Mathematical-Reasoning · Unit 25 · Video 3 · Interactive Practice

Images, Preimages and Set Operations: Six Conjectures

IKey Formulas

Conjecture	Modulo $8$ example	Real example
$f(A \cap B) \subseteq f(A) \cap f(B)$	$\{2\} \subsetneq \{2, 3, 6\}$	$[2, 3] \subsetneq [2, 6]$
$f(A \cup B) = f(A) \cup f(B)$	equal	equal
$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$	equal	equal
$f^{-1}(C \cup D) = f^{-1}(C) \cup f^{-1}(D)$	equal	equal
$A \subseteq f^{-1}(f(A))$	$\{1, 2, 4\} \subsetneq R_8$	$[0, 3] \subsetneq [-3, 3]$
$f(f^{-1}(C)) \subseteq C$	$\{2, 3\} \subsetneq \{1, 2, 3\}$	$[2, 6] = [2, 6]$

Key Insight: One example with a proper subset proves that equality can fail. No number of examples proves that a relation holds for every function; that takes a general proof.

IITesting the Six Conjectures Modulo 8

For $f(x) = (x^2 + 2) \bmod 8$ on $R_8$ , which comparisons stay equalities whatever the four sets are?

Compare Clicking edits

💡 Challenge: choose $A$ and $B$ so that $f(A \cap B) = f(A) \cap f(B)$ while $A \cap B \ne A$ and $A \cap B \ne B$ ; then try to make $f(A \cup B)$ differ from $f(A) \cup f(B)$ .

IIIRound Trips on the Real Line

For $f(x) = x^2 + 2$ , a round trip can pull in negative partners or return exactly what it started with.

Round trip a 0 b 3 c 2 d 6

IVQuiz Questions

Problem 1 · Intersect First or Apply f First?

Given: $R_6 = \{0, 1, 2, 3, 4, 5\}$ and $f: R_6 \to R_6$ , $f(x) = 2x \bmod 6$ , so the outputs of $0, 1, 2, 3, 4, 5$ are $0, 2, 4, 0, 2, 4$ . Let $A = \{1, 2\}$ and $B = \{4, 5\}$ . Find both sets.

What is f(A ∩ B)?

What is f(A) ∩ f(B)?

✅ Correct!

A \cap B = \emptyset

, so

f(A \cap B) = \emptyset

, while

f(A) = f(B) = \{2, 4\}

: the inclusion is proper, as extreme as it gets.

❌ Check $A \cap B$ first.

A

and

B

share no element, and the image of the empty set is empty.

❌ Apply $f$ to each set first.

f(1) = 2

and

f(2) = 4

;

f(4) = 2

and

f(5) = 4

Show solution

Intersect first: $A \cap B = \emptyset$ , and $f(\emptyset) = \emptyset$ , since there is no input whose output could be collected.

Apply $f$ first: $f(A) = \{f(1), f(2)\} = \{2, 4\}$ and $f(B) = \{f(4), f(5)\} = \{2, 4\}$ , so $f(A) \cap f(B) = \{2, 4\}$ .

Here $f(A \cap B) = \emptyset \subsetneq \{2, 4\} = f(A) \cap f(B)$ . The outputs $2$ and $4$ come from different inputs in $A$ and in $B$ .

Problem 2 · A Preimage of a Union

Given: the same $f(x) = 2x \bmod 6$ on $R_6$ , with $C = \{0, 1\}$ and $D = \{1, 2\}$ . Find $f^{-1}(C \cup D)$ .

✅ Correct!

C \cup D = \{0, 1, 2\}

; the inputs with output

0

are

0, 3

and those with output

2

are

1, 4

. It also equals

f^{-1}(C) \cup f^{-1}(D) = \{0, 3\} \cup \{1, 4\}

❌ Not quite. That is

C \cup D

itself, a set of outputs. The preimage is the set of inputs whose outputs lie in it.

❌ Not quite. That is

f^{-1}(C)

alone. The inputs

1

and

4

have output

2 \in D

❌ Not quite.

f(2) = 4

and

f(5) = 4

, and

4 \notin C \cup D

Show solution

$C \cup D = \{0, 1, 2\}$ . Test each input: $f(0) = 0$ , $f(1) = 2$ , $f(2) = 4$ , $f(3) = 0$ , $f(4) = 2$ , $f(5) = 4$ . The outputs in $\{0, 1, 2\}$ come from $0, 1, 3, 4$ ; no input has output $1$ .

f^{-1}(C \cup D) = \{0, 1, 3, 4\} = \{0, 3\} \cup \{1, 4\} = f^{-1}(C) \cup f^{-1}(D).

Problem 3 · A Round Trip That Loses Elements

Given: the same $f(x) = 2x \bmod 6$ on $R_6$ , and $C = \{1, 2, 3\}$ . Find $f(f^{-1}(C))$ .

✅ Correct!

f^{-1}(C) = \{1, 4\}

, and both go to

2

. The odd outputs

1

and

3

are never reached, so they drop out.

❌ Not quite. Image and preimage need not cancel: no input has output

1

3

, so those cannot come back.

❌ Close, but… that is

f^{-1}(C)

, the halfway point. Apply

f

once more.

❌ Not quite.

f^{-1}(C)

contains only

1

and

4

, and

f(1) = f(4) = 2

; nothing in it reaches

4

Show solution

Preimage first: the outputs are $0, 2, 4, 0, 2, 4$ , and only $2 \in C$ occurs, from the inputs $1$ and $4$ . So $f^{-1}(C) = \{1, 4\}$ .

Then the image: $f(\{1, 4\}) = \{2\}$ .

So $f(f^{-1}(C)) = \{2\} \subsetneq \{1, 2, 3\} = C$ , missing exactly the elements of $C$ that are not outputs of $f$ .

Problem 4 · What Two Examples Establish

Given: in both worked examples, $f(A \cap B)$ came out a proper subset of $f(A) \cap f(B)$ . Which conclusion do the examples justify?

✅ Correct! A single example with a proper subset is a counterexample to "equality always holds", so it proves this existence statement.

❌ Not quite. This inclusion is true, but two examples cannot establish a claim about every function; it needs a general proof (the next video gives one).

❌ Not quite. Equality can hold: take

A = B

, or the identity function on any set.

❌ Not quite. An example cannot prove a "for all" statement, but one counterexample does disprove it, which proves the corresponding "there exists" statement.

Show solution

Each example is one function with particular sets. In the modulo $8$ example, $f(A \cap B) = \{2\}$ and $f(A) \cap f(B) = \{2, 3, 6\}$ , so it witnesses (c).

(a) is a universal statement; it happens to be true, but finitely many examples cannot prove it. (b) is false: when $A = B$ , both sides equal $f(A)$ . (d) ignores that a counterexample settles a "for all" claim in the negative.

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