Mathematical-Reasoning · Unit 17 · Video 2 · Interactive Practice

Laws of Union and Intersection: Proving A ∩ B = B ∩ A

IKey Formulas

Statement	For subsets of $U$	Name
$A \cap B \subseteq A$ and $A \subseteq A \cup B$	any $A$ , $B$	Theorem, Part 1
$A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$	whenever $A \subseteq B$	Theorem, Part 2
$A \cap B = B \cap A$ and $A \cup B = B \cup A$	any $A$ , $B$	Commutative laws
$S = T \iff \big(\text{for each } x \in U,\ x \in S \iff x \in T\big)$	any $S$ , $T$	Equality, element by element

Key Insight: A chain $x \in S \iff \cdots \iff x \in T$ proves $S = T$ only when every link holds in both directions, as a definition or a logical equivalence does; a single one-way step $\Rightarrow$ (read from $S$ toward $T$ ) leaves only $S \subseteq T$ .

IIIntersect or Unite Both Sides with the Same $C$

If $A \subseteq B$ , then $A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$ ; the converses can fail.

IIIOne Chain, Every Element

Wherever $x$ lies in $U$ , all four statements in the chain have the same truth value.

Operation

IVTwo-Way Links and One-Way Steps

A link may sit in an equality chain only if it holds both ways in all four regions.

Link

💡 The four regions are the four rows of a truth table for " $x \in A$ " and " $x \in B$ ", so a link that agrees in all four holds for every $x \in U$ .

VQuiz Questions

Problem 1 · Combine the Two Claims of Part 1

Given: $A$ , $B$ and $C$ are subsets of a universal set $U$ . Which statement is true for every choice of $A$ , $B$ and $C$ ?

✅ Correct!

A \cap B \subseteq A

and

A \subseteq A \cup C

, and subset is transitive.

❌ That inclusion points the wrong way. A union contains each of its sets:

A = \emptyset

B = \{1\}

gives

A \cup B = \{1\} \not\subseteq A

❌ That inclusion points the wrong way. An intersection sits inside each of its sets:

A = \{1\}

B = \emptyset

gives

A \not\subseteq A \cap B = \emptyset

❌ Not quite. Take

A = \{1\}

and

C = \{2\}

A \cup C = \{1, 2\}

but

A \cap C = \emptyset

, so a union is generally not inside an intersection.

Show solution

By Part 1 of the theorem, $A \cap B \subseteq A$ (an intersection is contained in each set it is formed from) and $A \subseteq A \cup C$ (a union contains each of its sets). Element by element: let $x \in A \cap B$ . Then $x \in A$ and $x \in B$ , so $x \in A$ . Hence " $x \in A$ or $x \in C$ " is true, so $x \in A \cup C$ . Therefore $A \cap B \subseteq A \cup C$ .

The other three fail for some sets:

(b) $A = \emptyset$ , $B = \{1\}$ : $A \cup B = \{1\} \not\subseteq \emptyset$ .
(c) $A = \{1\}$ , $B = \emptyset$ : $A = \{1\} \not\subseteq A \cap B = \emptyset$ .
(d) $A = \{1\}$ , $C = \{2\}$ : $A \cup C = \{1, 2\} \not\subseteq A \cap C = \emptyset$ .

Problem 2 · What the Converse Does Not Give

Given: $A$ , $B$ , $C \subseteq U$ and $A \cap C \subseteq B \cap C$ . Which conclusion always follows?

✅ Correct! Each element of

A \cap C

lies in

B \cap C

, and

B \cap C \subseteq B

❌ That is the converse of Part 2, and it is false. With

C = \emptyset

, both intersections are empty for any

A

and

B

; for example

A = \{1\}

B = \emptyset

❌ A nonempty $C$ does not rescue the converse. Take

A = \{1\}

B = \emptyset

C = \{2\}

A \cap C = \emptyset \subseteq B \cap C = \emptyset

, yet

1 \in A

and

1 \notin B

❌ The hypothesis is only an inclusion. Take

A = \emptyset

B = C = \{1\}

A \cap C = \emptyset \subseteq \{1\} = B \cap C

, but the two intersections are not equal.

❌ Not quite. The hypothesis says nothing about elements of

A

that are not in

C

Show solution

(c) holds. Let $x \in A \cap C$ . By the hypothesis, $x \in B \cap C$ , so $x \in B$ and $x \in C$ ; in particular $x \in B$ . Hence $A \cap C \subseteq B$ .

The others fail. The hypothesis only restricts elements of $A$ that also lie in $C$ .

(a) $C = \emptyset$ , $A = \{1\}$ , $B = \emptyset$ : $A \cap C = \emptyset \subseteq \emptyset = B \cap C$ , but $A \not\subseteq B$ . This is the video's counterexample to the converse.
(b) $C = \{2\}$ , $A = \{1\}$ , $B = \emptyset$ : again both intersections are empty and $A \not\subseteq B$ .
(d) $A = \emptyset$ , $B = C = \{1\}$ : $A \cap C = \emptyset \subseteq \{1\} = B \cap C$ , but $A \cap C \neq B \cap C$ .

Problem 3 · Audit a Chain

Given: A student writes, for $x \in U$ , $x \in A \cap B \overset{(1)}{\iff} x \in A \text{ and } x \in B \overset{(2)}{\iff} x \in A \overset{(3)}{\iff} x \in A \cup B,$ and concludes $A \cap B = A \cup B$ . Which assessment is right?

✅ Correct! Two one-way steps; chaining the forward directions still gives

x \in A \cap B \Rightarrow x \in A \cup B

❌ Not quite. Take

A = \{1\}

B = \emptyset

A \cap B = \emptyset

but

A \cup B = \{1\}

, so the conclusion is false and some link must fail.

❌ Link (3) fails backward too. From

x \in A \cup B

we cannot get

x \in A

: an element of

B

outside

A

is in

A \cup B

❌ Link (1) is the definition of intersection, which holds in both directions.

Show solution

Write $P$ for " $x \in A$ " and $Q$ for " $x \in B$ ".

(1) $x \in A \cap B \iff P \wedge Q$ : the definition of $\cap$ , two-way.
(2) $P \wedge Q \Rightarrow P$ is true, but $P \Rightarrow P \wedge Q$ fails when $P$ is true and $Q$ false ( $x \in A$ , $x \notin B$ ).
(3) $P \Rightarrow P \vee Q$ is true, but $P \vee Q \Rightarrow P$ fails when $P$ is false and $Q$ true ( $x \in B$ , $x \notin A$ ).

Chaining only the forward directions gives: if $x \in A \cap B$ , then $x \in A \cup B$ . Since $x$ was arbitrary, the chain proves $A \cap B \subseteq A \cup B$ and nothing more. Equality is false in general: $A = \{1\}$ , $B = \emptyset$ gives $\emptyset \ne \{1\}$ .

Problem 4 · A Chain for a New Law

Given: for $x \in U$ , $x \in A \cap (A \cup B) \iff x \in A \text{ and } (x \in A \text{ or } x \in B) \iff \;?$ With $P$ = " $x \in A$ " and $Q$ = " $x \in B$ ", which logical equivalence closes the chain in one two-way link (check each candidate with a truth table), and what set equality does the chain prove?

Which equivalence is valid?

Which equality does the chain prove?

✅ Correct!

x \in A \cap (A \cup B) \iff P \wedge (P \vee Q) \iff P \iff x \in A

, so

A \cap (A \cup B) = A

❌ Check the truth table. When

P

is false,

P \wedge (P \vee Q)

is false whatever

Q

is; when

P

is true, it is true whatever

Q

is.

❌ Not quite. The last statement of the chain is

x \in A

, so the set on the right is the set of elements making that statement true.

Show solution

Truth table for $P \wedge (P \vee Q)$ :

$P$	$Q$	$P \vee Q$	$P \wedge (P \vee Q)$
T	T	T	T
T	F	T	T
F	T	T	F
F	F	F	F

The last column equals the $P$ column, so $P \wedge (P \vee Q) \equiv P$ . The other three fail: $P \vee Q$ differs from it in row 3, and $P \wedge Q$ and $Q$ differ from it in row 2. Then $x \in A \cap (A \cup B) \iff x \in A \text{ and } (x \in A \text{ or } x \in B) \iff x \in A,$ using the definitions of $\cap$ and $\cup$ , then the equivalence. Every link is two-way and $x \in U$ was arbitrary, so $A \cap (A \cup B) = A$ .

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Laws of Union and Intersection: Proving A ∩ B = B ∩ A

IKey Formulas

IIIntersect or Unite Both Sides with the Same CCC

IIIOne Chain, Every Element

IVTwo-Way Links and One-Way Steps

VQuiz Questions

IIIntersect or Unite Both Sides with the Same $C$