Summary and Quick Revision

The Chapter in One Idea

Everything in this chapter is one statement wearing different costumes:

A set is a well-defined collection — membership is a pure yes/no question — and every concept here (subset, equality, union, intersection, difference, complement) is defined by asking that yes/no question in a particular pattern.

From this: representations (Section 1), types of sets (Section 2), subsets and intervals (Section 3), and the four operations with De Morgan's laws (Section 4).

Exam weight at a glance: JEE Main — one question from the Sets-Relations-Functions unit in nearly every session; the sets favourites are power-set counting, laws of set algebra, and cardinality/survey problems (all in the JEE Corner). This chapter is also the language of Relations and Functions (Chapter 2) and Probability (Chapter 14) — fluency here is not optional.

Representations and Types — Formula Card

• Set: a well-defined collection. Membership: $a \in A$, $b \notin A$. Opinion words (best, most talented) disqualify a collection.
• Roster form: {2, 4, 6} — order immaterial, no repeats. Set-builder form: $\{x : x \text{ has property P}\}$ — colon reads "such that".
• Standard sets: $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$; irrationals $T = \mathbb{R} - \mathbb{Q}$ (contains $\sqrt{2}$, $\pi$).
• Empty set $\phi$ = { }: no elements, $n(\phi) = 0$, finite, subset of every set. Distinguish: $\phi \neq \{0\} \neq \{\phi\}$ (sizes 0, 1, 1).
• Finite/infinite: finite = empty or a definite count. Careful cases: $\phi$ finite; any interval of reals infinite; the set of animals on earth finite.
• Equal sets: exactly the same elements. Roster-convert, ignore order/repeats, compare. Watch for the extra negative root: $\{x : x - 5 = 0\} = \{5\}$ but $\{x : x^2 = 25\} = \{-5, 5\}$.

Key Point: Domain filters decide everything: $\{x \in \mathbb{N} : x^2 = 4\} = \{2\}$, over $\mathbb{Z}$ it is {$-2$, 2}, and $\{x \in \mathbb{N} : 2x = 1\} = \phi$.

Subsets and Intervals — Formula Card

• Subset: $A \subset B$ iff $a \in A \Rightarrow a \in B$. Always: $A \subset A$ and $\phi \subset A$. Equality via double inclusion: $A \subset B$ and $B \subset A \iff A = B$.
• Proper subset: $A \subset B$, $A \neq B$. Singleton: {a}.
• Transitivity: $A \subset B, B \subset C \Rightarrow A \subset C$. But $\in$ does NOT chain: $A \in B$, $B \subset C$ gives $A \in C$, not $A \subset C$.
• $\in$ vs $\subset$: element-to-set vs set-to-set. In A = {1, 2, {3, 4}, 5}: $\{3,4\} \in A$ but $\{3,4\} \not\subset A$; $\{\{3,4\}\} \subset A$.
• Intervals ($a < b$): $(a,b)$ open, $[a,b]$ closed, $[a,b)$ and $(a,b]$ semi-open; length $= b - a$ for all four. Infinity side always open: $[0, \infty)$, $(-\infty, \infty) = \mathbb{R}$.
• Translation: $<$ ↔ round bracket, $\leq$ ↔ square bracket. $x^2 \leq a^2 \iff x \in [-a, a]$.
• Universal set U: the fixed background set of the context; drawn as a rectangle in Venn diagrams, subsets as circles.

Four equivalent faces of one fact: $A \subset B \iff A - B = \phi \iff A \cup B = B \iff A \cap B = A$.

Operations — Formula Card

• Union: $A \cup B = \{x : x \in A \text{ or } x \in B\}$ (inclusive or). Intersection: $A \cap B = \{x : x \in A \text{ and } x \in B\}$. Disjoint: $A \cap B = \phi$.
• Difference: $A - B = \{x : x \in A, x \notin B\} = A \cap B'$. NOT commutative. $\mathbb{R} - \mathbb{Q} = T$.
• Complement: $A' = U - A$; $(A')' = A$; $A \cup A' = U$; $A \cap A' = \phi$; $\phi\,' = U$; $U' = \phi$.
• Laws: commutative, associative, idempotent ($A \cup A = A$), identity ($A \cup \phi = A$, $A \cap U = A$), distributive (both ways), absorption ($A \cup (A \cap B) = A = A \cap (A \cup B)$).
• De Morgan: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$ — priming swaps $\cup$ and $\cap$.
• Decompositions: $A = (A - B) \cup (A \cap B)$; $A \cup B = (A - B) \cup (A \cap B) \cup (B - A)$, the three pieces mutually disjoint; $A \cup (B - A) = A \cup B$.
• If $B \subset A$: $A \cup B = A$ and $A \cap B = B$ — union picks the bigger, intersection the smaller.

JEE Standard Results — Quick Card

Power set (JEE Main syllabus, beyond the rationalised textbook):

• $n(A) = m \Rightarrow n(P(A)) = 2^m$; proper subsets $2^m - 1$; non-empty proper $2^m - 2$.
• Subsets containing a fixed element: $2^{m-1}$; even-size (or odd-size) subsets: $2^{m-1}$.
• $X \in P(A) \iff X \subset A$. $P(\phi) = \{\phi\}$; $n(P(P(\phi))) = 2$.
• $2^m - 2^k = N$: write $N$ = (power of 2) × (odd); match $2^k$ and $2^{m-k} - 1$.

Cardinality (inclusion-exclusion):

• $n(A \cup B) = n(A) + n(B) - n(A \cap B)$; disjoint case drops the last term.
• $n(A - B) = n(A) - n(A \cap B)$; neither: $n(A' \cap B') = n(U) - n(A \cup B)$.
• Exactly one: $n(A \triangle B) = n(A) + n(B) - 2\,n(A \cap B)$.
• Three sets: $n(A \cup B \cup C) = \Sigma n(A) - \Sigma n(A \cap B) + n(A \cap B \cap C)$; fill the 8-region Venn from the centre outward.
• Bounds: $n(A) + n(B) - n(U) \leq n(A \cap B) \leq \min(n(A), n(B))$.

Symmetric difference: $A \triangle B = (A - B) \cup (B - A) = (A \cup B) - (A \cap B)$; commutative, associative, $A \triangle A = \phi$, $A \triangle \phi = A$, and $A \triangle B = A \triangle C \Rightarrow B = C$ (cancellation!).

Named JEE sets: $\{4^n - 3n - 1\} \subset \{9(n-1)\}$ and $\{8^n - 7n - 1\} \subset \{49(n-1)\}$ — via binomial expansion of $(1+3)^n$, $(1+7)^n$.

Last-Minute Mistake Checklist

Before the exam, scan this list — each item is a real mark lost by thousands of students every year:

1. Apply the domain filter after solving — $\{x \in \mathbb{N} : x^2 = 4\}$ is {2}, not {$-2$, 2}.
2. $\phi$, {0} and {$\phi$} are three different sets — sizes 0, 1, 1. The empty set is finite.
3. $\in$ relates element to set; $\subset$ relates set to set — {a} $\subset$ {a, b} but {a} $\notin$ {a, b}.
4. Packet elements stay sealed: in {1, {3, 4}}, the numbers 3 and 4 are NOT elements.
5. $\in$ does not chain with $\subset$ — only $\subset$ is transitive.
6. Brackets by inequality: $<$ round, $\leq$ square; the $\infty$ side is always round.
7. $A - B \neq B - A$ — compute each from its own left-hand set; rewrite $A - B = A \cap B'$ to simplify safely.
8. De Morgan swaps the operation — $(A \cup B)' = A' \cap B'$, never $A' \cup B'$.
9. No cancellation for $\cup$ or $\cap$ alone — $A \cap B = A \cap C$ does not give B = C (both equations together do).
10. In survey problems, handle "neither" first ($n(U) -$ neither = union), then inclusion-exclusion; check that all Venn regions sum to $n(U)$.

Done revising? Take the Section 7 mock drill under exam timing — that is the real test of readiness.