Step 1: Formal Definition of the Set

The given set is defined in set-builder notation as:
$A = \{x : x \in \mathbb{N}, x < 5 \text{ and } x > 7\}$
Where $\mathbb{N}$ denotes the set of all natural numbers $\{1, 2, 3, \dots\}$.

Step 2: Logical Decomposition of Conditions

The set $A$ is defined by the conjunction (logical "and") of two distinct inequalities acting upon the same variable $x$:

• Condition 1 ($P$): $x < 5$
• Condition 2 ($Q$): $x > 7$

For an element $x$ to be a member of set $A$, it must satisfy $P \land Q$ simultaneously [Per the Principle of Conjunction in Set Theory].

Step 3: Interval and Domain Analysis

We analyze the feasibility of the intersection of the two conditions:

There exists no real number $x$, and consequently no natural number $x$, that is strictly less than 5 and strictly greater than 7 at the same time [By the Law of Contradiction in Classical Logic].

Step 4: Visualization of the Number Line

Conclusion

Since the intersection of the sets representing the conditions is null, the set $A$ contains no elements. By definition, a set that contains no elements is termed a null set or empty set, denoted by $\emptyset$ or $\{\}$.

Final Solution: The set $\{x : x \in \mathbb{N}, x < 5 \text{ and } x > 7\}$ is an example of a null set.

Step 1: Formal Definition of the Set

To determine if the set is finite or infinite, we must first express the set in Roster Form. Let the set be denoted by $S$.

$S = \{ \text{January, February, March, April, May, June, July, August, September, October, November, December} \}$

Step 2: Defining Finiteness and Infiniteness

According to Set Theory [Per Cantor’s Set Theory Foundation]:

• Finite Set: A set is finite if it is empty or consists of a definite, countable number of elements such that the process of counting terminates.
• Infinite Set: A set is infinite if it is not finite, meaning the number of elements is unlimited and cannot be represented by a natural number $n \in \mathbb{N}$.

Step 3: Cardinality Analysis

The cardinality of a set, denoted by $n(S)$ or $|S|$, represents the total number of distinct elements within the set.

By enumerating the elements of $S$:
1. January
2. February
3. March
4. April
5. May
6. June
7. July
8. August
9. September
10. October
11. November
12. December

Since the counting process terminates at the integer $12$, the cardinality is $n(S) = 12$.

Visual Representation of Set Cardinality

Step 4: Logical Conclusion

Since the number of elements in the set is fixed and corresponds to a specific natural number ($12$), the set satisfies the condition for a finite set [By the definition of finiteness in Zermelo-Fraenkel Set Theory].

Final Solution: The set of months of a year is a Finite Set.

Step 1: Formal Definition of Set Equality

Two sets $A$ and $B$ are defined as equal ($A = B$) if and only if they contain exactly the same elements. That is, every element of $A$ is an element of $B$, and every element of $B$ is an element of $A$ [Per the Axiom of Extensionality].

Step 2: Representation of Set A

Set $A$ is provided in roster form:

$A = \{2, 3\}$

Step 3: Solving for Elements of Set B

Set $B$ is defined by the quadratic equation $x^2 + 5x + 6 = 0$. To determine the elements of $B$, we solve for the roots of this equation [By the Fundamental Theorem of Algebra]:

The quadratic equation is: $x^2 + 5x + 6 = 0$

Factorizing the middle term (splitting the middle term):

$x^2 + 3x + 2x + 6 = 0$

$x(x + 3) + 2(x + 3) = 0$

$(x + 2)(x + 3) = 0$

Setting each factor to zero:

$x + 2 = 0 \implies x = -2$

$x + 3 = 0 \implies x = -3$

Therefore, the solution set $B$ in roster form is: $B = \{-2, -3\}$.

Step 4: Comparative Analysis

Step 5: Logical Justification

Comparing the elements of both sets:

• Element $2 \in A$ but $2 \notin B$.
• Element $3 \in A$ but $3 \notin B$.
• Element $-2 \in B$ but $-2 \notin A$.
• Element $-3 \in B$ but $-3 \notin A$.

Since the sets do not contain the same elements, the condition for set equality is not satisfied [Per the definition of Set Equality].

Final Solution: The sets $A$ and $B$ are not equal because their elements are distinct ($A = \{2, 3\}$ and $B = \{-2, -3\}$).

Step 1: Formal Definition of the Set

The set is defined by the set-builder notation $S = \{ y : y \text{ is a point common to any two parallel lines} \}$. To determine if this is a null set (an empty set, denoted as $\emptyset$ or $\{\}$), we must evaluate the existence of any element $y$ that satisfies the given condition.

Step 2: Geometric Analysis of Parallel Lines

By Euclidean geometry axioms, two lines are defined as parallel if and only if they lie in the same plane and do not intersect at any point, regardless of how far they are extended in either direction [Per Euclid's Fifth Postulate and the definition of Parallelism].

• Let $L_1$ and $L_2$ be two distinct lines in a plane.
• If $L_1 \parallel L_2$, then for every point $P$ such that $P \in L_1$, it is guaranteed that $P \notin L_2$.
• Conversely, for every point $Q$ such that $Q \in L_2$, it is guaranteed that $Q \notin L_1$.

Step 3: Visual Representation

Step 4: Logical Evaluation

Since the distance $d$ between two parallel lines is constant and $d > 0$, the set of intersection points must be empty. Mathematically:

$L_1 \cap L_2 = \emptyset$

Because there exists no point $y$ such that $y \in L_1$ and $y \in L_2$, the set $S$ contains no elements.

Conclusion

By the definition of a null set (a set containing zero elements), since the condition "point common to any two parallel lines" is impossible to satisfy, the set is indeed a null set.

Final Solution: The set $\{ y : y \text{ is a point common to any two parallel lines} \}$ is an example of a null set ($\emptyset$).

Step 1: Formal Definition of the Set

Let $A$ be the set of odd natural numbers divisible by $2$. By definition, the set of natural numbers $\mathbb{N} = \{1, 2, 3, 4, ...\}$. An odd natural number $n$ is defined by the form $n = 2k - 1$, where $k \in \mathbb{N}$.

For a number to be divisible by $2$, it must satisfy the condition of being an even number. Mathematically, an integer $x$ is divisible by $2$ if $x = 2m$ for some integer $m$ [Per the definition of divisibility].

Step 2: Logical Contradiction Analysis

We test the existence of an element $x \in A$ such that $x$ satisfies two mutually exclusive conditions:

• Condition 1: $x$ is an odd natural number ($x \in \{1, 3, 5, ...\}$).
• Condition 2: $x$ is divisible by $2$ ($x \in \{2, 4, 6, ...\}$).

By the Parity Lemma, for any integer $x$, it is either even or odd, but never both. If $x$ is odd, $x \equiv 1 \pmod{2}$. If $x$ is divisible by $2$, $x \equiv 0 \pmod{2}$. Since $1 \not\equiv 0 \pmod{2}$, there is no integer $x$ that satisfies both conditions simultaneously.

Step 3: Visual Representation (Venn Diagram of Disjoint Sets)

Step 4: Classification

Since the set $A = \{x : x \in \mathbb{N}, x \text{ is odd}, \text{ and } 2|x\}$ contains no elements, it satisfies the definition of an empty set (or null set) $\emptyset$ [Definition: A set with no elements is called a null set].

Final Solution: The set of odd natural numbers divisible by 2 is an example of a null set, denoted as $\emptyset$ or $\{\}$.

Definition of Null Set (Empty Set)

A set is defined as a null set (or empty set), denoted by $\phi$ or $\{\}$, if it contains no elements [Per the Axiom of Extensionality].

Step 1: Identifying the Mathematical Constraints

The given condition defines the set $A$ as: $A = \{x : x \text{ is an even prime number}\}$

To evaluate if this is a null set, we must identify the set of prime numbers and apply the parity filter.

Step 2: Analysis of Prime Numbers

A prime number $p$ is a natural number greater than $1$ that has exactly two distinct positive divisors: $1$ and $p$.

• The set of the first few prime numbers is $\{2, 3, 5, 7, 11, 13, \dots\}$.
• Parity check: A number is even if it is divisible by $2$.
• In the set of prime numbers, $2$ is the only even prime number because all other even numbers are divisible by $2$, and thus possess more than two divisors [By the Fundamental Theorem of Arithmetic].

Step 3: Verification of the Set Composition

Since the number $2$ satisfies both conditions (being prime and being even), the set $A$ can be explicitly listed using the Roster Method:

$A = \{2\}$

Step 4: Cardinality Comparison

The cardinality of set $A$, denoted as $n(A)$, is the number of elements in the set.

Because $n(A) = 1 \neq 0$, the set is not empty.

Final Solution: The set of even prime numbers is not a null set, as it contains exactly one element: $\{2\}$. Therefore, it is a singleton set.

Step 1: Formal Definition of the Given Set

The set is defined as $A = \{1, 2, 3, \dots\}$. This notation indicates the set of all natural numbers, often denoted by the symbol $\mathbb{N}$. [By the definition of Roster Form and Ellipsis usage in Set Theory].

Step 2: Identifying the Cardinality

To determine if a set is finite or infinite, we evaluate its cardinality, $n(A)$, which represents the number of elements in the set.

• A set is finite if the process of counting its elements terminates at a specific natural number $n \in \mathbb{N}$.
• A set is infinite if the process of counting its elements does not terminate.

In the set $A = \{1, 2, 3, \dots\}$, the inclusion of the ellipsis ($\dots$) following the integer $3$ indicates that for every natural number $n$, there exists a successor $n+1$ that is also contained within the set. [Per the Axiom of Infinity in Zermelo-Fraenkel set theory].

Step 3: Visual Representation

Step 4: Logical Conclusion

Since the set $A$ contains an unlimited sequence of natural numbers and cannot be put into a one-to-one correspondence with the set $\{1, 2, \dots, n\}$ for any finite $n$, the set is classified as infinite. [By the Principle of Mathematical Induction, if a property holds for $k$, it holds for $k+1$, implying no upper bound exists].

Final Solution: The set $\{1, 2, 3, \dots\}$ is an Infinite Set.

Step 1: Formal Definition of Equality of Sets

Two sets $A$ and $B$ are defined as equal ($A = B$) if and only if they contain exactly the same elements. Formally, this is expressed as:

• $\forall x, x \in A \implies x \in B$
• $\forall x, x \in B \implies x \in A$

Step 2: Analysis of Set A

The set $A$ is provided as:

$A = \{ 4, 8, 12, 16 \}$

Elements of $A = \{ 4, 8, 12, 16 \}$

Step 3: Analysis of Set B

The set $B$ is provided as:

$B = \{ 8, 4, 16, 18 \}$

Elements of $B = \{ 4, 8, 16, 18 \}$ [Reordered for comparative clarity; order of elements does not affect set identity].

Step 4: Comparative Element Verification

Step 5: Conclusion via Logical Contradiction

To satisfy $A = B$, every element in $A$ must be in $B$. However, $12 \in A$ but $12 \notin B$. Furthermore, $18 \in B$ but $18 \notin A$. Since the condition of equality is not met, the sets are distinct.

Final Solution: $A \neq B$ because the set of elements in $A$ and $B$ are not identical.

Step 1: Theoretical Definition of Equality of Sets

Two sets $A$ and $B$ are defined as equal if and only if they contain the exact same elements. Order of the elements within the set is immaterial, and repetition of elements does not change the identity of the set [Per the Extensionality Axiom of Set Theory]. Formally: $A = B \iff (\forall x \in A, x \in B) \land (\forall x \in B, x \in A)$.

Step 2: Analysis of Set A

The set $A$ is given as: $A = \{a, b, c, d\}$ The cardinality of set $A$, denoted as $|A|$, is the number of distinct elements, which is $4$.

Step 3: Analysis of Set B

The set $B$ is given as: $B = \{d, c, b, a\}$ The cardinality of set $B$, denoted as $|B|$, is also $4$.

Step 4: Comparative Element Mapping

We verify the presence of every element of $A$ in $B$ and vice versa:

Step 5: Visual Representation of Set Equality

Conclusion of Analysis

Since every element of $A$ is an element of $B$ and every element of $B$ is an element of $A$, the condition for set equality is satisfied [By the definition of Equality of Sets].

Final Solution: $A = B$

Step 1: Roster Form Representation of Set A

Given set $A = \{ x : x \text{ is a letter in the word FOLLOW} \}$. To represent set $A$ in roster form, we list the distinct elements of the word "FOLLOW". By the definition of a set, each element must be unique [Per the Principle of Extensionality, a set is determined solely by its members].

• Letters in "FOLLOW": F, O, L, L, O, W.
• Distinct letters: F, O, L, W.

Thus, $A = \{F, O, L, W\}$.

Step 2: Roster Form Representation of Set B

Given set $B = \{ y : y \text{ is a letter in the word WOLF} \}$. To represent set $B$ in roster form, we list the distinct elements of the word "WOLF".

• Letters in "WOLF": W, O, L, F.
• Distinct letters: W, O, L, F.

Thus, $B = \{W, O, L, F\}$.

Step 3: Analytical Comparison

Two sets are considered equal if and only if they contain the exact same elements, regardless of the order in which they are written [Per the Axiom of Equality for Sets: $A = B \iff (\forall x, x \in A \implies x \in B) \land (\forall y, y \in B \implies y \in A)$].

Comparing the sets:

• Every element in $A$ is present in $B$.
• Every element in $B$ is present in $A$.
• The cardinality of both sets is identical: $n(A) = 4$ and $n(B) = 4$.

Step 4: Visual Representation

Final Solution: Yes, the sets A and B are equal because they contain the exact same elements $\{F, O, L, W\} = \{W, O, L, F\}$.

Step 1: Analyzing Set A

The set $A$ is defined by the set-builder form $A = \{ x : x \text{ is a multiple of 10} \}$. By applying the definition of multiples of an integer (where $x = 10n$ for $n \in \mathbb{Z}^+$), we can list the elements of set $A$ in roster form:

$A = \{10, 20, 30, 40, 50, \dots\}$

[Justification: Per the definition of multiples, any element $x \in A$ must satisfy $x \equiv 0 \pmod{10}$.]

Step 2: Analyzing Set B

The set $B$ is given in roster form as:

$B = \{10, 15, 20, 25, 30, \dots\}$

[Justification: The elements of $B$ are defined by the arithmetic progression $a_n = 10 + 5(n-1)$ for $n \in \{1, 2, 3, \dots\}$.]

Step 3: Comparative Analysis

To determine if $A = B$, we must verify the condition for the equality of sets: $A = B$ if and only if every element of $A$ is an element of $B$ and every element of $B$ is an element of $A$.

Step 4: Logical Conclusion

Because there exists an element $15$ such that $15 \in B$ but $15 \notin A$, the criterion for set equality is violated.

[Mathematical Axiom: Two sets are equal if and only if they contain exactly the same elements. By the Law of Extensionality, if $\exists x (x \in A \land x \notin B)$ or $\exists x (x \in B \land x \notin A)$, then $A \neq B$.]

Final Solution: $A \neq B$, because $15 \in B$ but $15 \notin A$.

Step 1: Analyzing Set A

Set $A$ is given in Roster Form as:
$A = \{2, 4, 6, 8, 10\}$

The elements of $A$ are clearly identified as the first five positive even integers.

Step 2: Analyzing Set B

Set $B$ is given in Set-Builder Form as:
$B = \{x : x \text{ is a positive even integer and } x \le 10\}$

To determine the contents of set $B$, we identify all positive even integers $x$ such that $x \le 10$.
The positive even integers are $2, 4, 6, 8, 10, 12, 14, \dots$
Applying the constraint $x \le 10$, we select the elements that satisfy this condition:
$B = \{2, 4, 6, 8, 10\}$

Step 3: Verification of Set Equality

Definition of equal sets: Two sets $A$ and $B$ are said to be equal if and only if every element of $A$ is an element of $B$, and every element of $B$ is an element of $A$ [Per the Axiom of Extensionality].

Since all elements of $A$ are contained within $B$, and all elements of $B$ are contained within $A$, the sets are identical.

Final Solution: Yes, $A = B$ because both sets contain exactly the same elements $\{2, 4, 6, 8, 10\}$.

Step 1: Definition of Equal Sets

Two sets $X$ and $Y$ are defined as equal if and only if they contain the exact same elements. Order of elements does not affect the equality of sets [Per the Axiom of Extensionality]. Mathematically, $X = Y \iff (\forall x \in X, x \in Y) \land (\forall y \in Y, y \in X)$.

Step 2: Comparative Analysis of Given Sets

We perform an element-by-element verification for all provided sets:

Step 3: Verification of Set Equality

• Comparison of $B$ and $D$:
  Set $B = \{1, 2, 3, 4\}$ and Set $D = \{3, 1, 4, 2\}$.
  Rearranging $D$ in ascending order: $\{1, 2, 3, 4\}$.
  Since every element of $B$ is in $D$ and every element of $D$ is in $B$, $B = D$.
• Comparison of $E$ and $G$:
  Set $E = \{-1, 1\}$ and Set $G = \{1, -1\}$.
  Since the sets consist of the same elements $\{-1, 1\}$, $E = G$.
• Analysis of remaining sets:
  Set $A = \{2, 4, 8, 12\}$: No other set contains the element $8$ and $12$ simultaneously while excluding $14$. Thus, $A$ has no equal counterpart.
  Set $C = \{4, 8, 12, 14\}$: Contains $14$, which is not in any other set. Thus, $C$ has no equal counterpart.
  Set $F = \{0, a\}$ and $H = \{0, 1\}$: Unless $a = 1$, $F \neq H$. Given the information, they are not strictly equal.

Visual Representation of Set Relationships

Final Solution: The equal sets are $B = D$ and $E = G$.

Step 1: Formal Definition of the Set

To determine the nature of the set, we first identify the set $A$ representing the English alphabet. By enumeration, the set is defined as:

$A = \{x \mid x \text{ is a letter in the English alphabet}\}$

$A = \{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z\}$

Step 2: Cardinality Analysis

The cardinality of a set, denoted by $n(A)$ or $|A|$, represents the number of elements contained within the set [Per Set Theory Axioms].

• The set begins at 'a' and terminates at 'z'.
• Counting the elements chronologically: $1, 2, 3, \dots, 26$.
• Thus, $n(A) = 26$.

Step 3: Logical Classification

In accordance with the definitions of set theory:

• Finite Set: A set is finite if the process of counting its elements comes to an end, resulting in a non-negative integer as its cardinality. [Definition of Finite Sets]
• Infinite Set: A set is infinite if the number of elements is unlimited and cannot be represented by a natural number.

Since the cardinality of set $A$ is $26$, which is a definite, fixed natural number, the set is classified as finite.

Step 4: Visual Representation of Cardinality

Final Solution: The set of letters in the English alphabet is a finite set, as it contains a countable and definite number of elements ($n=26$).

Step 1: Formal Definition of Finite and Infinite Sets

In set theory, a set $S$ is defined as finite if the cardinality of the set, denoted by $|S|$ or $n(S)$, is a non-negative integer. That is, the elements can be counted and the process of counting terminates [Per Zermelo-Fraenkel Set Theory Axioms]. Conversely, a set is infinite if it is not finite; it cannot be put into a one-to-one correspondence with the set $\{1, 2, 3, \dots, n\}$ for any natural number $n$.

Step 2: Analysis of the Given Set

The set under consideration is $A = \{x \mid x \text{ is an animal living on the earth}\}$. To determine the nature of $A$, we evaluate the constraints imposed by biological and physical reality:

• Biological Constraint: The number of living organisms at any specific discrete time $t$ is determined by the total biomass and population dynamics of all extant species.
• Temporal Constraint: While the population of animals fluctuates due to birth, death, and extinction rates, at any given moment in time, the number of individual animals inhabiting Earth is a finite, albeit large, integer.
• Mathematical Limit: Even if the exact count is difficult to calculate due to ecological complexity, the set does not satisfy the requirements of an infinite set (it does not contain an infinite sequence of distinct elements). It is bounded by the physical space and resources of the planet.

Step 3: Logical Deduction

Since the set of animals living on Earth at any time $t$ has a cardinal number $N$, where $N \in \mathbb{N}_0$, and there exists a definite upper bound to the number of biological entities supported by the biosphere, we conclude that the elements can be enumerated, theoretically, in finite time [By the Finite Set definition].

Step 4: Final Classification

Given that the population of animals on Earth is a bounded value, the set is finite.

Final Solution: The set of animals living on the earth is a finite set.

Step 1: Formal Definition of Finite and Infinite Sets

To determine the classification of the set $A = \{1, 2, 3, \dots, 99, 100\}$, we must apply the foundational definitions of Set Theory:

• Finite Set: A set is finite if it is either empty or consists of a definite, countable number of elements such that the process of counting the elements comes to an end [Per Cantor’s Set Theory].
• Infinite Set: A set is infinite if it is not finite; that is, the process of counting its elements never terminates.

Step 2: Analysis of the Cardinality of Set A

The set is defined as $A = \{x \in \mathbb{N} : 1 \leq x \leq 100\}$.

We represent the set in roster form and identify its boundaries:

The cardinality of a finite arithmetic sequence is calculated using the formula: $n = \frac{a_n - a_1}{d} + 1$.

Substituting the given values:
$n = \frac{100 - 1}{1} + 1$
$n = 99 + 1$
$n = 100$

Step 3: Logical Deduction

Since the cardinality $|A| = 100$, and $100$ is a fixed, non-negative integer, the set $A$ contains a definite number of elements. The counting process for this set is guaranteed to terminate at the 100th element [By the Principle of Finiteness].

Visual Representation

Final Solution: The set {1, 2, 3, . . .99, 100} is a Finite Set because it possesses a finite cardinality of 100.

Step 1: Formal Definition of the Set

Let $A$ be the set defined by the condition that its elements are positive integers ($x \in \mathbb{Z}^+$) such that $x > 100$. Mathematically, this set is expressed in set-builder notation as:

$A = \{x : x \in \mathbb{Z}^+, x > 100\}$

[By the definition of positive integers, $\mathbb{Z}^+ = \{1, 2, 3, \dots\}$]

Step 2: Enumeration of Elements

By applying the roster method to the condition $x > 100$, we identify the members of the set sequentially:

$A = \{101, 102, 103, 104, 105, \dots\}$

[By the Well-Ordering Principle, we identify the least element as $101$, but there is no maximum element in the set of positive integers.]

Step 3: Analytical Determination of Cardinality

A set is defined as finite if its elements can be counted and terminated, or if it is an empty set. A set is infinite if the process of counting its elements never reaches an end [Per the definition of Infinite Sets in Zermelo-Fraenkel set theory].

Visual Representation of the Infinite Progression

Step 4: Conclusion

Since the set $A$ continues indefinitely without a supremum, it is impossible to list or count all elements. Therefore, it satisfies the criteria for an infinite set.

Final Solution: The set of positive integers greater than 100 is an infinite set.

Step 1: Formal Definition of the Set

Let the set be denoted by $S$. By the definition provided in the problem statement, the set contains all positive integers that are multiples of $5$. We can express this set in set-builder notation as follows:

$S = \{x : x = 5n, n \in \mathbb{N}\}$

[Per the definition of natural numbers $\mathbb{N} = \{1, 2, 3, 4, ...\}$, where $\mathbb{N}$ is an infinite set.]

Step 2: Roster Form Representation

To analyze the cardinality of the set, we list the elements in roster form by substituting values of $n$ from the set of natural numbers:

$S = \{5(1), 5(2), 5(3), 5(4), ..., 5(n), ...\}$

$S = \{5, 10, 15, 20, ..., 5n, ...\}$

Step 3: Logical Analysis of Cardinality

A set is defined as finite if it consists of a definite number of elements that can be counted and eventually terminated. A set is defined as infinite if it is not finite, meaning the process of listing elements does not terminate.

• Since the set of natural numbers $\mathbb{N}$ has no upper bound, there is no largest natural number $n$.
• For every element $5n \in S$, there exists a successor $5(n+1) \in S$ such that $5(n+1) > 5n$.
• [By the Axiom of Infinity in Zermelo-Fraenkel set theory], because there is no last element, the cardinality of set $S$ is denoted as $\aleph_0$ (aleph-null).

Step 4: Visualizing the Infinite Progression

Step 5: Conclusion

Because the set of multiples of $5$ is equivalent to the set of natural numbers scaled by a factor of $5$, and the set of natural numbers is infinite, the set of multiples of $5$ must also be infinite.

Final Solution: The set of numbers which are multiples of 5 is an infinite set.

Step 1: Formal Definition of the Set

Let the set be denoted by $P$. According to the problem statement, the set $P$ consists of all prime numbers $p$ such that $1 < p < 99$.

Mathematically, we define the set as: $P = \{x \in \mathbb{N} \mid x \text{ is prime and } x < 99\}$

Step 2: Listing the Elements of the Set

To determine if the set is finite or infinite, we identify the elements that satisfy the criteria. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself [Definition of Prime Numbers].

Listing the prime numbers less than 99 sequentially:

$P = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}$

Step 3: Cardinality Analysis

The cardinality of a set, denoted by $n(P)$ or $|P|$, represents the number of elements contained within the set [Set Theory Axiom of Cardinality].

• Total count of elements in $P$: 25
• Since 25 is a finite natural number, the process of counting the elements of the set terminates [Definition of Finite Set].

Step 4: Theoretical Justification

A set is defined as finite if it is empty or consists of a definite number of elements such that the process of counting ends. Conversely, a set is infinite if it is not possible to list all its elements in a sequence that terminates [Axiom of Finitude]. Since the set $P$ contains a fixed, countable number of elements (25), it adheres to the criteria of a finite set.

Final Solution: The set of prime numbers less than 99 is a Finite Set because it contains exactly 25 elements, which is a fixed and countable quantity.

Step 1: Theoretical Foundation of Sets and Cartesian Geometry

To classify the given set, we first define the geometric parameters of the Cartesian plane $\mathbb{R}^2$. The set $S$ consists of all lines $L$ that are parallel to the x-axis. In a Cartesian coordinate system, any line parallel to the x-axis can be represented by the equation:

$L_c = \{ (x, y) \in \mathbb{R}^2 : y = c \}$

where $c$ is a constant representing the y-intercept of the line. [Per the Cartesian coordinate system axioms].

Step 2: Mapping the Set to the Real Number Line

Since the set $S$ includes all lines parallel to the x-axis, for every real number $c \in \mathbb{R}$, there exists a unique line $L_c$. Therefore, we can establish a bijection $f: S \to \mathbb{R}$ defined by $f(L_c) = c$. [By the definition of a bijection in Set Theory].

Step 3: Evaluating Cardinality

Because the set of all real numbers $\mathbb{R}$ is an uncountable infinite set, and we have established a one-to-one correspondence between the set of lines $S$ and the set $\mathbb{R}$, the set $S$ must also be infinite. [Per Cantor's Theorem on the cardinality of infinite sets].

Visual Representation of the Set

Step 4: Comparison Table of Set Properties

Final Solution: The set of lines which are parallel to the x-axis is an infinite set.