty.
$C \subseteq B$ : regions 4 and 7 are empty.
$B \subseteq A \cup C$ : $A \cup C$ is 1, 2, 4, 5, 6, 7, so only region 3 is empty.
$A \cap B \subseteq C$ : $A \cap B$ is 2, 5, so only region 2 is empty.

Only $B \subseteq C$ empties exactly regions 2 and 3.

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