Operations on Sets and Complement of a Set

Operations on Sets — Arithmetic for Collections

Just as addition and multiplication take two numbers and produce a number, set operations take two sets and produce a set. There are four to master: union, intersection, difference and complement. From here on, all our sets are subsets of some universal set U.

Union: $A \cup B$

Key Point (Definition): The union of A and B is the set of all elements which are in A or in B (including those in both): $A \cup B = \{x : x \in A \text{ or } x \in B\}$.

Example: A = {2, 4, 6, 8} and B = {6, 8, 10, 12} give $A \cup B$ = {2, 4, 6, 8, 10, 12} — the common elements 6 and 8 are taken only once.

The "or" here is the inclusive or of mathematics: x may be in A, in B, or in both.

If $B \subset A$, then $A \cup B = A$ — pouring a subset back into its parent adds nothing.

Properties of union:

• $A \cup B = B \cup A$ (commutative law)
• $(A \cup B) \cup C = A \cup (B \cup C)$ (associative law)
• $A \cup \phi = A$ (identity law — $\phi$ is the identity of $\cup$)
• $A \cup A = A$ (idempotent law)
• $U \cup A = U$ (law of U)

Intersection and Disjoint Sets

Key Point (Definition): The intersection of A and B is the set of all elements common to both: $A \cap B = \{x : x \in A \text{ and } x \in B\}$.

Example: with A = {2, 4, 6, 8} and B = {6, 8, 10, 12}, $A \cap B$ = {6, 8}.

If $B \subset A$, then $A \cap B = B$ — intersecting with a parent returns the subset.

Disjoint sets

If $A \cap B = \phi$ — no common element at all — then A and B are called disjoint sets. Example: {2, 4, 6, 8} and {1, 3, 5, 7} are disjoint. In a Venn diagram, disjoint sets are drawn as non-overlapping circles.

Properties of intersection:

• $A \cap B = B \cap A$ (commutative law)
• $(A \cap B) \cap C = A \cap (B \cap C)$ (associative law)
• $\phi \cap A = \phi$, $U \cap A = A$ (laws of $\phi$ and U)
• $A \cap A = A$ (idempotent law)
• $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ (distributive law — $\cap$ distributes over $\cup$)

The dual distributive law also holds: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

[JEE Tip] The distributive laws work exactly like $a(b + c) = ab + ac$ in algebra — but in sets, BOTH distributions are valid, which has no arithmetic analogue ($a + bc \neq (a+b)(a+c)$ for numbers!). Verifying either law by shading two Venn diagrams is a classic 3-mark question.

Difference of Sets

Key Point (Definition): The difference $A - B$ (read "A minus B") is the set of elements which belong to A but not to B: $A - B = \{x : x \in A \text{ and } x \notin B\}$.

Example: A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}:

• $A - B$ = {1, 3, 5} — in A, not in B.
• $B - A$ = {8} — in B, not in A.

Notice immediately: $A - B \neq B - A$. The difference of sets is not commutative — order matters, exactly as with subtraction of numbers.

The picture shows something worth memorising: the three sets $A - B$, $A \cap B$ and $B - A$ are mutually disjoint, and together they tile the whole of $A \cup B$:

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

Also useful: $A = (A - B) \cup (A \cap B)$ — a set splits cleanly into "private part" and "shared part".

A famous instance: $\mathbb{R} - \mathbb{Q}$ = the set of irrational numbers T. The difference operation defines the irrationals.

[Board Important] When computing $A - B$ in roster form, walk through A's elements one at a time and strike out those appearing in B. Do not touch B's extra elements — they are irrelevant to $A - B$.

Complement of a Set

Fix a universal set U. For a subset A of U:

Key Point (Definition): The complement of A is the set of all elements of U which are not elements of A: $A' = \{x : x \in U \text{ and } x \notin A\}$. Clearly $A' = U - A$.

Example: U = {1, 2, …, 10}, A = {1, 3, 5, 7, 9} gives A' = {2, 4, 6, 8, 10}.

Example: U = all students of Class XI, A = the girls; then A' = the boys.

Taking complement twice returns the original set: $(A')' = A$ — flipping membership two times restores it.

The laws of complement

1. Complement laws: $A \cup A' = U$ and $A \cap A' = \phi$ — a set and its complement tile U with no overlap.
2. De Morgan's laws: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$.
3. Double complementation: $(A')' = A$.
4. Laws of $\phi$ and U: $\phi' = U$ and $U' = \phi$.

De Morgan in words: the complement of a union is the intersection of the complements; the complement of an intersection is the union of the complements. Priming flips $\cup$ and $\cap$.

Quick verification with U = {1, 2, 3, 4, 5, 6}, A = {2, 3}, B = {3, 4, 5}: $A \cup B$ = {2, 3, 4, 5}, so $(A \cup B)'$ = {1, 6}. Separately A' = {1, 4, 5, 6} and B' = {1, 2, 6}, whose intersection is {1, 6}. Match! ✓

[JEE Tip] De Morgan's laws are the single most examined fact of this chapter in MCQs. An everyday translation helps: "NOT (rich or famous)" means "not rich AND not famous". The negation of an or is an and of negations.

Solved Examples

Example 1. Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find $A \cup B$.

Solution.

Step 1 — pool every element of both sets. From A: 2, 4, 6, 8. From B: 6, 8, 10, 12.

Step 2 — drop duplicates. 6 and 8 appear in both lists; a set records each element once.

Step 3 — write the union. $A \cup B = \{2, 4, 6, 8, 10, 12\}$.

Step 4 — sanity check. $n(A \cup B) = 6 = 4 + 4 - 2$: total minus the 2 shared elements — exactly the counting rule $n(A) + n(B) - n(A \cap B)$.

Takeaway: union merges lists; shared elements appear once, never twice.

Example 2. Let A = {a, e, i, o, u} and B = {a, i, u}. Show that $A \cup B = A$.

Solution.

Step 1 — observe the subset relation. Every element of B (a, i, u) already lies in A, so $B \subset A$.

Step 2 — compute the union. Pooling contributes nothing new from B: $A \cup B = \{a, e, i, o, u\} = A$.

Step 3 — extract the general rule. Whenever $B \subset A$, we get $A \cup B = A$ — pouring a subset back into its parent changes nothing.

Takeaway: with a subset, union just picks the bigger set — a rule you may quote directly in exams.

Example 3. X = {Ram, Geeta, Akbar} is the hockey team and Y = {Geeta, David, Ashok} the football team. Find $X \cup Y$ and $X \cap Y$, and interpret.

Solution.

Step 1 — the union. Pool both teams, listing Geeta once: $X \cup Y$ = {Ram, Geeta, Akbar, David, Ashok}. Interpretation: students who play hockey OR football (or both).

Step 2 — the intersection. Scan for names on BOTH lists: only Geeta. $X \cap Y$ = {Geeta}. Interpretation: students who play both games.

Step 3 — read the sizes. $5 = 3 + 3 - 1$: the one double-counted player (Geeta) is subtracted once ✓.

Takeaway: in real-world language, union = "atleast one of the two", intersection = "both".

Example 4. Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}. Find $A \cap B$ and show that $A \cap B = B$.

Solution.

Step 1 — find the common elements. Check B's elements against A: 2 ✓, 3 ✓, 5 ✓, 7 ✓ — every element of B lies in A.

Step 2 — write the intersection. $A \cap B = \{2, 3, 5, 7\}$, which is exactly B.

Step 3 — extract the general rule. Since $B \subset A$, intersecting with the parent returns the subset: $A \cap B = B$.

Takeaway: the companion of Example 2 — with a subset, union picks the bigger set and intersection picks the smaller.

Example 5. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}, find: (i) $A \cap B$ (ii) $B \cap C$ (iii) $A \cap C \cap D$ (iv) $A \cap (B \cup C)$ (v) $(A \cup D) \cap (B \cup C)$.

Solution.

Step 1 — work strictly inside-out: brackets first.

(i) Step 2. Common to A and B: 7, 9, 11 → $A \cap B = \{7, 9, 11\}$.

(ii) Step 3. Common to B and C: 11, 13 → $B \cap C = \{11, 13\}$.

(iii) Step 4. First $A \cap C = \{11\}$; then intersect with D = {15, 17}: 11 is absent → $A \cap C \cap D = \phi$.

(iv) Step 5. Inner bracket: $B \cup C = \{7, 9, 11, 13, 15\}$. Intersect with A: keep 7, 9, 11 → $\{7, 9, 11\}$.

(v) Step 6. $A \cup D = \{3, 5, 7, 9, 11, 15, 17\}$ and $B \cup C = \{7, 9, 11, 13, 15\}$; the common elements are 7, 9, 11, 15 → $\{7, 9, 11, 15\}$.

Step 7 — check (iv) against the distributive law. $(A \cap B) \cup (A \cap C) = \{7, 9, 11\} \cup \{11\} = \{7, 9, 11\}$ ✓ — matches the bracket-first computation.

Takeaway: never distribute mentally in one go — compute the inner bracket, then apply the outer operation.

Example 6. With A = naturals, B = even naturals, C = odd naturals, D = primes, find $A \cap B$, $A \cap C$, $A \cap D$, $B \cap C$, $B \cap D$, $C \cap D$.

Solution.

Step 1 — intersections with the parent set. B, C, D are all subsets of A, so intersecting with A returns each unchanged: $A \cap B = B$, $A \cap C = C$, $A \cap D = D$.

Step 2 — $B \cap C$. A number even AND odd would leave remainders 0 and 1 on division by 2 simultaneously — impossible: $B \cap C = \phi$.

Step 3 — $B \cap D$. An even prime is divisible by 2, and primality then forces it to BE 2: $B \cap D = \{2\}$.

Step 4 — $C \cap D$. The odd primes: every prime except 2, i.e. $\{x : x \text{ is prime and } x \neq 2\} = \{3, 5, 7, 11, \ldots\}$.

Takeaway: $B \cap D = \{2\}$ — the lone even prime — is the most quoted intersection in mathematics; it decides dozens of empty-set questions.

Example 7. Let V = {a, e, i, o, u} and B = {a, i, k, u}. Find V − B and B − V.

Solution.

Step 1 — V − B: walk through V, striking anything found in B. a ∈ B strike; e stays; i ∈ B strike; o stays; u ∈ B strike. Result: $V - B = \{e, o\}$.

Step 2 — B − V: walk through B, striking anything found in V. a strike; i strike; k stays; u strike. Result: $B - V = \{k\}$.

Step 3 — compare. $\{e, o\} \neq \{k\}$: the two differences are different sets, even of different sizes.

Takeaway: difference is NOT commutative — each difference is computed from its own left-hand set, and the shared elements (a, i, u) vanish from both answers.

Example 8. If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, find (i) A − B (ii) B − C (iii) C − A (iv) B − A.

Solution.

Step 1 — identify each pair's overlap first (the elements that will be struck). $A \cap B = \{12\}$; $B \cap C = \{4, 8, 12, 16\}$; $C \cap A = \{6, 12\}$.

(i) Step 2. Strike 12 from A: $A - B = \{3, 6, 9, 15, 18, 21\}$.

(ii) Step 3. Strike 4, 8, 12, 16 from B: $B - C = \{20\}$.

(iii) Step 4. Strike 6, 12 from C: $C - A = \{2, 4, 8, 10, 14, 16\}$.

(iv) Step 5. Strike 12 from B: $B - A = \{4, 8, 16, 20\}$.

Step 6 — observe. The overlap $A \cap B = \{12\}$ appears in NEITHER $A - B$ nor $B - A$ — shared elements never survive a difference.

Takeaway: compute the intersection first; the difference is then a clean strike-out with no second-guessing.

Example 9. If $\mathbb{R}$ is the set of real numbers and $\mathbb{Q}$ the set of rational numbers, what is $\mathbb{R} - \mathbb{Q}$?

Solution.

Step 1 — apply the definition of difference. $\mathbb{R} - \mathbb{Q} = \{x : x \in \mathbb{R} \text{ and } x \notin \mathbb{Q}\}$: the reals that are not rational.

Step 2 — recognise the set. Real-but-not-rational is precisely the definition of an IRRATIONAL number, so $\mathbb{R} - \mathbb{Q} = T$, the set of irrationals — home of $\sqrt{2}$, $\sqrt{5}$, $\pi$, e.

Takeaway: the irrationals are not defined by a property of their own — they are literally a set difference. A one-line question that appears year after year.

Example 10. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find (i) A' (ii) B' (iii) $(A \cup C)'$ (iv) $(A \cup B)'$ (v) (A')' (vi) $(B - C)'$.

Solution.

Step 1 — the method: every complement is U minus the set. Compute the inner set first, then strike its elements from U.

(i) Step 2. $A' = U - A = \{5, 6, 7, 8, 9\}$.

(ii) Step 3. $B' = U - B = \{1, 3, 5, 7, 9\}$.

(iii) Step 4. Inner set: $A \cup C = \{1, 2, 3, 4, 5, 6\}$. Complement: $\{7, 8, 9\}$.

(iv) Step 5. Inner set: $A \cup B = \{1, 2, 3, 4, 6, 8\}$. Complement: $\{5, 7, 9\}$.

(v) Step 6. $(A')' = U - \{5, 6, 7, 8, 9\} = \{1, 2, 3, 4\} = A$ ✓ — double complementation restores the set.

(vi) Step 7. Inner set: $B - C = \{2, 8\}$ (strike 4, 6 from B). Complement: $\{1, 3, 4, 5, 6, 7, 9\}$.

Takeaway: always resolve the inner operation to an explicit roster before priming — complements of unevaluated expressions invite errors.

Example 11. Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Verify that $(A \cup B)' = A' \cap B'$.

Solution.

Step 1 — compute the left side. $A \cup B = \{2, 3, 4, 5\}$, so $(A \cup B)' = U - \{2, 3, 4, 5\} = \{1, 6\}$.

Step 2 — compute the right side independently. $A' = \{1, 4, 5, 6\}$ and $B' = \{1, 2, 6\}$. Their common elements: 1 and 6 → $A' \cap B' = \{1, 6\}$.

Step 3 — compare. Both sides equal $\{1, 6\}$ ✓ — De Morgan's first law verified for this data.

Takeaway: the verification template is always: compute each side separately, then match. The same three steps verify the twin law $(A \cap B)' = A' \cup B'$.

Example 12. Taking $\mathbb{N}$ as the universal set, write the complements of: (i) $\{x : x \text{ is an even natural number}\}$ (ii) $\{x : x \in \mathbb{N} \text{ and } 2x + 1 > 10\}$ (iii) $\{x : x \text{ is a perfect square}\}$.

Solution.

Step 1 — the method for infinite universes. Listing is impossible; instead NEGATE the defining condition.

(i) Step 2. Not even (within $\mathbb{N}$) = odd: complement $= \{x : x \text{ is an odd natural number}\}$.

(ii) Step 3. Solve the inequality first: $2x + 1 > 10 \Rightarrow x > \frac{9}{2}$, so the set is {5, 6, 7, …}. Negate: $x \leq \frac{9}{2}$ within $\mathbb{N}$ gives $\{1, 2, 3, 4\}$ — here the complement happens to be listable.

(iii) Step 4. Complement $= \{x : x \in \mathbb{N} \text{ and } x \text{ is not a perfect square}\} = \{2, 3, 5, 6, 7, 8, 10, \ldots\}$.

Takeaway: in an infinite universe, the complement is described by negating the condition — solve inequalities BEFORE negating to avoid sign slips.

Example 13. Using properties of sets, show that $A \cup (A \cap B) = A$.

Solution.

Step 1 — plan a two-way containment. Set equality is proved by showing each side contains the other.

Step 2 — easy direction. $A \subset A \cup (A \cap B)$: any set sits inside its union with anything.

Step 3 — other direction. The union has two parts: A itself, and $A \cap B$. But $A \cap B \subset A$, so BOTH parts lie inside A, hence $A \cup (A \cap B) \subset A$.

Step 4 — conclude. Containment both ways gives $A \cup (A \cap B) = A$. ∎ The twin identity $A \cap (A \cup B) = A$ is proved the same way.

Takeaway: these are the absorption laws — a set absorbs its union or intersection with anything built from it, and the proof is pure containment-chasing.

Example 14. Show that $A \cup B = A \cap B$ implies A = B.

Solution.

Step 1 — recall the squeeze. For any sets, $A \cap B \subset A \subset A \cup B$ — every set sits between the intersection and the union.

Step 2 — show $A \subset B$. Take $a \in A$. Then $a \in A \cup B$. Using the hypothesis $A \cup B = A \cap B$: $a \in A \cap B$, and in particular $a \in B$.

Step 3 — show $B \subset A$. Symmetric: $b \in B \Rightarrow b \in A \cup B = A \cap B \Rightarrow b \in A$.

Step 4 — conclude. Both containments hold, so A = B. ∎

Takeaway: when the two ends of the squeeze $A \cap B \subset A, B \subset A \cup B$ coincide, everything trapped between them collapses to equality.

Example 15. Show that for any sets A and B: $A = (A \cap B) \cup (A - B)$.

Solution.

Step 1 — show every element of A lands in the right side. Take $x \in A$. Exactly one of two cases holds: $x \in B$, putting $x \in A \cap B$; or $x \notin B$, putting $x \in A - B$. Either way $x$ belongs to the union: $A \subset (A \cap B) \cup (A - B)$.

Step 2 — show the right side stays inside A. $A \cap B \subset A$ ✓ and $A - B \subset A$ ✓, so their union is contained in A.

Step 3 — conclude. Equality holds: A splits as shared part $\cup$ private part, and the two parts are disjoint. ∎

Takeaway: "shared plus private rebuilds the set" — this disjoint split is what powers the counting rule $n(A) = n(A \cap B) + n(A - B)$ used heavily in the JEE Corner.