Menu Top
Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Union of Sets Intersection of Sets Difference of Sets
Complement of a Set Venn Diagrams for Set Operations


Set Operations and Venn Diagrams



Union of Sets

Set operations allow us to combine sets or extract subsets based on specific rules. The first fundamental operation is the union, which combines the elements from two or more sets.

Definition of Union

The union of two sets $A$ and $B$ is the set containing all the elements that belong to set $A$, or belong to set $B$, or belong to both $A$ and $B$. When forming the union, as with any set, we list each distinct element only once.

Symbol and Reading:

The symbol for union is $\cup$. The union of sets $A$ and $B$ is written as $A \cup B$, and is read as "A union B".

Set-Builder Notation:

The union of $A$ and $B$ is formally defined as:

$A \cup B = \{x \mid x \in A \text{ or } x \in B \}$

The logical connective "or" (specifically, the inclusive or) is key here. An element is in the union $A \cup B$ if it is in $A$, or it is in $B$, or it is in both. Elements present in both sets are included only once in the union set.


Example 1. Let $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$. Find $A \cup B$.

Answer:

Given: $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5, 6\}$.

To Find: $A \cup B$.

Solution:

We collect all the elements from set A and all the elements from set B, listing each distinct element only once.

Elements in A are 1, 2, 3, 4.

Elements in B are 3, 4, 5, 6.

The distinct elements from both sets combined are 1, 2, 3, 4, 5, 6.

$A \cup B = \{1, 2, 3, 4, 5, 6\}$


Example 2. Let $V = \{a, e, i, o, u\}$ be the set of vowels and $C = \{a, b, c, d, e\}$ be a set of the first five letters of the English alphabet. Find $V \cup C$.

Answer:

Given: $V = \{a, e, i, o, u\}$, $C = \{a, b, c, d, e\}$.

To Find: $V \cup C$.

Solution:

We list all elements in V and all elements in C, including common elements only once.

Elements in V: a, e, i, o, u.

Elements in C: a, b, c, d, e.

The distinct elements from both sets combined are a, e, i, o, u, b, c, d.

$V \cup C = \{a, b, c, d, e, i, o, u\}$

(The order does not matter).

Properties of the Union Operation

Let $A, B$, and $C$ be any sets, and let $U$ be the universal set.

  1. Commutative Law:

    The order in which we take the union of two sets does not affect the result.

    $A \cup B = B \cup A$

  2. Associative Law:

    When taking the union of three or more sets, the grouping of the sets does not affect the result.

    $(A \cup B) \cup C = A \cup (B \cup C)$

    This allows us to write the union of multiple sets unambiguously as $A \cup B \cup C$.

  3. Identity Law (Law of $\emptyset$):

    The union of any set $A$ with the empty set $\emptyset$ is the set $A$ itself. The empty set acts as the identity element for the union operation.

    $A \cup \emptyset = A$

  4. Idempotent Law:

    The union of a set with itself is the set itself.

    $A \cup A = A$

  5. Law of $U$:

    The union of any set $A$ with the universal set $U$ is the universal set $U$.

    $A \cup U = U$

  6. Subset Property:

    Every set involved in a union is a subset of the resulting union.

    $A \subseteq A \cup B$ and $B \subseteq A \cup B$

  7. Subset Condition:

    If set $A$ is a subset of set $B$ ($A \subseteq B$), then their union is simply set $B$.

    If $A \subseteq B$, then $A \cup B = B$


Intersection of Sets

Another fundamental set operation is intersection, which identifies the elements that are common to two or more sets.

Definition of Intersection

The intersection of two sets $A$ and $B$ is the set containing all the elements that are simultaneously present in both set $A$ and set $B$.

Symbol and Reading:

The symbol for intersection is $\cap$. The intersection of sets $A$ and $B$ is written as $A \cap B$, and is read as "A intersection B".

Set-Builder Notation:

The intersection of $A$ and $B$ is formally defined as:

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

The logical connective "and" is critical here. An element belongs to the intersection $A \cap B$ only if it is a member of $A$ AND a member of $B$.


Example 3. Let $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$. Find $A \cap B$.

Answer:

Given: $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5, 6\}$.

To Find: $A \cap B$.

Solution:

We need to identify the elements that are common to both set A and set B.

Elements in A: 1, 2, 3, 4

Elements in B: 3, 4, 5, 6

The elements that appear in both lists are 3 and 4.

$A \cap B = \{3, 4\}$


Example 4. Let $E = \{x \in \mathbb{Z} \mid x \text{ is even}\}$ and $O = \{x \in \mathbb{Z} \mid x \text{ is odd}\}$. Find $E \cap O$.

Answer:

Given: $E$ = set of even integers, $O$ = set of odd integers.

To Find: $E \cap O$.

Solution:

We are looking for integers that are both even and odd. By definition, an integer cannot be simultaneously even and odd. There are no elements common to the set of even integers and the set of odd integers.

The intersection is the empty set.

$E \cap O = \emptyset$

Disjoint Sets

Based on the concept of intersection, we define disjoint sets:

Two sets $A$ and $B$ are said to be disjoint sets if their intersection is the empty set. This means they have no elements in common.

A and B are disjoint $\iff A \cap B = \emptyset$

Example 4 illustrates disjoint sets $E$ and $O$.


Properties of the Intersection Operation

Let $A, B$, and $C$ be any sets, and let $U$ be the universal set.

  1. Commutative Law:

    The order in which we take the intersection of two sets does not affect the result.

    $A \cap B = B \cap A$

  2. Associative Law:

    When taking the intersection of three or more sets, the grouping of the sets does not affect the result.

    $(A \cap B) \cap C = A \cap (B \cap C)$

    This allows us to write the intersection of multiple sets unambiguously as $A \cap B \cap C$.

  3. Identity Law (Law of $U$):

    The intersection of any set $A$ with the universal set $U$ is the set $A$ itself. The universal set acts as the identity element for the intersection operation.

    $A \cap U = A$

  4. Law of $\emptyset$:

    The intersection of any set $A$ with the empty set $\emptyset$ is the empty set.

    $A \cap \emptyset = \emptyset$

  5. Idempotent Law:

    The intersection of a set with itself is the set itself.

    $A \cap A = A$

  6. Subset Property:

    The intersection of two sets is always a subset of both sets.

    $A \cap B \subseteq A$ and $A \cap B \subseteq B$

  7. Subset Condition:

    If set $A$ is a subset of set $B$ ($A \subseteq B$), then their intersection is simply set $A$.

    If $A \subseteq B$, then $A \cap B = A$

  8. Distributive Laws:

    These laws show how union and intersection interact when combined.
    • Intersection distributes over union:

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

    • Union distributes over intersection:

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


Summary for Competitive Exams

Set Operations: Ways to combine or compare sets based on element membership.

Union ($A \cup B$): Elements in A OR B (or both). $A \cup B = \{x \mid x \in A \text{ or } x \in B \}$.

  • Properties: Commutative, Associative, $A \cup \emptyset = A$, $A \cup A = A$, $A \cup U = U$.

Intersection ($A \cap B$): Elements in A AND B. $A \cap B = \{x \mid x \in A \text{ and } x \in B \}$.

  • Properties: Commutative, Associative, $A \cap U = A$, $A \cap \emptyset = \emptyset$, $A \cap A = A$.
  • Distributive Laws: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ and $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Disjoint Sets: $A$ and $B$ are disjoint if $A \cap B = \emptyset$.



Difference of Sets

The difference of two sets is an operation that results in a new set containing elements from one set that are not present in the other. It helps isolate the unique elements of a set relative to another.

Definition of Difference of Sets

The difference of set $A$ and set $B$ (in that order) is the set containing all those elements which belong to set $A$ but do not belong to set $B$.

Symbol and Reading:

The difference is denoted by $A - B$ or $A \setminus B$. It is read as "A minus B" or "the difference of A and B".

Set-Builder Notation:

The difference $A - B$ is formally defined as:

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

Similarly, the difference of set $B$ and set $A$ is $B - A = \{x \mid x \in B \text{ and } x \notin A \}$.

The operation $A - B$ essentially removes any elements from $A$ that are also present in $B$.

Important Note:

The set difference operation is generally not commutative. In most cases, $A - B \neq B - A$. The order of the sets matters.


Example 1. Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{4, 5, 6, 7\}$. Find $A - B$ and $B - A$.

Answer:

Given: $A = \{1, 2, 3, 4, 5\}$, $B = \{4, 5, 6, 7\}$.

To Find: $A - B$ and $B - A$.

Solution:

Finding $A - B$:

We identify elements that are in set A but not in set B.

Elements in A: 1, 2, 3, 4, 5.

Elements in B: 4, 5, 6, 7.

Elements that are in A and also in B are 4 and 5. We remove these from A.

Elements in A but not in B are 1, 2, 3.

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

Finding $B - A$:

We identify elements that are in set B but not in set A.

Elements in B: 4, 5, 6, 7.

Elements in A: 1, 2, 3, 4, 5.

Elements that are in B and also in A are 4 and 5. We remove these from B.

Elements in B but not in A are 6, 7.

$B - A = \{6, 7\}$

As clearly shown, $A - B = \{1, 2, 3\}$ and $B - A = \{6, 7\}$, so $A - B \neq B - A$.


Properties of Set Difference

Let $A$ and $B$ be any sets, and let $U$ be the universal set relevant to the context.

  1. Subset Property:

    The difference $A - B$ is always a subset of the first set, $A$.

    $A - B \subseteq A$

    Similarly, $B - A \subseteq B$.

  2. Difference with Self:

    The difference of a set with itself is the empty set.

    $A - A = \emptyset$

  3. Difference with Empty Set:

    The difference of a set $A$ and the empty set $\emptyset$ is the set $A$.

    $A - \emptyset = A$

  4. Difference of Empty Set:

    The difference of the empty set $\emptyset$ and any set $A$ is the empty set.

    $\emptyset - A = \emptyset$

  5. Relationship with Intersection and Complement:

    The difference $A - B$ can be expressed using intersection and the complement of $B$ relative to $U$ (assuming $A$ and $B$ are subsets of $U$).

    $A - B = A \cap B'$

    where $B'$ is the complement of $B$ with respect to the universal set $U$ ($B' = U - B$). This property is often used in proofs and set theory identities.

    Derivation:

    Let $x \in A - B$. By definition, $x \in A$ and $x \notin B$. If $x \in U$ (which is assumed for complements), then $x \notin B \implies x \in B'$. So, $x \in A$ and $x \in B'$, which means $x \in A \cap B'$. Thus $A - B \subseteq A \cap B'$.

    Let $y \in A \cap B'$. By definition, $y \in A$ and $y \in B'$. By definition of complement, $y \in U$ and $y \notin B$. Since $y \in A$ and $y \notin B$, by definition of set difference, $y \in A - B$. Thus $A \cap B' \subseteq A - B$.

    Since both containments hold, $A - B = A \cap B'$.

  6. Disjoint Parts:

    The sets $A - B$, $B - A$, and $A \cap B$ are mutually disjoint. That is, the intersection of any pair of these three sets is the empty set.
    • $(A - B) \cap (B - A) = \emptyset$
    • $(A - B) \cap (A \cap B) = \emptyset$
    • $(B - A) \cap (A \cap B) = \emptyset$
  7. Union of Disjoint Parts:

    The union of the three disjoint sets $A - B$, $B - A$, and $A \cap B$ gives the union of the original sets $A$ and $B$.

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


Symmetric Difference ($\Delta$)

The symmetric difference of two sets $A$ and $B$ is the set of elements that are in either $A$ or $B$, but not in their intersection. It consists of the elements that are in one set or the other, but not in both.

Symbol and Reading:

The symmetric difference is denoted by $A \Delta B$ or $A \ominus B$. It is read as "A symmetric difference B".

Definitions:

The symmetric difference can be defined in two equivalent ways:

Example:

Using the sets from Example 1, $A = \{1, 2, 3, 4, 5\}$ and $B = \{4, 5, 6, 7\}$.

We found $A - B = \{1, 2, 3\}$ and $B - A = \{6, 7\}$.

Using the first definition of symmetric difference:

$A \Delta B = (A - B) \cup (B - A) = \{1, 2, 3\} \cup \{6, 7\} = \{1, 2, 3, 6, 7\}$

Let's verify using the second definition:

$A \cup B = \{1, 2, 3, 4, 5, 6, 7\}$

$A \cap B = \{4, 5\}$

$A \Delta B = (A \cup B) - (A \cap B) = \{1, 2, 3, 4, 5, 6, 7\} - \{4, 5\} = \{1, 2, 3, 6, 7\}$.

Both definitions yield the same result.

Property of Symmetric Difference:

The symmetric difference operation is commutative:

$A \Delta B = B \Delta A$


Complement of a Set

The complement of a set is defined relative to a larger set, usually the universal set, and includes all elements from the universal set that are not found in the given set. It's like everything "outside" the set but still within the scope of the discussion.

Definition of Complement

Let $U$ be the universal set for a particular discussion, and let $A$ be any set within this context (i.e., $A \subseteq U$). The complement of set $A$ with respect to $U$ is the set of all elements in $U$ that are not elements of $A$.

Symbols and Reading:

The complement of $A$ is commonly denoted by $A'$, $A^c$, or $\bar{A}$. It is read as "A complement", "A prime", or "complement of A".

Set-Builder Notation:

The complement of $A$ is formally defined as:

$A' = \{x \mid x \in U \text{ and } x \notin A \}$

Using the set difference notation, the complement of $A$ with respect to $U$ is simply $U - A$.

$A' = U - A$

The concept of a universal set is essential for defining the complement; a set's complement is meaningless without a defined universal set.


Example 2. Let the universal set be $U = \{1, 2, 3, ..., 10\}$ (first 10 natural numbers). Let $A = \{x \mid x \text{ is an even number in } U\}$ and $B = \{x \mid x \text{ is a prime number in } U\}$. Find $A'$ and $B'$.

Answer:

Given: $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

First, let's list the elements of A and B.

$A = \{x \in U \mid x \text{ is even}\} = \{2, 4, 6, 8, 10\}$.

$B = \{x \in U \mid x \text{ is prime}\}$. Prime numbers are natural numbers greater than 1 divisible only by 1 and themselves. The prime numbers in U are 2, 3, 5, 7.

$B = \{2, 3, 5, 7\}$.

To Find: $A'$ and $B'$.

Solution:

Finding $A'$:

$A'$ is the set of elements in $U$ that are not in $A$.

$U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

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

Elements in $U$ but not in $A$ are 1, 3, 5, 7, 9.

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

Finding $B'$:

$B'$ is the set of elements in $U$ that are not in $B$.

$U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

$B = \{2, 3, 5, 7\}$

Elements in $U$ but not in $B$ are 1, 4, 6, 8, 9, 10.

$B' = \{1, 4, 6, 8, 9, 10\}$


Properties of Complement

Let $A$ and $B$ be subsets of a universal set $U$.

  1. Complement Laws:

    • The union of a set and its complement is the universal set:

      $A \cup A' = U$

    • The intersection of a set and its complement is the empty set:

      $A \cap A' = \emptyset$

  2. Law of Double Complementation (Involution Law):

    Taking the complement of the complement of a set returns the original set.

    $(A')' = A$

  3. Laws of Empty Set and Universal Set:

    • The complement of the empty set is the universal set:

      $\emptyset' = U$

    • The complement of the universal set is the empty set:

      $U' = \emptyset$

  4. De Morgan's Laws:

    These two laws describe how complement interacts with union and intersection. They are named after Augustus De Morgan.
    • De Morgan's First Law: The complement of the union of two sets is the intersection of their complements.

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

    • De Morgan's Second Law: The complement of the intersection of two sets is the union of their complements.

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

  5. Subset Property:

    If $A$ is a subset of $B$, then the complement of $B$ is a subset of the complement of $A$.

    If $A \subseteq B$, then $B' \subseteq A'$

Proof of De Morgan's First Law: $(A \cup B)' = A' \cap B'$

To prove that two sets are equal, we show that each is a subset of the other.

Part 1: Show $(A \cup B)' \subseteq A' \cap B'$

Let $x$ be an arbitrary element such that $x \in (A \cup B)'$.

By the definition of complement, this means $x$ is in the universal set $U$ and $x$ is not in the union $A \cup B$.

$x \in U \text{ and } x \notin (A \cup B)$

... (i)

If an element is not in the union of A and B, it means it is neither in A nor in B.

$x \notin A \text{ and } x \notin B$

... (ii)

From (i) and (ii), we have $x \in U$ and $x \notin A$, which by the definition of complement means $x \in A'$.

$x \in A'$

... (iii)

Also from (i) and (ii), we have $x \in U$ and $x \notin B$, which by the definition of complement means $x \in B'$.

$x \in B'$

... (iv)

From (iii) and (iv), by the definition of intersection, we conclude that $x \in A' \cap B'$.

$x \in A' \cap B'$

... (v)

So, starting with an arbitrary element $x$ in $(A \cup B)'$ and logically deducing that it must be in $A' \cap B'$, we have shown that $(A \cup B)'$ is a subset of $A' \cap B'$.

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

... (vi)

Part 2: Show $A' \cap B' \subseteq (A \cup B)'$

Let $y$ be an arbitrary element such that $y \in A' \cap B'$.

By the definition of intersection, this means $y$ is in $A'$ and $y$ is in $B'$.

$y \in A' \text{ and } y \in B'$

... (vii)

By the definition of complement, $y \in A'$ means $y \in U$ and $y \notin A$.

$y \notin A$

... (viii)

And $y \in B'$ means $y \in U$ and $y \notin B$.

$y \notin B$

... (ix)

From (viii) and (ix), if an element is not in A and not in B, it cannot be in their union $A \cup B$.

$y \notin (A \cup B)$

... (x)

Also, since $y \in A'$ and $y \in B'$, it must be that $y \in U$. (The universal set is the same for both complements).

$y \in U$

... (xi)

From (x) and (xi), by the definition of complement, we conclude that $y$ is in the complement of $A \cup B$.

$y \in (A \cup B)'$

So, starting with an arbitrary element $y$ in $A' \cap B'$ and logically deducing that it must be in $(A \cup B)'$, we have shown that $A' \cap B'$ is a subset of $(A \cup B)'$.

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

... (xii)

Since we have shown both $(A \cup B)' \subseteq A' \cap B'$ [from (vi)] and $A' \cap B' \subseteq (A \cup B)'$ [from (xii)], by the definition of set equality, the two sets must be equal.

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

$\textsf{(Hence Proved)}$

The proof for De Morgan's Second Law $(A \cap B)' = A' \cup B'$ follows a similar logical structure.


Summary for Competitive Exams

Set Difference ($A - B$ or $A \setminus B$): Elements in A but NOT in B. $A - B = \{x \mid x \in A \text{ and } x \notin B \}$.

  • Not commutative: $A - B \neq B - A$ (generally).
  • Properties: $A - B \subseteq A$, $A - A = \emptyset$, $A - \emptyset = A$, $\emptyset - A = \emptyset$.
  • Relationship: $A - B = A \cap B'$ (where $B'$ is complement w.r.t. Universal Set U).
  • $A \cup B = (A - B) \cup (B - A) \cup (A \cap B)$ (Disjoint union).

Symmetric Difference ($A \Delta B$): Elements in (A or B) BUT NOT (A and B). $A \Delta B = (A - B) \cup (B - A) = (A \cup B) - (A \cap B)$. Commutative.

Complement ($A'$ or $A^c$ or $\bar{A}$): Elements in the Universal Set U that are NOT in A. $A' = \{x \mid x \in U \text{ and } x \notin A \} = U - A$. Requires a Universal Set U.

  • Properties: $A \cup A' = U$, $A \cap A' = \emptyset$, $(A')' = A$, $\emptyset' = U$, $U' = \emptyset$.
  • De Morgan's Laws: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$.
  • If $A \subseteq B$, then $B' \subseteq A'$.


Venn Diagrams for Set Operations

Venn diagrams are powerful visual tools used to represent sets and the relationships between them, including the results of set operations. They provide an intuitive way to understand complex set identities and solve problems involving counts of elements in sets.

Representation in Venn Diagrams

In a standard Venn diagram:

Visualizing Set Operations with Venn Diagrams

Venn diagrams make the results of set operations visually clear:

Venn diagrams are not just for visualization; they are invaluable for verifying set identities and deriving formulas related to the number of elements in sets.

Cardinality Formulas and Venn Diagrams (Principle of Inclusion-Exclusion)

Venn diagrams help us understand formulas for the number of elements (cardinality) in the union and intersection of finite sets. The Principle of Inclusion-Exclusion provides these formulas.

For Two Finite Sets A and B:

Consider the Venn diagram for $A \cup B$. The union is composed of three disjoint regions: elements only in A ($A-B$), elements only in B ($B-A$), and elements in both ($A \cap B$).

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

Since these three sets are disjoint, the number of elements in their union is the sum of their cardinalities:

$|A \cup B| = |A - B| + |B - A| + |A \cap B|$

... (i)

Also, notice from the Venn diagram that set $A$ consists of the disjoint regions $(A-B)$ and $(A \cap B)$. Similarly, set $B$ consists of the disjoint regions $(B-A)$ and $(A \cap B)$.

$|A| = |A - B| + |A \cap B|$

... (ii)

$|B| = |B - A| + |A \cap B|$

... (iii)

From (ii), $|A - B| = |A| - |A \cap B|$. From (iii), $|B - A| = |B| - |A \cap B|$.

Substitute these expressions for $|A - B|$ and $|B - A|$ into equation (i):

$|A \cup B| = (|A| - |A \cap B|) + (|B| - |A \cap B|) + |A \cap B|$

Simplifying the right side:

$|A \cup B| = |A| + |B| - |A \cap B|$

[Principle of Inclusion-Exclusion for 2 sets]

This is a fundamental formula for the cardinality of the union of two sets. It's called the Principle of Inclusion-Exclusion because you include the sizes of A and B, but then exclude the size of their intersection (which was counted twice - once in |A| and once in |B|).

If $A$ and $B$ are disjoint, $A \cap B = \emptyset$, so $|A \cap B| = 0$. The formula simplifies to $|A \cup B| = |A| + |B|$.

From this formula, we can also deduce others:

$|A \cap B| = |A| + |B| - |A \cup B|$

And using $|A-B| = |A| - |A \cap B|$ from above:

$|A - B| = |A| - (|A| + |B| - |A \cup B|) = |A| + |B| - |A| - |B| + |A \cup B|$

$|A - B| = |A \cup B| - |B|$

Similarly, $|B - A| = |A \cup B| - |A|$.

Also, the cardinality of the complement of A relative to U is:

$|A'| = |U| - |A|$

For Three Finite Sets A, B, and C:

The principle of inclusion-exclusion extends to three sets. Visualizing this requires a Venn diagram with three overlapping circles. The formula accounts for elements counted multiple times in the different regions:

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$

[Principle of Inclusion-Exclusion for 3 sets]

You sum the sizes of individual sets (include), subtract the sizes of pairwise intersections (exclude what was counted twice), and finally add back the size of the intersection of all three sets (include what was excluded three times in the pairwise subtractions).


Summary for Competitive Exams

Venn Diagrams: Pictorial tool (Rectangle for U, Circles for sets) to represent sets and operations.

  • Union ($A \cup B$): Shaded area covering A or B.
  • Intersection ($A \cap B$): Shaded overlapping area of A and B.
  • Difference ($A - B$): Shaded part of A outside B.
  • Complement ($A'$): Shaded area inside U but outside A.
  • Disjoint Sets: Non-overlapping circles.
  • Subset ($A \subseteq B$): Circle A inside circle B.

Cardinality Formulas (Inclusion-Exclusion): For finite sets.

  • $|A \cup B| = |A| + |B| - |A \cap B|$
  • If $A, B$ are disjoint, $|A \cup B| = |A| + |B|$.
  • $|A - B| = |A| - |A \cap B|$
  • $|A'| = |U| - |A|$
  • $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$

These formulas are essential for solving word problems involving counting elements in overlapping sets.