Sets

The chapter opens with the definition of a set and its two notations — roster form and set-builder form — together with the membership symbols $\in$ and $\notin$. It then builds a ladder of classifications: the empty set, finite versus infinite sets, and equal sets, each pinned down by a precise definition and tested against short examples. The middle portion turns to comparison: subsets and proper subsets, the special subsets of $\mathbb{R}$ (including $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$ and the irrationals $T$), intervals as subsets of the real line, and the power set of a set. A universal set $U$ and Venn diagrams then supply the picture used for everything that follows: the four operations on sets — union, intersection, difference and complement — each stated as a formal definition, each illustrated by a Venn diagram, and each obeying a family of algebraic laws (commutative, associative, distributive, idempotent, and De Morgan's laws for the complement). The chapter closes by turning these pictures into numbers, with the counting formula for $n(A \cup B)$ and its three-set extension, which is exactly the tool competitive exams love to test in "survey of students" problems.

Describing a set: roster form and set-builder form

A set is a well-defined collection of objects — "well-defined" meaning that for any given object, you can say definitely whether it belongs to the collection or not. Sets are conventionally named with capital letters ($A, B, X, \ldots$) and their elements with lowercase letters ($a, b, x, \ldots$). If $a$ is an element of set $A$ we write $a \in A$ (read "$a$ belongs to $A$"); if it is not, we write $a \notin A$. There are exactly two standard ways to describe a set.

where: in roster form every element is listed once, separated by commas, inside braces — order is immaterial and repetition changes nothing, so $\{1,2,3\}$ and $\{3,1,2,2\}$ describe the same set. In set-builder form, the colon reads "such that" and the braces read "the set of all", so $A = \{x : x \text{ is a natural number and } 3 < x < 10\}$ reads "the set of all $x$ such that $x$ is a natural number between $3$ and $10$", giving $A = \{4,5,6,7,8,9\}$ (this is Example 1 territory — NCERT's own Examples 1 to 5 build exactly this translation skill in both directions). Some sets — like $\{1,3,5,7,\ldots\}$, the odd naturals — use three dots to signal a pattern that continues indefinitely; not every infinite set can be listed this way, since a set like $\mathbb{R}$ follows no simple pattern at all.

Empty, finite and infinite sets

A set that contains no elements at all is the empty set (or null set, or void set), written $\phi$ or $\{\}$ — for instance $\{x : x \in \mathbb{N},\ 1 < x < 2\}$ is empty because no natural number lies strictly between $1$ and $2$ (NCERT Example 6 works through five such cases). If a set's number of elements $n(S)$ is a natural number (or zero, for $\phi$), the set is called finite; otherwise it is infinite. The set of natural numbers $\mathbb{N}$, for example, is infinite, while the set of days of the week is finite with $n(S) = 7$. Because you cannot list every element of an infinite set inside braces, infinite sets in roster form use the three-dot convention — $\{1,2,3,\ldots\}$ for $\mathbb{N}$, $\{\ldots,-2,-1,0,1,2,\ldots\}$ for $\mathbb{Z}$ — but a set like the real numbers, which follows no such listable pattern, cannot be written in roster form at all.

Equal sets

Two sets $A$ and $B$ are equal, written $A = B$, precisely when they contain exactly the same elements — every element of $A$ lies in $B$ and every element of $B$ lies in $A$. Order of listing and repeated elements never affect this: $\{1,2,3\}$ and $\{2,2,1,3,3\}$ are the same set, and the set of letters in "CATARACT" equals the set of letters in "TRACT" (both reduce to $\{C, A, T, R\}$, NCERT's Example 23). Equality is therefore a double-inclusion idea — $A = B$ exactly when $A \subset B$ and $B \subset A$ — a technique you will use repeatedly to prove two sets coincide, as in the first worked derivation below.

Subsets, intervals and the power set

$A$ is a subset of $B$, written $A \subset B$, if every element of $A$ is also an element of $B$ — formally, $A \subset B$ if $a \in A \implies a \in B$. Every set is a subset of itself ($A \subset A$), and the empty set is a subset of every set ($\phi \subset A$). When $A \subset B$ but $A \neq B$, $A$ is a proper subset of $B$, and $B$ is called a superset of $A$. A set with exactly one element is a singleton set. Among the number systems, $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$, while the irrationals $T = \{x : x \in \mathbb{R} \text{ and } x \notin \mathbb{Q}\}$ also satisfy $T \subset \mathbb{R}$, with $\mathbb{Q}$ and $T$ sharing no elements.

where: $a, b \in \mathbb{R}$ with $a < b$; a round bracket excludes that endpoint and a square bracket includes it, so $(a,b)$ is the open interval, $[a,b]$ is the closed interval, and $[a,b)$, $(a,b]$ are half-open (or half-closed) intervals. Every one of these is a subset of $\mathbb{R}$ containing infinitely many points, and its length is $b - a$ regardless of which endpoints are included. Special unbounded intervals extend the same notation: $[0,\infty)$ is the non-negative reals, $(-\infty,0)$ the negative reals, and $(-\infty,\infty)$ is all of $\mathbb{R}$.

Alongside subsets, the power set of $A$, written $P(A)$, is the set of all subsets of $A$ — every subset, including $\phi$ and $A$ itself, is one of its elements. For $A = \{1,2\}$, $P(A) = \{\phi, \{1\}, \{2\}, \{1,2\}\}$, which has $4 = 2^2$ elements. In general, if $n(A) = n$, then $n(P(A)) = 2^n$ — the first worked derivation below proves this precisely. Because the members of $P(A)$ are themselves sets, membership and subset-hood must be written with care: $\{1\} \in P(A)$ (it is one of the four listed elements) but $1 \in A$, not $1 \in P(A)$.

Universal sets and Venn diagrams

In a given context we usually work inside one basic set relevant to that context, called the universal set, denoted $U$; every set under discussion is then treated as a subset of $U$. While studying number systems, $U$ might be $\mathbb{R}$; in a human-population study, $U$ is the set of all people in the world. A Venn diagram, named after the logician John Venn, pictures $U$ as a rectangle and its subsets as circles inside it, with elements marked as points — an immediately readable way to see relationships such as $B \subset A$.

Union, intersection and difference of sets

The union of $A$ and $B$ collects every element that lies in $A$ or in $B$ (common elements counted only once); the intersection collects only the elements common to both; the difference $A - B$ collects elements of $A$ that are not in $B$.

where: "or" is used inclusively (an element in both sets still counts once in the union). If $A \cap B = \phi$, $A$ and $B$ are called disjoint — for example $\{2,4,6,8\}$ and $\{1,3,5,7\}$ share nothing. NCERT's own Examples 12 to 19 work these definitions on concrete sets; one useful special case worth remembering is that if $B \subset A$ then $A \cup B = A$ and $A \cap B = B$ (Examples 13 and 17). These three operations obey a family of algebraic laws that mirror ordinary arithmetic: union and intersection are each commutative ($A \cup B = B \cup A$, $A \cap B = B \cap A$) and associative ($(A \cup B) \cup C = A \cup (B \cup C)$, similarly for $\cap$), each is idempotent ($A \cup A = A$, $A \cap A = A$), $\phi$ is the identity for union ($A \cup \phi = A$) while $U$ is the identity for intersection ($A \cap U = A$), and intersection distributes over union — $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ — exactly as multiplication distributes over addition. Note carefully that difference is not commutative: $A - B \neq B - A$ in general (Examples 18 and 19 show this directly), and the three pieces $A - B$, $A \cap B$, $B - A$ are always mutually disjoint — a fact the third derivation below turns into the counting formula.

Complement of a set and De Morgan's laws

Once a universal set $U$ is fixed, the complement of $A \subset U$ is the set of everything in $U$ that is not in $A$.

where: $A'$ is itself a subset of $U$. Complementation satisfies the complement laws $A \cup A' = U$ and $A \cap A' = \phi$, the law of double complementation $(A')' = A$, and the boundary cases $\phi' = U$, $U' = \phi$. The single most important pair of results about complements are De Morgan's laws, named after Augustus De Morgan:

where: in words, the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements — "not (A or B)" means "not A, and not B", exactly as in ordinary logic. NCERT's Example 22 verifies both laws numerically on $U = \{1,\ldots,6\}$, $A = \{2,3\}$, $B = \{3,4,5\}$; the second derivation below proves the first law in full generality by the double-inclusion method introduced with equal sets.

Counting elements: the union formula for finite sets

For finite sets, the number of elements in a union is not simply $n(A) + n(B)$, because any element common to both sets would be counted twice. The correction is one of the most quoted formulas in this chapter, since it converts a Venn diagram directly into an exam answer.

where: the second formula, read in full, is $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$. When $A$ and $B$ are disjoint, $n(A \cap B) = 0$ and the formula reduces to plain addition, $n(A \cup B) = n(A) + n(B)$. This "add, then correct for double-counting" pattern — called the inclusion-exclusion principle — is exactly the tool behind every "how many students like both subjects" survey problem, worked out in full in the third derivation and the Competitive-level worked example below.

Board level - listing all subsets of a three element set and checking the power set count (Example 24 pattern)

List all the subsets of the set $A = \{-1, 0, 1\}$, and verify that the total count matches the power-set formula.

Step 1. Since $n(A) = 3$, the power-set formula from equation $(2)$ region predicts $n(P(A)) = 2^3 = 8$ subsets before any listing is done — a target to check the final list against.

Step 2. List subsets systematically by size, from smallest to largest. Size $0$: $\phi$. Size $1$: $\{-1\}, \{0\}, \{1\}$. Size $2$: $\{-1,0\}, \{-1,1\}, \{0,1\}$. Size $3$: $\{-1,0,1\}$, which is $A$ itself.

Step 3. Count the list: $1 + 3 + 3 + 1 = 8$ subsets in total, matching the prediction from Step 1 exactly.

Step 4. Note the pattern $1, 3, 3, 1$ is the row of Pascal's triangle for $n=3$, since choosing a subset of size $k$ from $3$ elements can be done in $\binom{3}{k}$ ways, and $\binom{3}{0}+\binom{3}{1}+\binom{3}{2}+\binom{3}{3} = 2^3$. Answer: P(A) = {φ, {−1}, {0}, {1}, {−1,0}, {−1,1}, {0,1}, {−1,0,1}}, eight subsets in total.

Board level - proving A union B equals A intersection B forces A equal to B (Example 25 pattern)

Show that $A \cup B = A \cap B$ implies $A = B$.

Step 1. To prove $A = B$, use the double-inclusion method: show $A \subset B$ first. Let $a$ be an arbitrary element of $A$. Since $A \subset A \cup B$ always (every element of a set is trivially in that set's union with anything), $a \in A \cup B$.

Step 2. By the given hypothesis, $A \cup B = A \cap B$, so $a \in A \cap B$ as well. By the definition of intersection, membership in $A \cap B$ requires membership in both sets, so in particular $a \in B$. Since $a$ was an arbitrary element of $A$, this shows every element of $A$ lies in $B$, i.e. $A \subset B$.

Step 3. Now show the reverse inclusion $B \subset A$ by the identical argument with the roles swapped. Let $b \in B$; then $b \in A \cup B$ (since $B \subset A \cup B$), so by the hypothesis $b \in A \cap B$, which by definition of intersection gives $b \in A$. Hence $B \subset A$.

Step 4. Combining Steps 2 and 3, $A \subset B$ and $B \subset A$, which is exactly the definition of set equality. Answer: A = B.

Competitive level (JEE / NEET) - a three-subject survey solved with the inclusion-exclusion formula

In a survey of $100$ students, $42$ like Mathematics, $40$ like Physics, $38$ like Chemistry, $20$ like both Mathematics and Physics, $18$ like both Mathematics and Chemistry, $15$ like both Physics and Chemistry, and $8$ like all three subjects. Find how many students like none of the three subjects, and how many like exactly one subject.

Step 1. Apply the three-set inclusion-exclusion formula from equation $(6)$: $n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(M \cap C) - n(P \cap C) + n(M \cap P \cap C)$.

Step 2. Substitute the given values: $n(M \cup P \cup C) = 42 + 40 + 38 - 20 - 18 - 15 + 8$. Summing the additions first, $42+40+38 = 120$; summing the pairwise subtractions, $20+18+15=53$; so $n(M \cup P \cup C) = 120 - 53 + 8 = 75$.

Step 3. Students liking none of the three subjects lie outside $M \cup P \cup C$ within the universal set of $100$ surveyed students, so that count is $100 - n(M \cup P \cup C) = 100 - 75 = 25$.

Step 4. For exactly one subject, subtract both pairwise overlaps from each single count but add back the triple overlap once (since it was subtracted twice by the two pairwise terms): exactly-Mathematics $= 42 - 20 - 18 + 8 = 12$; exactly-Physics $= 40 - 20 - 15 + 8 = 13$; exactly-Chemistry $= 38 - 18 - 15 + 8 = 13$. Adding these disjoint counts, $12+13+13 = 38$. Answer: 25 students like none of the three subjects, and 38 students like exactly one subject. Timing note: budget about three to four minutes. The efficient route is to compute $n(M \cup P \cup C)$ once with the full inclusion-exclusion formula and reuse it for the "none" count, rather than drawing and labelling all seven Venn-diagram regions by hand — reserve the region-by-region breakdown (as in Step 4) only for questions that explicitly ask for an "exactly one" or "exactly two" count.

The most common slip in this chapter is confusing $\in$ (element of) with $\subset$ (subset of): for $A = \{1, 2, \{3,4\}, 5\}$, the object $\{3,4\}$ is itself an element of $A$, so $\{3,4\} \in A$ is correct, while $\{3,4\} \subset A$ is false, since $3 \notin A$ and $4 \notin A$ individually — only the singleton $\{\{3,4\}\} \subset A$ is a valid subset statement. A closely related error treats an element and the singleton set containing it as interchangeable: $1 \in A$ is correct but $1 \subset A$ is meaningless, since an element can never be a subset of itself unless it happens to be a set already containing that element. Students also frequently forget that $\phi \subset A$ for every set $A$, including $\phi$ itself, yet confuse $\phi$ (a set with zero elements) with $\{\phi\}$ (a set with one element, namely $\phi$) — these are never equal. In roster form, forgetting that order and repetition are immaterial leads to wrongly calling equal sets "different" when written differently. With set operations, difference is the one operation that is not commutative — $A - B \neq B - A$ in general — and inclusion-exclusion is routinely botched by adding $n(A)+n(B)$ without subtracting the overlap, or, for three sets, forgetting to add back $n(A \cap B \cap C)$ after subtracting all three pairwise intersections. Finally, in interval notation, mixing up round and square brackets silently changes whether an endpoint counts, which matters enormously in inequalities and later in domains of functions.

• Translate a set fluently between roster form and set-builder form in both directions.
• Classify a given set as empty, finite or infinite, and justify the classification.
• Prove two sets are equal using the double-inclusion method: show each is a subset of the other.
• Distinguish $\in$ (element of) from $\subset$ (subset of), especially when a set contains another set as an element.
• Write and interpret all four interval types $(a,b)$, $[a,b]$, $[a,b)$, $(a,b]$ as subsets of $\mathbb{R}$, and convert between interval and set-builder notation.
• State the power-set formula $n(P(A)) = 2^n$ and justify it by the independent include/exclude argument or by induction.
• Compute $A \cup B$, $A \cap B$, $A - B$ and $A'$ for concrete sets, and identify disjoint sets.
• State and apply De Morgan's laws $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$, and prove either one from the definitions.
• Apply $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ and its three-set extension to solve survey-style counting problems, including "exactly one" and "none" counts.

Ask the tutor: Want to work through proving $(A \cap B)' = A' \cup B'$ yourself, step by step, using the same double-inclusion method?