Step 1: Analyzing the Set Composition

The given set is defined as $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$. To evaluate the subset relationship correctly, we must identify the distinct elements belonging to set $A$. These are:

• $1 \in A$
• $2 \in A$
• $\{3, 4\} \in A$ (Note: The set $\{3, 4\}$ is treated as a single atomic element of $A$)
• $5 \in A$

Step 2: Defining the Subset Relation ($\subset$)

Per the formal definition of a subset, if $X \subset A$, then every element $x \in X$ must also be an element of $A$. That is, $\forall x (x \in X \implies x \in A)$ [By the Axiom of Extensionality and Definition of Subsets].

Step 3: Verification of the Statement $\{1, 2, 3\} \subset A$

Let $X = \{1, 2, 3\}$. To determine if $X \subset A$, we check each element of $X$ against the members of $A$ identified in Step 1:

Although $1 \in A$ and $2 \in A$, the value $3$ is not an element of $A$. While $3$ is a constituent part of the set $\{3, 4\}$, it does not exist as an independent element within set $A$ [By the Principle of Set Membership].

Visual Representation of Elements

Step 4: Conclusion of Logical Validity

Because there exists at least one element in the set $\{1, 2, 3\}$ (specifically the element $3$) that is not contained within set $A$, the condition for the subset relation fails [Per the logical definition of subset non-inclusion].

Final Solution: The statement $\{1, 2, 3\} \subset A$ is incorrect because $3$ is not an element of $A$. $3$ only exists as part of the element $\{3, 4\}$.

Step 1: Analyzing the Set Notation

The given set is defined as $A = \{x : x \in \mathbb{R}, 3 \le x \le 4\}$. The notation $\{x : x \in \mathbb{R}, a \le x \le b\}$ represents a subset of real numbers that includes all values between $a$ and $b$, inclusive of both endpoints.

Step 2: Defining Interval Notation Standards

In real analysis, intervals are categorized based on whether their endpoints are included or excluded:

• Closed Interval $[a, b]$: Represents the set $\{x \in \mathbb{R} : a \le x \le b\}$. This includes both endpoints $a$ and $b$. [By Definition of Closed Intervals in Real Analysis]
• Open Interval $(a, b)$: Represents the set $\{x \in \mathbb{R} : a < x < b\}$. This excludes endpoints $a$ and $b$.

Step 3: Logical Mapping

Comparing the given expression $3 \le x \le 4$ with the standard definition of a closed interval:
The inequality $3 \le x$ implies that $3$ is included in the set.
The inequality $x \le 4$ implies that $4$ is included in the set.
Since both inequality signs include equality ($\le$), the interval is closed at both ends. [By the Law of Set-Interval Correspondence]

Step 4: Visual Representation

Step 5: Final Formulation

Mapping the logical constraints to the interval notation, we assign the lower bound $a = 3$ and the upper bound $b = 4$. Because the conditions are inclusive, we use square brackets to indicate the closure of the set.

Final Solution: [3, 4]

Step 1: Formal Definition of the Sets

To determine the correct subset relationship, we must define the given sets $A$ and $B$ in roster or descriptive notation based on the provided conditions:

• Let Set $A = \{x : x \text{ is a student of Class XI of your school}\}$.
• Let Set $B = \{x : x \text{ is a student of your school}\}$.

Step 2: Logical Analysis of Set Inclusion

According to the definition of a subset, $A \subset B$ if and only if every element $x$ that is a member of $A$ is also a member of $B$ [By the definition of subset relation].

• Consider an arbitrary element $a \in A$. By the definition of $A$, $a$ is a student in Class XI of your school.
• Since every student in Class XI of your school is, by necessity, a student of the school itself, it follows that $a \in B$.

Step 3: Evaluation of the Relationship

Since every element of $A$ is contained within $B$, the condition for a subset holds true. Specifically:

Step 4: Conclusion

Since all members of the set of Class XI students are inherently members of the set of all students in the school, the symbol $\subset$ is the mathematically correct operator.

Final Solution: {x : x is a student of Class XI of your school} $\subset$ {x : x student of your school}

Given Set Definition

The set $A$ is defined as $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$.

To determine the correctness of the statement $\{1, 2, 5\} \in A$, we must analyze the membership criteria of set $A$ based on the definition of the element-of relation ($\in$).

Step 1: Identifying Elements of Set $A$

The elements of set $A$ are those objects explicitly separated by commas within the set brackets. By inspection:

• $1$ is an element of $A$ ($1 \in A$)
• $2$ is an element of $A$ ($2 \in A$)
• $\{3, 4\}$ is an element of $A$ ($\{3, 4\} \in A$)
• $5$ is an element of $A$ ($5 \in A$)

[Per the definition of Set Membership, an object $x$ is an element of $A$ if and only if $x$ is listed as a primary member of the set.]

Step 2: Evaluating the Statement $\{1, 2, 5\} \in A$

The statement posits that the set $\{1, 2, 5\}$ is an element of $A$. For this to be true, the object $\{1, 2, 5\}$ must appear as an explicit entry within the curly braces of $A$. Comparing the set $\{1, 2, 5\}$ against the list of elements identified in Step 1, we observe that it is not present.

Step 3: Distinguishing Between Membership ($\in$) and Subset ($\subset$)

It is crucial to note the difference between being an element of a set and being a subset of a set:

[By the Principle of Set Inclusion, $B \subset A$ if $\forall x (x \in B \implies x \in A)$. Since $1, 2, 5 \in A$, the subset relation holds, not the element relation.]

Final Solution: The statement $\{1, 2, 5\} \in A$ is incorrect. Although $1, 2,$ and $5$ are individual elements of $A$, the set $\{1, 2, 5\}$ itself is not listed as an element within $A$. It is, however, a proper subset of $A$ ($\{1, 2, 5\} \subset A$).

Step 1: Theoretical Definition of Subset

According to set theory [Per the definition of the subset relation], a set $A$ is a subset of set $B$ (denoted as $A \subseteq B$) if and only if every element of $A$ is also an element of $B$. Conversely, $A \not\subset B$ signifies that there exists at least one element $x$ such that $x \in A$ but $x \notin B$.

Step 2: Analysis of the Given Sets

Let the two sets be defined as follows:

• Set $A = \{a, b\}$
• Set $B = \{b, c, a\}$

We examine the containment of the elements of $A$ within $B$:

1. Check element $a$: Is $a \in \{b, c, a\}$? Yes, $a$ is present in $B$.
2. Check element $b$: Is $b \in \{b, c, a\}$? Yes, $b$ is present in $B$.

Step 3: Logical Deduction

Since every element of set $A$ (specifically $a$ and $b$) is contained within set $B$, it follows by the definition of subset that $A \subseteq B$.

The original statement posits that $\{a, b\} \not\subset \{b, c, a\}$. However, since $A \subseteq B$ holds true, the assertion that $A \not\subset B$ is mathematically incorrect [By the Law of Contradiction].

Step 4: Conclusion

Because $\{a, b\} \subset \{a, b, c\}$, the statement $\{a, b\} \not\subset \{b, c, a\}$ contradicts the established membership criteria.

Final Solution: The statement is False.

Initial Definitions & Theoretical Foundation

An interval is a subset of the real number system $\mathbb{R}$. In set theory, an open interval denoted by $(a, b)$ consists of all real numbers $x$ such that $a < x < b$. The endpoints $a$ and $b$ are excluded from the set [Per the Definition of Open Intervals in Real Analysis].

Step 1: Analyzing the Given Interval

The given interval is $(-3, 0)$.

• The left endpoint $a = -3$.
• The right endpoint $b = 0$.
• Since the interval uses parentheses '()', it signifies an open interval where the boundary values are not included in the set.

Step 2: Constructing the Set-Builder Notation

To write this in set-builder form, we define a variable $x$ and establish the constraints provided by the interval bounds. The standard syntax for set-builder notation is $\{x : x \in \text{Domain, Constraint}\}$.

For the interval $(-3, 0)$, the condition is that $x$ must be greater than $-3$ and strictly less than $0$, where $x$ belongs to the set of all real numbers $\mathbb{R}$.

Mathematical expression: $\{x : x \in \mathbb{R}, -3 < x < 0\}$

Visual Representation of the Interval

Justification of Terms

Final Solution: \{x : x \in \mathbb{R}, -3 < x < 0\}

Step 1: Analyzing the Set Composition

The given set is $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$. To evaluate the validity of any subset relation, we must first identify the distinct elements of $A$ by examining the contents enclosed within the outermost braces separated by commas:

• Element 1: $1$
• Element 2: $2$
• Element 3: $\{ 3, 4 \}$ (This is an object treated as a single element of $A$)
• Element 4: $5$

[Per the definition of Set Membership: $x \in A$ if $x$ is one of the distinct items separated by commas in the roster form.]

Step 2: Defining the Subset Relationship

For a set $B$ to be a subset of $A$ (denoted as $B \subset A$), every element belonging to $B$ must also be an element of $A$. Specifically, if we propose a set $B = \{ 3, 4 \}$, the condition for $B \subset A$ requires that $3 \in A$ and $4 \in A$.

Step 3: Verification of Inclusion

We test the elements of the proposed set $\{ 3, 4 \}$:

[By Definition of Subset: $B \subset A \iff \forall x (x \in B \implies x \in A)$. Since $3 \notin A$ and $4 \notin A$, the implication fails.]

Step 4: Distinguishing Membership from Subset Relations

It is critical to note that $\{ 3, 4 \}$ is an element of $A$, not a subset of $A$.
Relation: $\{ 3, 4 \} \in A$ is a TRUE statement.
Relation: $\{ 3, 4 \} \subset A$ is a FALSE statement because the elements $3$ and $4$ are not individual members of $A$; they are contained within a nested set which itself is a member.

Final Solution: The statement $\{ 3, 4 \} \subset A$ is incorrect. Although $\{ 3, 4 \}$ is an element of $A$ ($\{ 3, 4 \} \in A$), the values $3$ and $4$ are not individual elements of $A$. For $\{ 3, 4 \}$ to be a subset, both $3$ and $4$ would need to be members of $A$.

1. Analysis of Given Sets

To determine the universal set $U$ for the given collection of sets, we must identify a set that contains all elements present in the constituent sets as its subsets. The constituent sets are:

• $A = \{1, 3, 5\}$
• $B = \{2, 4, 6\}$
• $C = \{0, 2, 4, 6, 8\}$

[Definition of Universal Set: A set that contains all the elements under consideration in a given context is called the universal set, denoted by $U$. Symbolically, $A \subseteq U$, $B \subseteq U$, and $C \subseteq U$.]

2. Determination of the Minimal Universal Set

The minimal universal set $U_{min}$ must be the union of all sets $A$, $B$, and $C$:

$U_{min} = A \cup B \cup C$

By listing all unique elements from $A, B,$ and $C$:

$U_{min} = \{1, 3, 5\} \cup \{2, 4, 6\} \cup \{0, 2, 4, 6, 8\}$

$U_{min} = \{0, 1, 2, 3, 4, 5, 6, 8\}$

3. Criteria for Universal Set Candidates

Any set $S$ can be considered a universal set for $A, B,$ and $C$ if and only if $U_{min} \subseteq S$. In standard textbook problems of this type, the candidates provided are typically:

4. Visualization of Set Inclusion

5. Conclusion

Based on the formal requirements of set theory, any set $U$ such that $\{0, 1, 2, 3, 4, 5, 6, 8\} \subseteq U$ qualifies as a universal set for the given inputs. In the context of the standard problem set often associated with this question, the set containing $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ is the correct candidate.

Final Solution: Any set containing all elements of $\{0, 1, 2, 3, 4, 5, 6, 8\}$ is a valid universal set for A, B, and C.

Step 1: Theoretical Definition of the Subset Relation

A set $A$ is defined as a subset of set $B$ (denoted as $A \subseteq B$ or simply $A \subset B$ in elementary contexts) if and only if every element of set $A$ is also an element of set $B$. Formally, this is expressed as:
$A \subset B \iff \forall x (x \in A \implies x \in B)$ [By the definition of Subset Relation in Set Theory].

Step 2: Identification of Elements

Let the given sets be defined as follows:

• Set $A = \{a, b, c\}$
• Set $B = \{b, c, d\}$

Step 3: Logical Verification of Membership

To determine if $A \subset B$, we must verify if every element of $A$ exists within $B$.

Step 4: Conclusion based on Logical Disproof

Since the element $a$ belongs to the first set but does not belong to the second set ($a \in \{a, b, c\}$ but $a \notin \{b, c, d\}$), the condition for the subset relation fails. Therefore, $A$ is not a subset of $B$ [By the negation of the universal quantifier: $\exists x (x \in A \land x \notin B) \implies A \not\subset B$].

Final Solution: \{a, b, c\} $\not\subset$ \{b, c, d\}

Step 1: Identifying the Elements of Set A

Given the set $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$, we must first define the individual members (elements) contained within the set. By inspection of the roster form:

• Element 1: $1$
• Element 2: $2$
• Element 3: $\{3, 4\}$ (This is a subset acting as a single element)
• Element 4: $5$

[Definition: An element $x$ is a member of set $A$ if it belongs to the collection defining $A$, denoted as $x \in A$.]

Step 2: Evaluating the Subset Condition ($\subset$)

The statement is $\{ \{3, 4\} \} \subset A$. For this statement to be true, the definition of a subset must be satisfied [Per the definition of Subset: $S \subset A$ if and only if every element of $S$ is also an element of $A$].

Let $S = \{ \{3, 4\} \}$. The only element in set $S$ is $\{3, 4\}$.

We check if this element is present in $A$:

• Is $\{3, 4\} \in A$? Yes, as established in Step 1, $\{3, 4\}$ is explicitly listed as an element of $A$.

Since the sole element of the subset $\{ \{3, 4\} \}$ is indeed a member of set $A$, the condition $S \subset A$ holds.

Step 3: Logical Analysis of the Statement

The statement $\{ \{3, 4\} \} \subset A$ is correct. Often, confusion arises regarding the nested braces. To clarify, we distinguish between membership and containment:

Visual Representation of Set Membership

Final Solution: The statement $\{ \{3, 4\} \} \subset A$ is correct. It is a valid subset because the element $\{3, 4\}$ is contained as an element within set $A$. Therefore, the set formed by this element is a proper subset of $A$.

Step 1: Theoretical Foundation of Power Sets

To determine the subsets of a set $A$, we utilize the power set definition. For any set with $n$ elements, the total number of subsets is given by the formula $2^n$ [Per the Power Set Cardinality Theorem].

Given the set $A = \{a, b\}$, the cardinality of the set is $n = 2$. Therefore, the total number of subsets is:

$2^n = 2^2 = 4$

Step 2: Systematic Enumeration of Subsets

We enumerate the subsets by considering combinations of elements of size $k$, where $0 \le k \le n$.

• Size $k=0$ (Empty Set): Every set contains the empty set as a subset [By the Axiom of the Empty Set].
  Subset: $\emptyset$
• Size $k=1$ (Singletons): We take each individual element as a distinct set.
  Subsets: $\{a\}$, $\{b\}$
• Size $k=2$ (The Improper Subset): The set itself is always a subset of itself [Per the Reflexive Property of Inclusion].
  Subset: $\{a, b\}$

Step 3: Visual Representation (Power Set Mapping)

Step 4: Final Compilation

The collection of all subsets, known as the Power Set $P(A)$, is expressed as:

$P(A) = \{ \emptyset, \{a\}, \{b\}, \{a, b\} \}$

Final Solution: The subsets of $\{a, b\}$ are $\emptyset, \{a\}, \{b\},$ and $\{a, b\}$.

Theoretical Foundation: Power Sets

The set under consideration is $A = \{1, 2, 3\}$. The number of elements in set $A$, denoted as $n(A)$, is $3$. According to the fundamental property of power sets, the total number of subsets for any set $A$ is given by the formula $2^{n(A)}$ [Per the Set Theory Axiom of Power Sets].

For $n(A) = 3$, the total number of subsets is:

$2^3 = 2 \times 2 \times 2 = 8$

Step 1: Systematic Enumeration of Subsets

To ensure 100% precision, subsets are categorized by their cardinality (number of elements):

• Subsets with 0 elements (The Empty Set):
  $\emptyset$ [By the definition of the empty set being a subset of every set]
• Subsets with 1 element (Singleton Sets):
  $\{1\}$, $\{2\}$, $\{3\}$
• Subsets with 2 elements:
  $\{1, 2\}$, $\{1, 3\}$, $\{2, 3\}$
• Subsets with 3 elements (The Set itself):
  $\{1, 2, 3\}$ [By the definition of improper subsets]

Step 2: Visualizing the Power Set Hierarchy

Step 3: Verification Table

Final Solution: The set of all subsets (Power Set) of $\{1, 2, 3\}$ is $\{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$.

Step 1: Analyzing the Set Composition

The given set $A$ is defined as $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$. To determine the validity of the statement $\{3, 4\} \in A$, we must first perform an explicit enumeration of all elements contained within the set $A$.

By inspection of the set brackets, the distinct elements ($x$) such that $x \in A$ are:

• $x_1 = 1$
• $x_2 = 2$
• $x_3 = \{3, 4\}$
• $x_4 = 5$

[Definition of Element Membership: An object $x$ is an element of set $A$ ($x \in A$) if and only if $x$ is explicitly listed as a member within the set definition.]

Step 2: Evaluating the Membership Relation

The question asks to verify the statement $\{3, 4\} \in A$.

From our enumeration in Step 1, we identify that the object $\{3, 4\}$ exists as a discrete entity within the set. Specifically, $\{3, 4\}$ serves as the third element of set $A$ ($x_3$).

Comparing the query statement to our identified set members:

Step 3: Addressing the Question's Logic

The prompt asks to identify if the statement is "incorrect." Based on the rigorous definition of the element-of relation ($\in$), the statement $\{3, 4\} \in A$ is mathematically correct because the set $\{3, 4\}$ is contained within set $A$ as a single element.

Final Solution: The statement $\{3, 4\} \in A$ is correct. It is not incorrect because $\{3, 4\}$ is indeed one of the four elements explicitly listed in the definition of set $A$.

Step 1: Formal Definition of the Sets

Let the given sets be represented as follows:

• Set $A = \{x : x \text{ is a circle in the plane}\}$
• Set $B = \{x : x \text{ is a circle in the same plane with radius } 1 \text{ unit}\}$

[Definition of a Subset: A set $A$ is a subset of set $B$ ($A \subset B$) if and only if every element of $A$ is also an element of $B$.]

Step 2: Logical Analysis of Elements

We analyze the criteria for inclusion in each set:

• Elements of $A$ consist of all circles of any radius $r \in (0, \infty)$.
• Elements of $B$ are restricted solely to circles where $r = 1$.

Consider an element $c \in A$ such that its radius $r_c = 5$ units. Since $r_c \neq 1$, it is logically impossible for this circle to be an element of $B$. [Per the Law of Excluded Middle, a circle cannot simultaneously have a radius of 5 and a radius of 1].

Step 3: Visual Representation

Step 4: Conclusion based on Subset Criteria

Since there exist elements in $A$ (e.g., circles with radii $r \neq 1$) that are not contained within $B$, the condition for the subset relation fails. [By the logical definition of proper subset negation, $\exists x : x \in A \land x \notin B \implies A \not\subset B$].

Final Solution: \{x : x \text{ is a circle in the plane}\} \not\subset \{x : x \text{ is a circle in the same plane with radius 1 unit}\}

Step 1: Identifying the Set Properties

The given set is the empty set (or null set), denoted by the symbol $\phi$ or $\{\}$. By definition, the empty set contains no elements. Therefore, the cardinality of the set, denoted as $n(A)$, is $0$.

Step 2: Calculating the Number of Subsets

According to the Power Set Theorem, if a set $A$ has $n$ elements, the total number of subsets of $A$ is given by the formula $2^n$ [Per Cantor’s Theorem].

For the set $\phi$:

• Number of elements ($n$) = $0$
• Total number of subsets = $2^0 = 1$

Step 3: Determining the Subsets

By the definition of a subset, a set $B$ is a subset of $A$ if every element of $B$ is also an element of $A$. Furthermore, the empty set $\phi$ is a subset of every set [Axiom of the Empty Set]. Since the set $\phi$ contains no elements, its only subset is itself.

Step 4: Formal Verification

Let $A = \phi$. The collection of all subsets of $A$ is called the Power Set, denoted by $P(A)$.

$P(\phi) = \{ \phi \}$

Since the cardinality of $P(\phi)$ is $1$, which matches our calculation in Step 2, the result is verified.

Final Solution: The only subset of $\phi$ is $\phi$.

Step 1: Analyzing the Set-Builder Notation

The given set is $A = \{x : x \in \mathbb{R}, -4 < x \le 6\}$. This notation defines a subset of the real number line, $\mathbb{R}$, bounded by specific endpoints. We must evaluate the nature of these boundaries to determine the interval type.

• Left Boundary ($-4$): The inequality symbol is $<$ (less than). This indicates that $-4$ is not included in the set. This represents an open boundary, which in interval notation is denoted by a parenthesis, "$($".
• Right Boundary ($6$): The inequality symbol is $\le$ (less than or equal to). This indicates that $6$ is included in the set. This represents a closed boundary, which in interval notation is denoted by a square bracket, "$]$".

Step 2: Defining Interval Classification

Based on the boundary analysis, the set is defined as a semi-open or half-open interval. Specifically, it is an interval open at the left and closed at the right.

Step 3: Visual Representation on the Real Number Line

The following SVG illustrates the interval on the real number line, where the open circle at $-4$ signifies exclusion and the closed dot at $6$ signifies inclusion.

Step 4: Mapping to Standard Interval Notation

Per the standard definitions in Real Analysis [Set Theory Fundamentals], an interval $\{x : a < x \le b\}$ is expressed as $(a, b]$. Substituting our specific values:

$-4 < x \le 6$ corresponds to $(-4, 6]$.

Final Solution: (-4, 6]

Step 1: Formal Definition of the Sets

To determine the relationship between the two given sets, we define them formally using set-builder notation:

• Set $A = \{x : x \text{ is a triangle in a plane}\}$
• Set $B = \{x : x \text{ is a rectangle in a plane}\}$

[Definition of a Set: A collection of well-defined distinct objects. A triangle is a polygon with three edges and three vertices, whereas a rectangle is a quadrilateral with four edges and four right angles.]

Step 2: Analyzing the Subset Criterion

A set $A$ is a subset of set $B$ ($A \subset B$) if and only if every element belonging to $A$ is also an element of $B$. Formally, $\forall x, x \in A \implies x \in B$.

We test the condition: If an object is a triangle, must it necessarily be a rectangle?

• A triangle has $3$ sides and the sum of its interior angles is $180^\circ$.
• A rectangle has $4$ sides and the sum of its interior angles is $360^\circ$.

[Geometric Axiom: By the definition of polygons, a polygon with $n$ sides cannot be a polygon with $m$ sides if $n \neq m$. Since $3 \neq 4$, the set of triangles and the set of rectangles are disjoint in the Euclidean plane.]

Step 3: Visual Representation

Step 4: Conclusion of Relationship

Since there exists no element $x$ such that $x$ is a triangle and $x$ is a rectangle, $A \not\subseteq B$. In fact, $A \cap B = \emptyset$. Therefore, the correct symbol to fill the blank space is $\not\subset$.

Final Solution: $\{x : x \text{ is a triangle in a plane}\} \not\subset \{x : x \text{ is a rectangle in the plane}\}$

Step 1: Analyzing the Set Composition

The given set $A$ is defined as $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$. To determine the validity of the statement $1 \in A$, we must identify the individual elements contained within the set $A$.

The elements of set $A$ are separated by commas:

• Element 1: $1$
• Element 2: $2$
• Element 3: $\{ 3, 4 \}$ (Note: This is a set acting as a single element of $A$)
• Element 4: $5$

Step 2: Defining the Membership Relation ($\in$)

The symbol $\in$ denotes the membership relation [Per the definition of set theory]. If an object $x$ is an element of set $A$, then $x \in A$ is a true statement. Conversely, if $x$ is not listed as a member within the curly braces defining set $A$, then $x \notin A$ [By the Law of Set Membership].

Step 3: Verification of the Statement

We evaluate the statement $1 \in A$ against the identified elements in Step 1:

Step 4: Evaluating the Correctness

The question asks to identify if the statement is incorrect. Since $1$ is explicitly present as an element in the set $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$, the statement $1 \in A$ is mathematically correct [By the definition of the set roster method].

Visual Representation of Set Membership

Final Solution: The statement $1 \in A$ is correct because $1$ is an element of set $A$.

Step 1: Theoretical Definition of a Subset

A set $A$ is said to be a subset of a set $B$ (denoted as $A \subset B$) if and only if every element belonging to set $A$ is also an element belonging to set $B$. Formally: $A \subset B \iff \forall x (x \in A \implies x \in B)$ [Per the Axiom of Extensionality and Definition of Subsets].

Step 2: Identification of Set Elements

Given the two sets:

• Set $A = \{2, 3, 4\}$
• Set $B = \{1, 2, 3, 4, 5\}$

Step 3: Membership Verification

We analyze the membership of each element $x \in A$ within set $B$:

Step 4: Geometric Representation (Venn Diagram)

Step 5: Logical Conclusion

Since every element of $\{2, 3, 4\}$ is contained within the set $\{1, 2, 3, 4, 5\}$, the condition $A \subset B$ is satisfied. Therefore, the correct symbol to fill in the blank is $\subset$.

Final Solution: $\{ 2, 3, 4 \} \subset \{ 1, 2, 3, 4, 5 \}$

Step 1: Theoretical Foundation of Subset Relation

To determine the validity of the statement $\{ a \} \subset \{ a, b, c \}$, we refer to the formal definition of a subset. By definition, a set $A$ is a subset of set $B$ (denoted as $A \subseteq B$ or $A \subset B$) if and only if every element present in $A$ is also an element of $B$. [Per the Axiomatic Definition of Subsets in Set Theory].

Step 2: Decomposition of Given Sets

We identify the sets provided in the expression:

• Let $A = \{ a \}$
• Let $B = \{ a, b, c \}$

Step 3: Verification of Element Inclusion

We analyze the elements of set $A$ to verify their presence in set $B$:

Since the only element of $A$, which is $a$, is explicitly contained within set $B$, the condition for subset relation is satisfied [By the Membership Property of Sets].

Step 4: Geometric/Venn Representation

Step 5: Logical Synthesis

Because $\{ a \} \subseteq \{ a, b, c \}$ and $\{ a \} \neq \{ a, b, c \}$, the statement utilizes the proper subset notation ($\subset$) correctly. Since all elements of the left set are contained in the right set, the statement is logically valid.

Final Solution: The statement { a } $\subset$ { a, b, c } is True.