Mathematical-Reasoning · Unit 25 · Video 4 · Interactive Practice

Proof: The Image of an Intersection Is Only a Subset

IKey Formulas

Statement	Kind	How it is proved
$y \in f(A) \iff$ there is an $x \in A$ with $f(x) = y$	Definition of the image	Used in every step below
$f(A \cap B) \subseteq f(A) \cap f(B)$	Inclusion, for every $f$	One witness $x \in A \cap B$ serves both sets
$f(A \cup B) = f(A) \cup f(B)$	Equality, for every $f$	Two inclusions, each with two cases

Key Insight: From $y \in f(A)$ and $y \in f(B)$ we get a witness in $A$ and a witness in $B$ , and nothing makes them the same element. That is why the reverse inclusion $f(A) \cap f(B) \subseteq f(A \cap B)$ can fail.

IIWitnesses for an Output

An output lies in $f(A \cap B)$ only when one of its witnesses lies in both $A$ and $B$ .

Clicking an input

💡 Challenge: keep $A = \{1, 2, 4\}$ and change $B$ so that $f(A \cap B) = f(A) \cap f(B)$ , with $B$ still containing an element outside $A$ .

IIIThe Two Proofs, Step by Step

Each inclusion starts from an arbitrary element of the left-hand set and unpacks the definition of the image.

Proof Step 1

Step 1 — Choose an element. Let

y \in f(A \cap B)

. We must show

y \in f(A) \cap f(B)

Step 2 — Unpack the image. By the definition of the image, there is an

x \in A \cap B

with

f(x) = y

Step 3 — Split the intersection. Since

x \in A \cap B

, we have

x \in A

and

x \in B

Step 4 — Use the same $x$ twice.

x \in A

and

f(x) = y

give

y \in f(A)

; the same

x \in B

gives

y \in f(B)

Step 5 — Conclude. So

y \in f(A) \cap f(B)

. Every element of

f(A \cap B)

lies in

f(A) \cap f(B)

, so

f(A \cap B) \subseteq f(A) \cap f(B)

Step 1 — First inclusion: choose an element. Let

y \in f(A \cup B)

. Then there is an

x \in A \cup B

with

f(x) = y

, so

x \in A

x \in B

Step 2 — The two cases. If

x \in A

, then

y = f(x) \in f(A)

. If

x \in B

, then

y \in f(B)

. Either way

y \in f(A) \cup f(B)

, so

f(A \cup B) \subseteq f(A) \cup f(B)

Step 3 — Reverse inclusion: choose an element. Let

y \in f(A) \cup f(B)

, so

y \in f(A)

y \in f(B)

Step 4 — The two cases. If

y \in f(A)

, there is an

x \in A

with

f(x) = y

; since

A \subseteq A \cup B

y \in f(A \cup B)

. If

y \in f(B)

, the same argument works with

B \subseteq A \cup B

Step 5 — Conclude. So

f(A) \cup f(B) \subseteq f(A \cup B)

. Each set is a subset of the other, hence

f(A \cup B) = f(A) \cup f(B)

IVQuiz Questions

Problem 1 · The Step After "Let y Be in the Image"

Given: a proof of $f(A \cap B) \subseteq f(A) \cap f(B)$ for $f: S \to T$ begins "Let $y \in f(A \cap B)$ ." Which statement follows directly from the definition of the image?

✅ Correct!

y \in f(X)

means exactly that some element of

X

is sent to

y

; here

X = A \cap B

❌ Not quite.

y

is an element of the codomain

T

, while

A \cap B \subseteq S

. The image consists of outputs, not inputs.

❌ Not quite. The definition gives some witness in

A \cap B

, not that all witnesses are there. In the modulo

8

example,

f(4) = 2

with

4 \in A \cap B

, but also

f(0) = 2

with

0 \notin A \cap B

❌ Not quite. That would make

f

constant on

A \cap B

. Only one element of

A \cap B

is promised to reach

y

Show solution

By definition, $f(A \cap B) = \{ f(x) \mid x \in A \cap B \}$ , so $y \in f(A \cap B)$ means $y = f(x)$ for at least one $x \in A \cap B$ . That is statement (b), the witness the rest of the proof uses twice.

Problem 2 · Find the Counterexample

Given: $f: \{1, 2, 3\} \to \{p, q\}$ with $f(1) = p$ , $f(2) = q$ and $f(3) = p$ . Which choice of $A$ and $B$ gives $f(A \cap B) \ne f(A) \cap f(B)$ ?

✅ Correct!

A \cap B = \emptyset

, so

f(A \cap B) = \emptyset

, but

f(A) = f(B) = \{p\}

: the witnesses

1

and

3

for

p

are different elements.

❌ Not quite. Compute both sides: for this choice they agree. A counterexample needs an output with a witness in

A

and a different witness in

B

, and none in

A \cap B

Show solution

(a) $f(\{2\}) = \{q\}$ and $\{p, q\} \cap \{q\} = \{q\}$ : equal.
(b) $f(\{3\}) = \{p\}$ and $\{p\} \cap \{p\} = \{p\}$ : equal.
(c) $f(\emptyset) = \emptyset$ and $\{q\} \cap \{p\} = \emptyset$ : equal.
(d) $f(\emptyset) = \emptyset$ but $\{p\} \cap \{p\} = \{p\}$ : not equal.

Only (d) has an output, $p$ , reached from $A$ by $1$ and from $B$ by $3$ with no witness in both sets.

Problem 3 · One Case of the Reverse Inclusion

Given: in proving $f(A) \cup f(B) \subseteq f(A \cup B)$ , we have reached the case $y \in f(B)$ . Which argument completes this case?

✅ Correct! One witness in

B

is automatically a witness in

A \cup B

, which is all the image of

A \cup B

asks for.

❌ Not quite.

y

is an output in

T

; it is the witness

x

that belongs to

B

, and the goal is

y \in f(A \cup B)

, not

y \in A \cup B

❌ Not quite.

y \in f(B)

promises only one witness in

B

; other inputs may also reach

y

. One witness is enough, so the claim about every witness is both unjustified and unnecessary.

❌ Not quite.

y \in f(B)

does not imply

y \in f(A)

: in the modulo

8

example with

A = \{2\}

and

B = \{3\}

3 \in f(B)

but

f(A) = \{6\}

. The case is handled by the same argument with

B

in place of

A

, not reduced to it.

Show solution

Assume $y \in f(B)$ . By the definition of the image there is an $x \in B$ with $f(x) = y$ . Every element of $B$ is in $A \cup B$ , so $x \in A \cup B$ , and therefore $y = f(x) \in f(A \cup B)$ .

Together with the case $y \in f(A)$ , this shows $f(A) \cup f(B) \subseteq f(A \cup B)$ .

Problem 4 · Two Witnesses on the Real Line

Given: $g: \mathbb{R} \to \mathbb{R}$ , $g(x) = x^2$ , $A = [-2, -1]$ and $B = [1, 3]$ . Which number lies in $g(A) \cap g(B)$ but not in $g(A \cap B)$ ?

✅ Correct!

2 = g(-\sqrt{2})

with

-\sqrt{2} \in A

, and

2 = g(\sqrt{2})

with

\sqrt{2} \in B

; but

A \cap B = \emptyset

, so

g(A \cap B) = \emptyset

❌ Not quite.

9 = g(3) \in g(B)

, but

g(A) = [1, 4]

, so

9 \notin g(A)

❌ Not quite.

0

is the square of

0

only, and

0

lies in neither

A

nor

B

❌ Not quite.

6 \in g(B) = [1, 9]

, but

g(A) = [1, 4]

, so

6 \notin g(A)

Show solution

On $A = [-2, -1]$ the squares run from $1$ to $4$ , so $g(A) = [1, 4]$ ; on $B = [1, 3]$ they run from $1$ to $9$ , so $g(B) = [1, 9]$ . Hence $g(A) \cap g(B) = [1, 4]$ .

But $A \cap B = \emptyset$ , so $g(A \cap B) = \emptyset$ . Every number in $[1, 4]$ answers the question; among the options only $2$ lies in $[1, 4]$ . Its witnesses $-\sqrt{2} \in A$ and $\sqrt{2} \in B$ are different elements.

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