Step 1: Formal Definition of Subset and Set Difference

To prove that $C - B \subseteq C - A$ given the condition $A \subseteq B$, we must rely on the fundamental definitions of set theory:

• Subset Definition: $A \subseteq B$ if and only if $x \in A \implies x \in B$ [Per the definition of subset inclusion].
• Set Difference Definition: $C - S = \{x : x \in C \text{ and } x \notin S\}$ [Per the definition of relative complement].

Step 2: Logical Derivation of the Inclusion

To prove the inclusion $C - B \subseteq C - A$, we must demonstrate that any arbitrary element $x$ belonging to $C - B$ also necessarily belongs to $C - A$.

Let $x$ be an arbitrary element such that:

$x \in (C - B)$

By the definition of set difference, this implies:

$x \in C$ and $x \notin B$ [Condition 1]

Step 3: Applying the Contrapositive of the Given Condition

We are given $A \subseteq B$. Applying the logical law of contraposition:

$A \subseteq B \iff (x \in A \implies x \in B)$

The contrapositive is:

$(x \notin B \implies x \notin A)$ [By the Rule of Contrapositive in propositional logic]

Step 4: Synthesizing the Argument

Since we established from Condition 1 that $x \notin B$, we can now invoke the result from Step 3:

1. We know $x \in C$ [From Step 2].
2. We know $x \notin B$, which implies $x \notin A$ [From Step 3].

Combining these, we have $x \in C$ and $x \notin A$.

By the definition of set difference, this exactly satisfies the condition for membership in $C - A$:

$x \in (C - A)$

Step 5: Visual Representation

Final Conclusion

Since every element $x \in (C - B)$ has been shown to be an element of $(C - A)$, by the definition of a subset, we conclude:

Final Solution: $C - B \subseteq C - A$

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: Theoretical Definition of the Set

The set $C$ is defined by the set-builder notation: $C = \{x : x \in \mathbb{N}, 10 \le x \le 99, \text{sum of digits of } x = 8\}$.
Let the two-digit number be represented as $10a + b$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $b \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The constraint given is $a + b = 8$ [Constraint based on problem statement].

Step 2: Systematic Identification of Elements

To find all possible values for $x$, we iterate through all possible values of the tens digit ($a$) and determine the corresponding units digit ($b$) such that $b = 8 - a$. Since $x$ is a natural number, $a$ must be at least 1.

Step 3: Verification of Constraints

We check if any other values exist. If $a=9$, $b = 8 - 9 = -1$, which is not a natural digit [By definition of decimal numeral system]. Thus, the set is complete.

Step 4: Representation in Roster Form

The roster form requires listing all elements of the set separated by commas and enclosed within curly braces [Per the definition of Roster/Tabular form].

Final Solution: $C = \{17, 26, 35, 44, 53, 62, 71, 80\}$

Step 1: Given Sets and Definitions

The Universal set and the two subsets are defined as follows:

• $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
• $A = \{2, 4, 6, 8\}$
• $B = \{2, 3, 5, 7\}$

[Definition: The complement of a set $X$ with respect to $U$, denoted as $X'$, is the set of all elements in $U$ that are not in $X$. Formally: $X' = \{x : x \in U \text{ and } x \notin X\}$.]

Step 2: Calculating Left-Hand Side (LHS) = $(A \cap B)'$

First, we find the intersection of $A$ and $B$:

$A \cap B = \{x : x \in A \text{ and } x \in B\} = \{2\}$

Next, we determine the complement of $(A \cap B)$:

$(A \cap B)' = U \setminus \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$

Step 3: Calculating Right-Hand Side (RHS) = $A' \cup B'$

First, calculate the complements of individual sets $A$ and $B$:

$A' = U \setminus A = \{1, 3, 5, 7, 9\}$

$B' = U \setminus B = \{1, 4, 6, 8, 9\}$

Next, perform the union of $A'$ and $B'$:

$A' \cup B' = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\}$

$A' \cup B' = \{1, 3, 4, 5, 6, 7, 8, 9\}$

Step 4: Verification via Venn Diagram Representation

The Venn diagram above shows the distribution of elements. The region $(A \cap B)'$ corresponds to everything outside the central intersection overlap. By computing the union of the regions outside $A$ and $B$, we observe identical element coverage.

Step 5: Comparative Analysis

[Per De Morgan's Law for sets: The complement of the intersection of two sets is equal to the union of their complements.]

Final Solution: Since LHS = RHS = $\{1, 3, 4, 5, 6, 7, 8, 9\}$, the identity $(A \cap B)' = A' \cup B'$ is verified.

Step 1: Theoretical Definition of Set Difference

The difference of two sets $Y$ and $X$, denoted by $Y - X$, is defined as the set of all elements that belong to $Y$ but do not belong to $X$. Mathematically, this is expressed as:

$Y - X = \{ y \mid y \in Y \text{ and } y \notin X \}$

[Per the Set-Theoretic Definition of Relative Complement]

Step 2: Listing Given Elements

• Set $X = \{ a, b, c, d \}$
• Set $Y = \{ f, b, d, g \}$

Step 3: Sequential Element-wise Evaluation

We evaluate each element $y \in Y$ to determine if it is present in $X$:

Step 4: Venn Diagrammatic Representation

Step 5: Final Set Construction

By collecting the elements identified in Step 3 that satisfy the condition ($y \in Y$ and $y \notin X$):

$Y - X = \{ f, g \}$

Final Solution: $Y - X = \{ f, g \}$

Step 1: Identifying the Given Sets

The problem provides four distinct sets defined by the roster method:

• $A = \{3, 5, 7, 9, 11\}$
• $B = \{7, 9, 11, 13\}$
• $C = \{11, 13, 15\}$
• $D = \{15, 17\}$

Step 2: Defining the Intersection Operation

The intersection of two sets $A$ and $D$, denoted by $A \cap D$, is defined as the set containing all elements $x$ such that $x \in A$ AND $x \in D$. [By the Set-Theoretic Definition of Intersection].

Step 3: Evaluating Elements of A and D

We perform a comparative analysis to identify common elements between the two sets:

Step 4: Visualizing the Disjoint Sets

Step 5: Formulating the Final Result

Since there are no elements that satisfy the condition $x \in A \cap D$, the resulting set is the empty set (or null set). [By the definition of Disjoint Sets].

Final Solution: $A \cap D = \emptyset$ or $\phi$

Definition and Theoretical Framework

The set $A$ is defined using the set-builder notation: $A = \{x : x \text{ is an odd natural number}\}$. To list the elements of this set, we invoke the following definitions:

• Natural Numbers ($\mathbb{N}$): The set of counting numbers starting from 1, i.e., $\{1, 2, 3, 4, 5, \dots\}$.
• Odd Numbers: Integers not divisible by 2. Formally, an odd number is defined by the expression $2n - 1$, where $n \in \mathbb{N}$.

Step 1: Identifying the Sequence

We generate the elements of the set by substituting consecutive values of $n$ into the expression for odd numbers:

Step 2: Visual Representation of the Set Elements

The set $A$ constitutes an infinite sequence along the number line representing discrete odd points.

Step 3: Constructing the Roster Form

By the Roster Method (or Tabular Form), we list all elements within curly braces, separated by commas. [Per the Axiom of Extensionality, a set is defined solely by its elements].

$A = \{1, 3, 5, 7, 9, \dots\}$

Final Solution: The set of odd natural numbers is represented in roster form as $A = \{1, 3, 5, 7, 9, \dots\}$.

Step 1: Identifying the Universal Set and the Given Subset

The universal set $U$ is defined as the set containing all elements under consideration for the given context. The subset $B$ contains specific elements from $U$.

• Universal Set: $U = \{a, b, c, d, e, f, g, h\}$
• Subset: $B = \{d, e, f, g\}$

Step 2: Defining the Complement of a Set

Per the definition of set theory, the complement of a set $B$ with respect to a universal set $U$, denoted as $B'$ or $B^c$, is the set of all elements that are members of $U$ but are not members of $B$.

Mathematically, this is expressed as:

$B' = \{x : x \in U \text{ and } x \notin B\}$

Step 3: Executing the Set Difference Operation

To find $B'$, we remove the elements present in $B$ from the collection $U$:

$B' = U \setminus B$

$B' = \{a, b, c, \color{red}{d, e, f, g}, h\} \setminus \{d, e, f, g\}$

Following the exclusion of elements $\{d, e, f, g\}$ from $U$, we retain the remaining elements:

$B' = \{a, b, c, h\}$

Visual Representation of the Complement

Conclusion

By subtracting the subset $B$ from the universal set $U$, we identify the remaining elements that constitute the complement set.

Final Solution: $B' = \{a, b, c, h\}$

Step 1: Theoretical Foundation of Set Difference

The set difference of two sets $B$ and $A$, denoted as $B - A$, is defined as the set of all elements that belong to set $B$ but do not belong to set $A$. Mathematically, this is expressed as:

$B - A = \{x : x \in B \text{ and } x \notin A\}$

[Per the Definition of Relative Complement in Set Theory]

Step 2: Identifying the Given Elements

We are provided with the following sets:

• $A = \{3, 6, 9, 12, 15, 18, 21\}$
• $B = \{4, 8, 12, 16, 20\}$

Step 3: Sequential Evaluation of Elements in B

To determine $B - A$, we evaluate each element of $B$ to check for its presence in $A$:

Step 4: Vector Visualization of Set Difference

Step 5: Final Construction of the Resulting Set

By removing the intersection $\{12\}$ from set $B$, we obtain the set containing the remaining elements.

$B - A = \{4, 8, 16, 20\}$

Final Solution: $B - A = \{4, 8, 16, 20\}$

Step 1: Theoretical Foundation

To represent the complement of the union of two sets, $(A \cup B)'$, in a Venn diagram, we must satisfy the following definitions:

• Universal Set ($U$): Represented by a rectangle, containing all elements under consideration.
• Union of Sets ($A \cup B$): The collection of all elements that are in $A$, in $B$, or in both.
• Complement $(A \cup B)'$: The set of all elements in the universal set $U$ that are not in the set $(A \cup B)$. [By the definition of Complement of a Set, $(A \cup B)' = \{x \in U : x \notin (A \cup B)\}]$.

Step 2: Logical Construction

1. Draw a rectangle to denote the Universal Set $U$.
2. Draw two intersecting circles, $A$ and $B$, within the rectangle.
3. Identify the region occupied by $(A \cup B)$, which constitutes the entire interior of both circles.
4. Shade the area outside both circles but inside the rectangle. This region represents the elements that belong to neither $A$ nor $B$.

Step 3: Visual Representation

Step 4: Algebraic Summary

The operation can be broken down using De Morgan's Law [Per De Morgan's Law]:

$(A \cup B)' = A' \cap B'$

This confirms that the region $(A \cup B)'$ is the intersection of elements outside $A$ ($A'$) and elements outside $B$ ($B'$).

Final Solution: The Venn diagram for $(A \cup B)'$ is the region of the universal set $U$ excluding the union of circles $A$ and $B$.

Step 1: Identifying the Universal Set and the Given Subset

Let $U$ be the universal set of all natural numbers. By definition, the set of natural numbers is represented as:

$U = \{1, 2, 3, 4, 5, 6, 7, 8, \dots \}$ [Per the definition of the set of natural numbers, $\mathbb{N}$]

Let $A$ be the given set of odd natural numbers:

$A = \{x : x \text{ is an odd natural number}\} = \{1, 3, 5, 7, 9, \dots \}$

Step 2: Defining the Complement of a Set

The complement of a set $A$, denoted as $A'$ or $A^c$, consists of all elements in the universal set $U$ that are not members of set $A$. Mathematically, this is expressed as:

$A' = \{x \in U : x \notin A\}$ [By the definition of Set Complement]

Step 3: Calculating the Complement

To find $A'$, we subtract the set of odd natural numbers from the set of all natural numbers:

$A' = \{x \in \mathbb{N} : x \text{ is not an odd natural number}\}$

Since any natural number that is not odd must necessarily be an even natural number, we evaluate the set as follows:

$A' = \{2, 4, 6, 8, 10, \dots \}$

This corresponds to the set of all even natural numbers.

Visual Representation of the Complement

Step 4: Formal Set-Builder Notation

The resulting set $A'$ can be expressed in set-builder notation as:

$A' = \{x : x \text{ is an even natural number}\}$

Final Solution: The complement of the set of odd natural numbers is the set of all even natural numbers, represented as $\{x : x \text{ is an even natural number}\}$.

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: Identifying the Given Sets

The problem provides four distinct sets containing discrete elements as follows:

• $A = \{1, 2, 3, 4\}$
• $B = \{3, 4, 5, 6\}$
• $C = \{5, 6, 7, 8\}$
• $D = \{7, 8, 9, 10\}$

Step 2: Defining the Operation

The union of two sets, denoted as $A \cup C$, is defined as the set of all elements that are members of set $A$, set $C$, or both. Mathematically, this is expressed as:

$A \cup C = \{x : x \in A \text{ or } x \in C\}$ [By the definition of Set Union]

Step 3: Execution of the Union Operation

To find the union, we combine the elements of set $A$ and set $C$, ensuring that duplicate elements are listed only once [Per the Property of Idempotency and Set Representation Standards].

List elements of $A$: $1, 2, 3, 4$

List elements of $C$: $5, 6, 7, 8$

Combining these sets: $\{1, 2, 3, 4, 5, 6, 7, 8\}$

Visual Representation (Venn Diagram)

Step 4: Verification of Elements

Final Solution: $A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

Step 1: Theoretical Foundation of Complementary Sets

By definition, the complement of a subset $A$ with respect to a universal set $U$, denoted by $A'$ or $A^c$, is the set of all elements that belong to the universal set $U$ but do not belong to the set $A$. [Per the Set Difference definition: $A^c = U \setminus A = \{x : x \in U \text{ and } x \notin A\}$].

Step 2: Identification of Given Data

• Universal Set ($U$): $\{a, b, c, d, e, f, g, h\}$
• Subset ($A$): $\{a, b, c\}$

Step 3: Systematic Logical Filtering

To determine $A^c$, we iterate through every element of $U$ and apply the exclusion criteria for set $A$:

Step 4: Visual Representation via Venn Diagram

Step 5: Final Derivation

The complement set $A^c$ comprises the elements remaining in the universal set after removing the intersection of $U$ and $A$.

$A^c = \{x : x \in \{a, b, c, d, e, f, g, h\} \text{ and } x \notin \{a, b, c\}\}$

$A^c = \{d, e, f, g, h\}$

Final Solution: The complement of set $A$ is $A^c = \{d, e, f, g, h\}$.

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.

Step 1: Identifying Given Sets

The universal sets provided for the calculation are defined as follows:

• $A = \{3, 5, 7, 9, 11\}$
• $B = \{7, 9, 11, 13\}$
• $C = \{11, 13, 15\}$

Step 2: Performing the Union Operation $(B \cup C)$

The union of two sets, $B$ and $C$, is defined as the set of all elements that are members of $B$, or $C$, or both. [Per the definition of Set Union: $B \cup C = \{x : x \in B \text{ or } x \in C\}$].

$B \cup C = \{7, 9, 11, 13\} \cup \{11, 13, 15\}$

By listing all unique elements from both sets and removing duplicates:

$B \cup C = \{7, 9, 11, 13, 15\}$

Step 3: Performing the Intersection Operation $A \cap (B \cup C)$

The intersection of set $A$ and the resulting set $(B \cup C)$ is defined as the set containing elements common to both. [Per the definition of Set Intersection: $A \cap (B \cup C) = \{x : x \in A \text{ and } x \in (B \cup C)\}$].

$A \cap (B \cup C) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15\}$

Analyzing the common elements:

• $7 \in A$ and $7 \in (B \cup C)$
• $9 \in A$ and $9 \in (B \cup C)$
• $11 \in A$ and $11 \in (B \cup C)$

Visual Representation of Set Intersection

Final Solution: $A \cap (B \cup C) = \{7, 9, 11\}$

Step 1: Identifying the Mathematical Operation

The problem requires the determination of the set difference, denoted as $A - D$. By definition, the difference of two sets $A$ and $D$ is the set of elements which belong to $A$ but do not belong to $D$.

Symbolically, $A - D = \{x : x \in A \text{ and } x \notin D\}$. [Per the Set Theoretic Definition of Relative Complement].

Step 2: Defining the Given Sets

• $A = \{3, 6, 9, 12, 15, 18, 21\}$
• $D = \{5, 10, 15, 20\}$

Step 3: Systematic Comparison of Elements

We evaluate each element of set $A$ to determine if it is present in set $D$:

Step 4: Constructing the Resultant Set

By collecting all elements of $A$ that are not in $D$, we obtain:

$A - D = \{3, 6, 9, 12, 18, 21\}$

Visual Representation of Set Difference

Final Solution: $A - D = \{3, 6, 9, 12, 18, 21\}$

Step 1: Theoretical Foundation of Set Intersection

The intersection of two sets $X$ and $Y$, denoted by $X \cap Y$, is defined as the set containing all elements that are members of both set $X$ AND set $Y$ simultaneously [Per the Definition of Intersection in Set Theory]. Formally:

$X \cap Y = \{ x : x \in X \text{ and } x \in Y \}$

Step 2: Identifying Given Set Elements

• Set $X = \{ a, b, c, d \}$
• Set $Y = \{ f, b, d, g \}$

Step 3: Sequential Element-by-Element Comparison

Step 4: Logical Construction of the Resultant Set

Extracting the common elements identified in Step 3, we observe that elements $b$ and $d$ satisfy the condition of belonging to both sets $X$ and $Y$.

$X \cap Y = \{ b, d \}$

Step 5: Visual Representation (Venn Diagram)

Final Solution: $X \cap Y = \{ b, d \}$