Mathematical-Reasoning · Unit 16 · Video 2 · Interactive Practice

Complements Reverse Subsets: A ⊆ B Implies Bᶜ ⊆ Aᶜ

IKey Formulas

Statement	Meaning	Name
$A \subseteq B$	for each $x \in U$ , if $x \in A$ , then $x \in B$	Definition of subset
$x \in A^c$	$x \in U$ and $x \notin A$	Definition of complement
$P \to Q \equiv \neg Q \to \neg P$	if $x \in A$ , then $x \in B$ $\;\equiv\;$ if $x \notin B$ , then $x \notin A$	Contrapositive
$A \subseteq B \;\Rightarrow\; B^c \subseteq A^c$	for any subsets $A$ , $B$ of $U$	The proposition

Key Insight: Written with complements, the contrapositive of "if $x \in A$ , then $x \in B$ " is "if $x \in B^c$ , then $x \in A^c$ ", which is exactly what $B^c \subseteq A^c$ asks of an arbitrary element $x$ .

IIChase One Element of the Complement

With $U = \{1, 2, \ldots, 12\}$ , $A = \{4, 8, 12\}$ and $B = \{2, 4, 6, 8, 10, 12\}$ , so $A \subseteq B$ , the same three steps send any chosen $x \in B^c$ into $A^c$ .

IIIOne Witness Breaks Both Inclusions

An element of $A$ outside $B$ is the same thing as an element of $B^c$ outside $A^c$ .

💡 A finite picture only tests these eight elements; the proof is what covers every pair of subsets $A \subseteq B$ of every universal set.

IVThe Know-Show Table, Built from Both Ends

The table fills from the top (hypothesis) and the bottom (conclusion) until the two ends meet.

Step 1

Step	Know	Reason
P	$A \subseteq B$	Hypothesis
P1	Let $x \in B^c$ .	Choose an arbitrary element of $B^c$
P2	If $x \in A$ , then $x \in B$ .	Definition of "subset"
P3	If $x \notin B$ , then $x \notin A$ .	Contrapositive of P2
P4	If $x \in B^c$ , then $x \in A^c$ .	Step P3 and definition of "complement"
Q2	The element $x$ is in $A^c$ .	Steps P1 and P4
Q1	Every element of $B^c$ is an element of $A^c$ .	Choose-an-element method with Steps P1 and Q2
Q	$B^c \subseteq A^c$	Definition of "subset"
Step	Show	Reason

Proof. PLet $A$ and $B$ be subsets of the universal set $U$ , and assume that $A \subseteq B$ . We will prove that $B^c \subseteq A^c$ by proving that each element of $B^c$ is an element of $A^c$ . P1So let $x \in B^c$ . P2, P3Since $A \subseteq B$ , we know that if $x \in A$ , then $x \in B$ . The contrapositive of this conditional statement is: if $x \notin B$ , then $x \notin A$ . P4, Q2Since $x \in B^c$ , we have $x \notin B$ , and so we conclude that $x \notin A$ . That is, $x \in A^c$ . Q1, QThis proves that every element of $B^c$ is an element of $A^c$ , and hence $B^c \subseteq A^c$ . ∎

VQuiz Questions

Problem 1 · Where the Proof Starts

Given: $A$ and $B$ are subsets of $U$ with $A \subseteq B$ . To prove $B^c \subseteq A^c$ with the choose-an-element method, which plan is correct?

✅ Correct!

B^c \subseteq A^c

means every element of

B^c

is in

A^c

, so the proof starts from an arbitrary

x \in B^c

and ends with

x \in A^c

❌ Not quite. That plan would prove

A^c \subseteq B^c

, the reverse inclusion. The element must be chosen from the set on the left of

\subseteq

, here

B^c

❌ Not quite. That swaps hypothesis and conclusion: it is a plan for the converse proposition. Here

A \subseteq B

is given and

B^c \subseteq A^c

is what must be shown.

❌ Not quite. Checking one element shows only that this element lies in

A^c

. The conclusion is a universal statement, so the chosen

x

must be arbitrary.

Show solution

By the definition of subset, $B^c \subseteq A^c$ means: for each $x \in U$ , if $x \in B^c$ , then $x \in A^c$ .

This is a universal statement, so the choose-an-element method applies: let $x$ be an arbitrary element of $B^c$ , use only the fact $x \in B^c$ (together with the hypothesis $A \subseteq B$ ), and prove $x \in A^c$ . The answer is (c).

Problem 2 · Which Conditional Is Available

Given: $A \subseteq B$ and $x \in B^c$ . Which conditional is true for every $x \in U$ because $A \subseteq B$ , and lets you conclude $x \notin A$ ?

✅ Correct! It is the contrapositive of "if

x \in A

, then

x \in B

", so it is logically equivalent to it, and its hypothesis

x \notin B

is exactly what

x \in B^c

says.

❌ Not quite. This is the converse of "if

x \in A

, then

x \in B

", and

A \subseteq B

does not give the converse: with

A = \{1\}

and

B = \{1, 2\}

, the element

2

is in

B

but not in

A

❌ Not quite. This is the contrapositive of the converse "if

x \in B

, then

x \in A

", so it is logically equivalent to the converse and does not follow from

A \subseteq B

. It also starts from

x \notin A

, the very fact we are trying to reach.

❌ Not quite. From

A \subseteq B

, an element of

A

is in

B

, so it is not in

B^c

: for each

x \in A

this conditional is false, so it is not true for every

x \in U

once

A

is nonempty. Its hypothesis

x \in A

does not match

x \in B^c

anyway.

Show solution

The hypothesis $A \subseteq B$ gives, for our $x$ : if $x \in A$ , then $x \in B$ . Its starting fact $x \in A$ does not match what we know, $x \in B^c$ .

The contrapositive is logically equivalent: if $x \notin B$ , then $x \notin A$ . Since $x \in B^c$ means $x \notin B$ , this conditional applies and gives $x \notin A$ , that is, $x \in A^c$ . The answer is (a).

Problem 3 · Compute Both Complements

Given: $U = \{1, 2, \ldots, 10\}$ , $A = \{2, 4, 6\}$ and $B = \{1, 2, 3, 4, 5, 6\}$ , so $A \subseteq B$ . Find $B^c$ , then decide which statement is true.

What is $B^c$ ?

Which statement is true?

✅ Correct!

B^c = \{7, 8, 9, 10\}

sits inside

A^c = \{1, 3, 5, 7, 8, 9, 10\}

, and

1

3

5

are in

A^c

but not in

B^c

❌ Not quite.

\{1, 3, 5\}

B - A

, the elements of

B

that are not in

A

B^c

is everything in

U

that is not in

B

❌ Not quite.

\{2, 4, 6\}

A

itself.

B^c

consists of the elements of

U

outside

B

❌ Not quite.

\{1, 3, 5, 7, 8, 9, 10\}

A^c

, the complement of

A

. Remove only the elements of

B

from

U

❌ Not quite.

1 \in A^c

but

1 \in B

, so

1 \notin B^c

. The inclusion reverses: it is

B^c

that sits inside

A^c

❌ Not quite.

B^c \subseteq A^c

holds, but the sets are not equal:

1

3

and

5

are in

A^c

and not in

B^c

❌ Not quite. Since

B^c \subseteq A^c

, we get

A^c \cap B^c = B^c = \{7, 8, 9, 10\}

, which is not empty.

Show solution

$B^c = U - B = \{7, 8, 9, 10\}$ and $A^c = U - A = \{1, 3, 5, 7, 8, 9, 10\}$ .

Each of $7$ , $8$ , $9$ , $10$ is in $A^c$ , so $B^c \subseteq A^c$ , as the proposition guarantees for $A \subseteq B$ . The element $1$ is in $A^c$ but not in $B^c$ , so $A^c \not\subseteq B^c$ and $B^c \neq A^c$ . The intersection $A^c \cap B^c = \{7, 8, 9, 10\}$ is nonempty.

Answers: $B^c = \{7, 8, 9, 10\}$ (d), and $B^c \subseteq A^c$ with $B^c \neq A^c$ (b).

Problem 4 · Complements of Divisibility Sets

Given: $U = \mathbb{Z}$ , $A = \{n \in \mathbb{Z} \mid 12 \text{ divides } n\}$ and $B = \{n \in \mathbb{Z} \mid 4 \text{ divides } n\}$ , where $a$ divides $n$ means $n = ak$ for some integer $k$ . Since $12k = 4(3k)$ , $A \subseteq B$ . Which statement follows from $B^c \subseteq A^c$ ?

✅ Correct!

B^c

is the set of integers not divisible by

4

and

A^c

the set not divisible by

12

, so

B^c \subseteq A^c

says exactly this.

❌ Not quite. This would mean

A^c \subseteq B^c

, the reversed inclusion. It is false:

12

does not divide

4

, but

4

divides

4

❌ Not quite. This says

B \subseteq A

, the converse of the hypothesis. It is false:

4

divides

4

, but

12

does not divide

4

❌ Not quite. This says

A^c \subseteq B

, which is false:

12

does not divide

1

, and

4

does not divide

1

either.

Show solution

If $n \in A$ , then $n = 12k$ for some integer $k$ , so $n = 4(3k)$ with $3k \in \mathbb{Z}$ , and $n \in B$ . Hence $A \subseteq B$ .

By the proposition, $B^c \subseteq A^c$ . Here $B^c = \{n \in \mathbb{Z} \mid 4 \text{ does not divide } n\}$ and $A^c = \{n \in \mathbb{Z} \mid 12 \text{ does not divide } n\}$ , so: for every integer $n$ , if $4$ does not divide $n$ , then $12$ does not divide $n$ . The answer is (d).

The integer $n = 4$ is a counterexample to (a) and (b), and $n = 1$ is a counterexample to (c).

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