Sets
Sets and Set Theory
Foundations of Set Theory
Definition and Representations of SetsTopic 1
A set is a well-defined collection of distinct objects, meaning that a clear mathematical rule determines whether any given element belongs to the set. Sets are conventionally represented in two forms: Roster (Tabular) Form, where all elements are explicitly listed inside curly brackets separated by commas, and Set-Builder Form, where elements are described by their defining mathematical property, written as $A = \{x : P(x)\}$. A common conceptual pitfall in JEE is overlooking the structural constraints placed on variables within set-builder notation (such as domain restrictions like $x \in \mathbb{Z}$ versus $x \in \mathbb{R}$), which dramatically changes the cardinality of the set.
Express the set $A = \{x \in \mathbb{Z} : |2x - 3| \le 5\}$ in roster form and find its cardinality.
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To determine the elements of set $A$, we must solve the absolute value inequality within the domain of integers ($\mathbb{Z}$): \[ |2x - 3| \le 5 \implies -5 \le 2x - 3 \le 5 \] Add $3$ across all parts of the compound inequality: \[ -2 \le 2x \le 8 \] Divide the inequality by $2$: \[ -1 \le x \le 4 \] Since $x \in \mathbb{Z}$, extract all integers that lie within this closed interval: \[ x \in \{-1, 0, 1, 2, 3, 4\} \] Listing these elements inside curly braces gives the roster form of the set: \[ A = \{-1, 0, 1, 2, 3, 4\} \] Counting the distinct elements, we find the cardinality $|A| = 6$. Final Answer: $A = \{-1, 0, 1, 2, 3, 4\}$ with cardinality $6$.
Write the set $B = \left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}\right\}$ in standard set-builder notation.
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- Term 1: $\frac{1}{2} = \frac{1}{1^2 + 1}$
- Term 2: $\frac{2}{5} = \frac{2}{2^2 + 1}$
- Term 3: $\frac{3}{10} = \frac{3}{3^2 + 1}$
- Term 4: $\frac{4}{17} = \frac{4}{4^2 + 1}$
- Term 5: $\frac{5}{26} = \frac{5}{5^2 + 1}$
Let $X = \{x \in \mathbb{R} : x^2 - 5x + 6 = 0\}$ and $Y = \{x \in \mathbb{Z} : x^2 - 5x + 6 = 0\}$. Determine whether these sets are identical or different under their respective domain rules.
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- For Set $X$: The constraint is $x \in \mathbb{R}$ (Real Numbers). Since both $2$ and $3$ are real numbers, $X = \{2, 3\}$.
- For Set $Y$: The constraint is $x \in \mathbb{Z}$ (Integers). Since both $2$ and $3$ are integers, $Y = \{2, 3\}$.
Types of Sets, Subsets and CardinalityTopic 2
Sets are structurally categorized based on their element counts. The Empty Set ($\emptyset$ or $\{\}$) contains $0$ elements. A Singleton Set contains exactly $1$ element. Finite Sets have a countable cardinality $n \in \mathbb{N}_0$, while Infinite Sets cannot be fully counted. Set $A$ is a Subset of $B$ ($A \subseteq B$) if every element in $A$ also belongs to $B$. If $A \subseteq B$ and $A \neq B$, then $A$ is a Proper Subset ($A \subset B$). A common mistake in advanced questions is confusing element membership ($\in$) with subset containment ($\subseteq$), especially when sets contain other sets as elements (e.g., if $A = \{1, \{2\}\}$, then $\{2\} \in A$ but $\{2\} \not\subseteq A$).
Let $A = \{1, 2, \{3, 4\}, 5\}$. Determine whether the following structural statements are true or false: (a) $\{3, 4\} \subseteq A$, (b) $\{3, 4\} \in A$, and (c) $\{\{3, 4\}\} \subseteq A$.
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- Statement (a): $\{3, 4\} \subseteq A$. For this to be true, the components $3$ and $4$ must be individual elements of $A$. Looking at our list, neither $3$ nor $4$ is an independent element of $A$. Therefore, this statement is False.
- Statement (b): $\{3, 4\} \in A$. This statement asserts that the set object $\{3, 4\}$ is an explicit element of $A$. This matches our list perfectly. Therefore, this statement is True.
- Statement (c): $\{\{3, 4\}\} \subseteq A$. This subset statement is true if the element inside the outer braces, which is the object $\{3, 4\}$, belongs to $A$. Since $\{3, 4\} \in A$ is true from part (b), this containment statement is True.
Find the total number of subsets and proper subsets of the set $X = \{x \in \mathbb{Z} : x^2 - x - 2 = 0\}$.
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- Total number of subsets $= 2^n = 2^2 = 4$. (The subsets are $\emptyset$, $\{-1\}$, $\{2\}$, and $\{-1, 2\}$).
- Total number of proper subsets $= 2^n - 1 = 4 - 1 = 3$. (We exclude the set $X$ itself).
Identify the type of set represented by $S = \{x \in \mathbb{R} : x^2 - 4x + 5 = 0\}$.
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Let us solve the defining quadratic condition over the real numbers domain: \[ x^2 - 4x + 5 = 0 \] Check the discriminant to determine the nature of the roots: \[ D = b^2 - 4ac = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0 \] Since the discriminant is negative, this equation has no real roots. It only produces complex conjugate solutions. Because the set-builder notation explicitly restricts the domain to real numbers ($x \in \mathbb{R}$), no values satisfy the condition. Therefore, the set contains zero elements. \[ S = \emptyset \] This is an Empty (Null) Set. Final Answer: Empty Set ($\emptyset$).
The Power Set and Its CardinalityTopic 3
The Power Set of a set $A$, denoted as $\mathcal{P}(A)$ or $2^A$, is the collection of all possible subsets of $A$. Mathematically, $\mathcal{P}(A) = \{X : X \subseteq A\}$. Crucially, both the empty set $\emptyset$ and the set $A$ itself are always elements of $\mathcal{P}(A)$. If the cardinality of a finite set $A$ is $|A| = n$, then the cardinality of its power set is given by the formula $|\mathcal{P}(A)| = 2^n$. A common mistake in multi-tier power set problems (e.g., finding $|\mathcal{P}(\mathcal{P}(\mathcal{P}(\emptyset)))|$) is miscounting how the layers scale exponentially.
Given the finite set $A = \{a, b\}$, write out the full power set $\mathcal{P}(A)$ explicitly in roster form.
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- The subset containing zero elements (null set): $\emptyset$
- Subsets containing exactly one element (singletons): $\{a\}$ and $\{b\}$
- The subset containing both elements (the set itself): $\{a, b\}$
Find the total number of elements inside the multi-tier nested power set expression $|\mathcal{P}(\mathcal{P}(\mathcal{P}(\emptyset)))|$.
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- Step 1: Let the base set be $A_0 = \emptyset$. The cardinality of the empty set is $|A_0| = 0$.
- Step 2: Find the cardinality of the first power set layer, $A_1 = \mathcal{P}(A_0)$: \[ |A_1| = 2^{|A_0|} = 2^0 = 1 \] (The set is $A_1 = \{\emptyset\}$).
- Step 3: Find the cardinality of the second power set layer, $A_2 = \mathcal{P}(A_1)$: \[ |A_2| = 2^{|A_1|} = 2^1 = 2 \] (The set is $A_2 = \{\emptyset, \{\emptyset\}\}$).
- Step 4: Find the cardinality of the final outermost power set layer, $A_3 = \mathcal{P}(A_2)$: \[ |A_3| = 2^{|A_2|} = 2^2 = 4 \]
If $X$ is a set such that $\mathcal{P}(X)$ has exactly $64$ elements, find the number of elements inside set $X$.
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Let the cardinality of set $X$ be $|X| = n$. According to the power set cardinality theorem, the total number of elements inside $\mathcal{P}(X)$ is given by $2^n$. Set up the equation using the given value: \[ 2^n = 64 \] Express $64$ as a base-2 exponential power: \[ 64 = 2^6 \implies 2^n = 2^6 \] Equating the exponents yields: \[ n = 6 \] Thus, set $X$ contains exactly $6$ elements. Final Answer: $6$.
Set Operations & Algebraic Laws
Primary Set Operations & Venn DiagramsTopic 1
Set operations allow us to combine or modify collections of objects. The Union ($A \cup B$) combines all elements belonging to $A$, $B$, or both. The Intersection ($A \cap B$) extracts only the shared elements that belong to both sets simultaneously. The Difference ($A - B$) removes all elements of $B$ from $A$. The Symmetric Difference ($A \triangle B = (A-B) \cup (B-A)$) captures elements that belong to exactly one of the sets, but not both. The Complement ($A'$) contains all elements in the Universal Set ($U$) that do not belong to $A$. Venn diagrams provide an excellent way to visualize these properties geometrically.
Given the universal set $U = \{1, 2, 3, \dots, 10\}$ and two subsets $A = \{1, 3, 5, 7, 9\}$, $B = \{2, 3, 5, 7\}$, evaluate the symmetric difference set $A \triangle B$.
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- Find $A - B$: Remove all elements of $B$ from $A$. The elements $3, 5, 7$ are shared, so removing them leaves: \[ A - B = \{1, 9\} \]
- Find $B - A$: Remove all elements of $A$ from $B$: \[ B - A = \{2\} \]
If $X$ and $Y$ are two sets such that $X \cap Y = X$, prove that $X \subseteq Y$ using element containment definitions.
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Let us use deductive element analysis to prove this property: To prove that $X$ is a subset of $Y$ ($X \subseteq Y$), we must show that any arbitrary element $x$ that belongs to $X$ must also belong to $Y$. \[ \text{Let } x \in X \] The problem provides the identity equation $X = X \cap Y$. Substitute this into our statement: \[ \text{Since } X = X \cap Y, \text{ then } x \in (X \cap Y) \] By the definition of a set intersection, an element belongs to $X \cap Y$ if and only if it belongs to both sets simultaneously: \[ x \in X \quad \text{and} \quad x \in Y \] This directly implies that $x \in Y$. Since choosing any arbitrary element $x \in X$ logically proves that $x \in Y$, then $X \subseteq Y$. Final Answer: Proved.
Let the universal set be the set of real numbers $\mathbb{R}$. If $A = (2, 5]$ and $B = [4, 7)$ are intervals, find the set difference $A - B$.
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Intervals are continuous sets of real numbers. Let us look at the overlapping sections of the two intervals: \[ A = \{x \in \mathbb{R} : 2 < x \le 5\} \] \[ B = \{x \in \mathbb{R} : 4 \le x < 7\} \] The operation $A - B$ removes all real numbers that belong to $B$ from the interval $A$. The values shared between the two sets lie in the interval $[4, 5]$. Removing these numbers from $A$ leaves the remaining section of the interval: The values start immediately after $2$ and stop just before $4$. Since $4 \in B$, it is removed from $A$, so the upper bound at $4$ becomes an open boundary. \[ A - B = (2, 4) \] Final Answer: $A - B = (2, 4)$.
Laws of Algebra of Sets & De Morgan's DualsTopic 2
Set operations follow algebraic laws that are structurally identical to propositional logic rules. These include Idempotent Laws ($A \cup A = A$), Identity Laws ($A \cap U = A$), and Distributive Laws, which allow an operation to distribute over another: \[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \] De Morgan's Laws describe how a complement distributes over a union or intersection, flipping the operation in the process: \[ (A \cup B)' = A' \cap B' \quad \text{and} \quad (A \cap B)' = A' \cup B' \] In the JEE Advanced exam, a common pitfall is dropping parenthetical brackets when applying distributive or complement laws, which leads to algebraic errors.
Simplify the complex set expression: $E = (A \cap B) \cup (A \cap B')$.
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We can simplify this expression efficiently by applying the Distributive Law in reverse. Notice that the term $A \cap$ is shared by both parts of the union: \[ (A \cap B) \cup (A \cap B') = A \cap (B \cup B') \] Now analyze the term inside the parentheses using the Complementation Law. The union of any set with its own complement always forms the complete Universal Set ($U$): \[ B \cup B' = U \] Substitute this back into our expression: \[ E = A \cap U \] Finally, apply the Identity Law. The intersection of any set with the universal set returns the original set unchanged: \[ A \cap U = A \] Final Answer: $A$.
Prove De Morgan's Law $(A \cup B)' = A' \cap B'$ using element-wise logical steps.
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To prove that these two expressions are identical, we must show that an arbitrary element $x$ belonging to the left-hand side logically implies it belongs to the right-hand side: \[ \text{Let } x \in (A \cup B)' \] By the definition of a set complement, if an element belongs to the complement of a set, it does not belong to the set itself: \[ x \notin (A \cup B) \] Using the logical definition of a set union, an element belongs to $A \cup B$ if it belongs to $A$ OR it belongs to $B$. For an element to *not* belong to the union, it must not belong to $A$ AND it must not belong to $B$: \[ x \notin A \quad \text{and} \quad x \notin B \] Convert these individual non-membership statements into membership statements using complements: \[ x \in A' \quad \text{and} \quad x \in B' \] By the definition of a set intersection, an element that belongs to both $A'$ and $B'$ at the same time belongs to their intersection: \[ x \in (A' \cap B') \] Since $x \in (A \cup B)' \implies x \in (A' \cap B')$, the identity is proven. Final Answer: Proved.
Simplify the expression $X = [A \cup (A \cap B)] \cap C$.
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Let us first look at the expression inside the square brackets: $A \cup (A \cap B)$.
According to the Absorption Law in set algebra, a set unioned with its intersection with another set simply absorbs the intersection, returning the original set:
\[ A \cup (A \cap B) = A \]
Let us verify this visually: $A \cap B$ is a subset of $A$. Therefore, unioning it with $A$ adds no new elements, leaving the set $A$ unchanged.
Substitute this simplified result back into the main expression:
\[ X = A \cap C \]
Final Answer: $A \cap C$.
Cardinality & Inclusion-Exclusion
Cardinality Formulas & Inclusion-Exclusion PrincipleTopic 1
Cardinality formulas allow us to count the number of elements in unions and intersections of sets. For two sets, the formula is $|A \cup B| = |A| + |B| - |A \cap B|$, where we subtract the intersection to avoid double-counting the shared elements. For three sets, we use the Principle of Inclusion-Exclusion (PIE): \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C| \] This principle scales to any finite number of sets by alternating between adding single sets, subtracting pairs, adding triples, and so on.
In a class of $60$ students, $35$ survey inputs state they like Mathematics, $20$ state they like Physics, and $12$ state they like both subjects. Find the number of students who like neither Mathematics nor Physics.
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Let $M$ represent the set of students who like Mathematics, and $P$ represent the set of students who like Physics. Identify the given values from the problem statement:
The total number of students is the universal set size: $|U| = 60$.
The individual category counts are: $|M| = 35$ and $|P| = 20$.
The number of students who like both subjects is the intersection size: $|M \cap P| = 12$.
Step 1: Calculate the number of students who like at least one of the two subjects ($|M \cup P|$) using the two-set cardinality formula:
\[ |M \cup P| = |M| + |B| - |M \cap P| \]
Substitute the given values into the equation:
\[ |M \cup P| = 35 + 20 - 12 = 55 - 12 = 43 \]
Step 2: Find the number of students who like neither subject. This matches the complement of the union, $(M \cup P)'$, which can be found by subtracting the union size from the total universal set size:
\[ \text{Neither} = |U| - |M \cup P| = 60 - 43 = 17 \]
Final Answer: 17 students.
Develop an explicit formula to find the number of elements that belong to exactly one of the two sets $A$ or $B$.
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The region corresponding to elements that belong to exactly one of the two sets is the symmetric difference, $A \triangle B = (A - B) \cup (B - A)$. Let us find its cardinality formula using standard unions and intersections.
From a Venn diagram, the total union $A \cup B$ is made of three disjoint regions: elements in $A$ only ($A-B$), elements in $B$ only ($B-A$), and elements in both ($A \cap B$).
\[ |A \cup B| = |A - B| + |B - A| + |A \cap B| \]
The symmetric difference excludes the shared intersection region:
\[ |A \triangle B| = |A - B| + |B - A| \]
Isolate this sum by substituting it into the first equation:
\[ |A \cup B| = |A \triangle B| + |A \cap B| \implies |A \triangle B| = |A \cup B| - |A \cap B| \]
Now expand the union term $|A \cup B|$ using the standard two-set formula $|A| + |B| - |A \cap B|$:
\[ |A \triangle B| = (|A| + |B| - |A \cap B|) - |A \cap B| = |A| + |B| - 2|A \cap B| \]
This is the required formula.
Final Answer: $|A| + |B| - 2|A \cap B|$.
A survey of $100$ students tracks language preferences: $50$ study Hindi ($H$), $40$ study English ($E$), and $30$ study Sanskrit ($S$). The intersections are logged as: $|H \cap E| = 15$, $|E \cap S| = 12$, and $|H \cap S| = 10$. If $5$ students study all three languages, find how many students study at least one language.
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To find the number of students who study at least one language, we must calculate the cardinality of the full union $|H \cup E \cup S|$. Apply the three-set Principle of Inclusion-Exclusion (PIE): \[ |H \cup E \cup S| = |H| + |E| + |S| - |H \cap E| - |E \cap S| - |H \cap S| + |H \cap E \cap S| \] Substitute the given values directly into the formula: \[ |H \cup E \cup S| = 50 + 40 + 30 - 15 - 12 - 10 + 5 \] Calculate the intermediate sums and differences systematically from left to right: \[ \text{Sum of singles} = 50 + 40 + 30 = 120 \] \[ \text{Subtract pairs} = 120 - 15 - 12 - 10 = 83 \] \[ \text{Add triple} = 83 + 5 = 88 \] Thus, exactly $88$ students study at least one language. Final Answer: 88 students.
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