Mathematical-Reasoning · Unit 4 · Video 2 · Interactive Practice

Element, Subset, Truth Set: The Notation That Turns Sentences into Sets

IKey Formulas

Notation	Read as	Meaning
$y \in A$ , $\ z \notin A$	$y$ is an element of $A$ ; $z$ is not	$y$ is one of the objects in $A$ ; $z$ is not
$A = B$	$A$ equals $B$	precisely the same elements; order and repeats do not matter
$A \subseteq B$	$A$ is a subset of $B$	each element of $A$ is an element of $B$ ; always $A \subseteq A$
truth set of $P(x)$	solution set, for an equation	the objects in $U$ that make $P(x)$ a true statement

Key Insight: $5 \in \{5, 10\}$ but $\{5\} \subseteq \{5, 10\}$ : an element is an object in the set, while a subset is itself a set.

IIEqual Sets and Subsets

Among the video's six sets, which are subsets of which, and which two are equal?

X Y

IIIBuilding a Truth Set

The truth set collects exactly the values from $U$ that turn the open sentence into a true statement.

P(x) U Substituted —

💡 The video takes $U = \mathbb{R}$ for the two equations. Over $\mathbb{R}$ their truth sets are $\{-2, 5\}$ and $\{6\}$ , the same as the sweep over $\mathbb{Z}$ finds, because the algebra (factoring, or solving for $x$ ), not a check of integers, rules out every other real number. Over $\mathbb{N} = \{1, 2, 3, \ldots\}$ the root $-2$ drops out, and the truth set of $x^2 - 3x - 10 = 0$ is $\{5\}$ .

IVElement or Subset?

The number $5$ is an element of $\{5, 10\}$ ; the set $\{5\}$ is a subset of it, not an element.

Left Relation Right

VQuiz Questions

Problem 1 · Membership in a Listed Set

Given: $A = \{-6, -3, 0, 3, 6, 9, \ldots\}$ , the list that starts at $-6$ and climbs by $3$ with no end. Which statement is true?

✅ Correct!

15 = -6 + 7 \cdot 3

continues the pattern, so

15 \in A

❌ Close, but…

-9

is a multiple of

3

, but the elements of

A

begin at

-6

, so

-9 \notin A

. It is the same trap as

-12

in the video.

❌ Not quite.

-3

appears in the list itself, so

-3 \in A

and the statement

-3 \notin A

is false.

❌ Not quite. Every element of

A

is a multiple of

3

, and

4

is not, so

4 \notin A

Show solution

The elements of $A$ are $-6, -6 + 3, -6 + 2 \cdot 3, \ldots$ , that is, the numbers $-6 + 3k$ for $k = 0, 1, 2, \ldots$ . Check each statement:

$-9 = -6 + 3 \cdot (-1)$ would need $k = -1$ , which comes before the start of the list, so $-9 \notin A$ . Statement (a) is false.
$4$ is not a multiple of $3$ , so $4 \notin A$ . Statement (b) is false.
$15 = -6 + 3 \cdot 7$ , so $15 \in A$ . Statement (c) is true.
$-3$ is listed, so $-3 \in A$ . Statement (d) is false.

Problem 2 · Repeats and Order

Given: $S = \{4, 8, 12, 16\}$ and $T = \{16, 4, 8, 4, 12\}$ . Which statement is true?

✅ Correct! Both sets have exactly the elements

4

8

12

and

16

; the repeated

4

and the new order add nothing.

❌ Close, but… listing

4

twice does not add a new element.

T

has four elements, the same four as

S

❌ Not quite.

S \subseteq T

is right, but every element of

T

(

16

4

8

12

) is also in

S

, so

T \subseteq S

as well.

❌ Not quite. Order does not matter in a set:

\{1, 3, 5\} = \{3, 5, 1\}

❌ Not quite. Compare the elements of the two sets, ignoring order and repeats.

Show solution

Two sets are equal when they have precisely the same elements.

Each element of $S$ — $4$ , $8$ , $12$ , $16$ — appears in the list for $T$ , so $S \subseteq T$ .
Each entry of $T$ — $16$ , $4$ , $8$ , $4$ , $12$ — is an element of $S$ , so $T \subseteq S$ .

So $S$ and $T$ have the same elements and $S = T$ . The repeated $4$ and the different order change only how $T$ is written, not which elements it has. That makes (a), (c) and (d) false.

Problem 3 · The Universal Set Matters

Given: the open sentence $P(x)$ : $x^2 \le 9$ . Find its truth set for two different universal sets.

Truth set when $U = \mathbb{Z}$

Truth set when $U = \mathbb{N} = \{1, 2, 3, \ldots\}$

✅ Correct! The same open sentence has truth set

\{-3, -2, -1, 0, 1, 2, 3\}

over

\mathbb{Z}

but only

\{1, 2, 3\}

over

\mathbb{N}

❌ Not quite. Negative integers are in

U = \mathbb{Z}

too:

(-3)^2 = 9 \le 9

, so

P(-3)

is true.

❌ Not quite.

0

and the negative integers are in

U = \mathbb{Z}

too:

0^2 = 0 \le 9

and

(-3)^2 = 9 \le 9

, so

P(0)

and

P(-3)

are true.

❌ Close, but…

\le

allows equality:

3^2 = 9 \le 9

is true, so

3

and

-3

belong to the truth set.

❌ Close, but… here

\mathbb{N} = \{1, 2, 3, \ldots\}

, so

0

is not in

U

and cannot be in the truth set.

❌ Not quite. A truth set holds only objects of

U

, and

0

and the negative integers are not natural numbers.

❌ Check the integers. Test each integer from

-3

3

, and one just outside.

❌ Check the universal set. Only natural numbers can be substituted.

Show solution

$U = \mathbb{Z}$ . The squares of $-3, -2, -1, 0, 1, 2, 3$ are $9, 4, 1, 0, 1, 4, 9$ , all at most $9$ , so each makes $P(x)$ true. An integer with $x \ge 4$ or $x \le -4$ has $x^2 \ge 16 > 9$ , so $P(x)$ is false there. The truth set is

\{-3, -2, -1, 0, 1, 2, 3\}.

$U = \mathbb{N}$ . Only $1, 2, 3, \ldots$ may be substituted. $1^2 = 1$ , $2^2 = 4$ and $3^2 = 9$ are at most $9$ , while $n \ge 4$ gives $n^2 \ge 16$ . The truth set is

\{1, 2, 3\}.

The open sentence is the same; the universal set decides which objects can be in its truth set.

Problem 4 · Comparing Two Truth Sets

Given: $U = \mathbb{R}$ . Let $A$ be the truth set of $x^2 - 5x + 6 = 0$ and let $B$ be the truth set of $(x - 2)(x - 3)(x + 1) = 0$ .

What is $A$ ?

Which statement is true?

✅ Correct!

A = \{2, 3\}

and

B = \{-1, 2, 3\}

, so both elements of

A

are in

B

A \subseteq B

❌ Close, but check the signs.

x^2 - 5x + 6 = (x - 2)(x - 3)

, which is

0

x = 2

and

x = 3

; substituting

-2

gives

4 + 10 + 6 = 20

❌ Check the factoring. Substitute your two values into

x^2 - 5x + 6

; both must give

0

❌ Not quite.

-1 \in B

, since

(-1 - 2)(-1 - 3)(-1 + 1) = 0

, but

-1 \notin A

❌ Close, but… the elements of

A

are the numbers

2

and

3

; the set

\{2\}

is neither of them. What is true is

2 \in A

and

\{2\} \subseteq A

❌ Not quite. Write out

A

and

B

by roster, then compare element by element.

Show solution

Step 1 — Find $A$ . Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$ , which is $0$ exactly when $x = 2$ or $x = 3$ .

A = \{2, 3\}

Step 2 — Find $B$ . A product is $0$ exactly when one of its factors is $0$ , so $x = 2$ , $x = 3$ or $x = -1$ .

B = \{-1, 2, 3\}

Step 3 — Compare. Both elements of $A$ , $2$ and $3$ , are elements of $B$ , so $A \subseteq B$ . The reverse fails because $-1 \in B$ and $-1 \notin A$ , so $B \nsubseteq A$ and $A \neq B$ . Finally, $\{2\} \in A$ is false: $A$ 's elements are numbers, and $\{2\}$ is a set.

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