Mathematical-Reasoning · Unit 19 · Video 1 · Interactive Practice

Union and Intersection of a Family: Some Versus All

IKey Formulas

Formula	Name	Meaning
$(A \cup B) \cup C = A \cup (B \cup C)$ $(A \cap B) \cap C = A \cap (B \cap C)$	Associative laws	A string of unions alone, or of intersections alone, needs no parentheses
$\displaystyle\bigcup_{X \in \mathcal{C}} X = \{x \in U \mid x \in X \text{ for some } X \in \mathcal{C}\}$	Union of a family	Elements in at least one member of $\mathcal{C}$
$\displaystyle\bigcap_{X \in \mathcal{C}} X = \{x \in U \mid x \in X \text{ for all } X \in \mathcal{C}\}$	Intersection of a family	Elements in every member of $\mathcal{C}$
$\displaystyle\Big(\bigcup_{X \in \mathcal{A}} X\Big)^c = \bigcap_{X \in \mathcal{A}} X^c$	De Morgan for a family	Checked here for $\mathcal{A} = \{A, B, C, D\}$ ; proved for every family later in the unit

Key Insight: The two definitions differ in a single word: some for the union, all for the intersection. With $A = \{1, \ldots, 5\}$ , $B = \{2, \ldots, 6\}$ , $C = \{3, \ldots, 7\}$ , $D = \{4, \ldots, 8\}$ in $U = \mathbb{N}$ , the family $\mathcal{A} = \{A, B, C, D\}$ has union $\{1, \ldots, 8\}$ and intersection $\{4, 5\}$ .

IIVisualization 1 — Some Versus All

A number is in the union of a family when it lies in some member, and in the intersection when it lies in all.

Family Element x 7

IIIVisualization 2 — Grouping a String of Sets

Regrouping a string of unions, or of intersections, never changes the result; regrouping a mixed string can.

Expression Step 0

IVVisualization 3 — The Complement of a Union

A natural number lies in no member exactly when it lies in every complement, so $\big(\bigcup X\big)^c = \bigcap X^c$ .

Number n 3

VQuiz Questions

Problem 1 · Intersection of a Family

Given: $B = \{2, \ldots, 6\}$ , $C = \{3, \ldots, 7\}$ , $S = \{5, \ldots, 9\}$ and the family $\mathcal{C} = \{B, C, S\}$ — find $\displaystyle\bigcap_{X \in \mathcal{C}} X$ .

✅ Correct! Only

5

and

6

lie in all three members:

B

ends at

6

and

S

begins at

5

❌ Not quite. That is the union, the numbers in some member. The intersection keeps only the numbers in every member.

❌ Close, but… Check

6

: it is in

B = \{2, \ldots, 6\}

, in

C

and in

S

, so it belongs as well.

❌ Not quite. The members overlap:

5

is in

B

, in

C

and in

S

, so the intersection is not empty.

❌ Not quite. List the numbers that are in

B

and in

C

and in

S

Show solution

An element is in $\bigcap_{X \in \mathcal{C}} X$ exactly when it is in every member of $\mathcal{C}$ , that is, in $B$ and in $C$ and in $S$ :

\bigcap_{X \in \mathcal{C}} X = B \cap C \cap S = \{2, \ldots, 6\} \cap \{3, \ldots, 7\} \cap \{5, \ldots, 9\}.

First $B \cap C = \{3, 4, 5, 6\}$ , and keeping only what is also in $S$ leaves

\bigcap_{X \in \mathcal{C}} X = \{5, 6\}.

For comparison, the union is $\{2, \ldots, 9\}$ : every number from the smallest element of $B$ to the largest element of $S$ .

Problem 2 · Members Versus Elements

Given: $A = \{1, \ldots, 5\}$ , $B = \{2, \ldots, 6\}$ , $C = \{3, \ldots, 7\}$ , $D = \{4, \ldots, 8\}$ and the family $\mathcal{A} = \{A, B, C, D\}$ — which statement is true?

✅ Correct!

3 \in A

and

A \in \mathcal{A}

, so

3

is in some member of

\mathcal{A}

, which is exactly membership in the union.

❌ Not quite. The elements of

\mathcal{A}

are the four sets

A

B

C

D

. The number

3

is an element of

A

, not of

\mathcal{A}

❌ Not quite.

\{4, 5\}

is the intersection of the family, but it is none of the four members

A

B

C

D

❌ Not quite.

A \subseteq \mathcal{A}

would need every element of

A

, such as

1

, to be an element of

\mathcal{A}

. The true statement is

A \in \mathcal{A}

Show solution

The family $\mathcal{A}$ has exactly four elements, and each is a set: $A$ , $B$ , $C$ and $D$ .

$3 \in \mathcal{A}$ is false: $3$ is a number, and no member of $\mathcal{A}$ is a number.
$\{4, 5\} \in \mathcal{A}$ is false: $\{4, 5\}$ is not equal to any of $A$ , $B$ , $C$ , $D$ .
$A \subseteq \mathcal{A}$ is false: $1 \in A$ but $1 \notin \mathcal{A}$ .
$3 \in \bigcup_{X \in \mathcal{A}} X$ is true: take $X = A$ ; then $X \in \mathcal{A}$ and $3 \in X$ , so $3$ is in some member.

To reach the numbers inside the members of a family, take its union or intersection.

Problem 3 · Mixed Operations Keep Their Parentheses

Given: $A = \{1, \ldots, 5\}$ , $C = \{3, \ldots, 7\}$ , $D = \{4, \ldots, 8\}$ — find $A \cup (C \cap D)$ and $(A \cup C) \cap D$ .

What is $A \cup (C \cap D)$ ?

What is $(A \cup C) \cap D$ ?

✅ Correct! The same sets and the same two operations give

\{1, \ldots, 7\} \ne \{4, 5, 6, 7\}

, so in a mixed expression the parentheses must stay.

❌ Not quite. That is the value of the other grouping. Here the intersection is inside:

C \cap D = \{4, 5, 6, 7\}

, and then the union with

A

adds

1

2

3

❌ Not quite. That is the value of the other grouping. Here the union is inside:

A \cup C = \{1, \ldots, 7\}

, and then intersecting with

D

keeps only what is also in

D

❌ Check the first expression. Evaluate the parentheses first:

C \cap D

is the numbers in both

C

and

D

❌ Check the second expression. Evaluate

A \cup C

first, then keep only the numbers that are also in

D = \{4, \ldots, 8\}

Show solution

First grouping. Inside the parentheses, $C \cap D = \{3, \ldots, 7\} \cap \{4, \ldots, 8\} = \{4, 5, 6, 7\}$ . Then

A \cup (C \cap D) = \{1, \ldots, 5\} \cup \{4, 5, 6, 7\} = \{1, \ldots, 7\}.

Second grouping. Inside the parentheses, $A \cup C = \{1, \ldots, 5\} \cup \{3, \ldots, 7\} = \{1, \ldots, 7\}$ . Then

(A \cup C) \cap D = \{1, \ldots, 7\} \cap \{4, \ldots, 8\} = \{4, 5, 6, 7\}.

The results differ ( $1 \in A \cup (C \cap D)$ but $1 \notin (A \cup C) \cap D$ ), so the expression $A \cup C \cap D$ has no single meaning. The associative laws only cover strings with one operation.

Problem 4 · A Family of Subsets

Given: $\mathcal{D}$ is the family of all subsets of $\{1, 2, 3\}$ that contain $1$ — find $\displaystyle\bigcup_{X \in \mathcal{D}} X$ and $\displaystyle\bigcap_{X \in \mathcal{D}} X$ .

What is the union of $\mathcal{D}$ ?

What is the intersection of $\mathcal{D}$ ?

✅ Correct! Every number

1

2

3

is in some member, but only

1

is in every member.

❌ Not quite.

\{1\}

is what every member shares. The union only needs some member:

2 \in \{1, 2\}

and

3 \in \{1, 3\}

❌ Not quite. That is the family

\mathcal{D}

itself, a set of sets. Its union is a set of numbers: the numbers that lie in some member.

❌ Check the union. Ask, for each of

1

2

3

, whether some member of

\mathcal{D}

contains it.

❌ Not quite. Every member of

\mathcal{D}

contains

1

by definition, so

1

is in all of them.

❌ Not quite.

2

is not in the member

\{1\}

, and

3

is not in

\{1\}

either, so neither is in every member.

❌ Not quite. The intersection collects numbers, not sets: its elements are the numbers lying in every member, so it is

\{1\}

, not

\{\{1\}\}

❌ Check the intersection. Ask, for each of

1

2

3

, whether every member of

\mathcal{D}

contains it.

Show solution

A subset of $\{1, 2, 3\}$ that contains $1$ is $\{1\}$ together with any choice of $2$ and $3$ , so

\mathcal{D} = \big\{\{1\}, \{1, 2\}, \{1, 3\}, \{1, 2, 3\}\big\}.

Union (some). $1 \in \{1\}$ , $2 \in \{1, 2\}$ and $3 \in \{1, 3\}$ , so each of $1$ , $2$ , $3$ is in some member:

\bigcup_{X \in \mathcal{D}} X = \{1, 2, 3\}.

Intersection (all). $1$ is in every member by the definition of $\mathcal{D}$ , while $2 \notin \{1\}$ and $3 \notin \{1\}$ :

\bigcap_{X \in \mathcal{D}} X = \{1\}.

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