NDA Maths · Sets & Relations

Set Fundamentals, Operations and Algebra

A set is a well-defined collection of distinct objects; the operations (union, intersection, complement, difference, symmetric difference) obey a fixed list of algebraic laws that the NDA tests by asking which identity is NOT correct.

All Sets & Relations notes NDA Maths strategy

Why this matters

Start here — everything else in the chapter is built on these operations. 23 PYQs live in this subtopic, and the dominant shape is identity verification: you are given four set identities and must spot the wrong one. That is pure law-recognition (distributive, De Morgan, absorption) plus careful region-counting on a Venn diagram. All EASY or MODERATE except a handful.

Test yourself — a quick 14-question Sets & Relations recall quiz, instant score.

Concept 1 of 4: Sets — notation, empty set, and equal vs equivalent

A set is just a collection of distinct objects. Two ideas trip students up and the NDA tests both: the empty set (a set with no elements, written ∅) is still a perfectly good set, and 'equal' sets are not the same as 'equivalent' sets — equal means identical elements, equivalent means merely the same NUMBER of elements.

Definition

Core vocabulary:

Set — a well-defined collection of distinct objects; $x \in A$ means x belongs to A.
Empty (null) set $\emptyset$ — has no elements; it is a subset of every set. A condition with no solutions defines $\emptyset$ (e.g. $\{x \in \mathbb{R} : x^2 + 1 = 0\} = \emptyset$ ).
Subset $A \subseteq B$ — every element of A is in B. Proper subset $A \subset B$ excludes $A = B$ .
Equal sets have exactly the same elements; equivalent sets only have the same number of elements (same cardinality). Equal $\Rightarrow$ equivalent, not the reverse.
Finite vs infinite — read the condition, not the number in it: $\{x\in\mathbb Z: x \text{ odd}, x<1000\}$ is infinite (unbounded below: $-1,-3,\dots$ ); any real interval, however short, contains infinitely many rationals AND infinitely many irrationals; $\{x\in\mathbb N: x<1000\}$ is finite. 'Less than $N$ ' bounds a set from above only.
Membership is not transitive, and $\in$ is not $\subseteq$ : $1\in\{1\}$ and $\{1\}\in\{\{1\},2\}$ do NOT give $1\in\{\{1\},2\}$ . For $A=\{\{1\},2\}$ : $\{1\}\in A$ is true, $\{1\}\subseteq A$ is false, $\{\{1\}\}\subseteq A$ is true. Subset-of is transitive; element-of is not.

Worked example

Are the sets

A = \{1, 2, 3\}

and

B = \{a, e, i\}

equal, equivalent, or neither?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q25Easy

Example 1 · Sets & Relations · Set Operations, Identities, and Cartesian Products of Sets

S = \{x : x^2 + 1 = 0,\, x \text{ is real}\}

, then

S

Equal vs equivalent

\{1,3,5\}

and

\{2,4,7\}

are equivalent (both size 3) but not equal (different elements). The NDA pairs these words deliberately — equal is about WHICH elements, equivalent is about HOW MANY.

Drill 3 more on sets — notation, empty set, and equal vs equivalent

Concept 2 of 4: Union, intersection, complement and difference

Five operations build every set expression. Union

A \cup B

is 'in A OR B'; intersection

A \cap B

is 'in A AND B'; complement

A'

is 'NOT in A'; difference

A - B

is 'in A but not B'. Picture them as regions of a Venn diagram and most questions become a region count.

Definition

The operations and the identities the bank leans on:

$A \cup B$ (union), $A \cap B$ (intersection), $A'$ (complement, relative to the universal set), $A - B = A \cap B'$ (difference).
Complement is an involution: $(A')' = A$ . A long nested complement like $E-(E-(E-A))$ collapses by cancelling in pairs.
A set can be defined by a condition — solving it gives the set: $(x-a)(x-b) > 0$ (with $a<b$ ) gives $x < a$ or $x > b$ ; multiples of 2 AND 3 are the multiples of 6.
Counting the difference: $|A-B|=|A|-|A\cap B|$ (the cardinality version of $A-B=A\cap B'$ ), so $|A-B|+|B-A|=|A|+|B|-2|A\cap B|=|A\triangle B|$ . And $A-B=\emptyset\iff A\subseteq B$ — NOT $A=B$ ; equality needs $B-A=\emptyset$ as well.
Absolute-value facts a set condition may hide: $|ab|=|a||b|$ , $|a+b|\le|a|+|b|$ , $|a-b|\ge\big||a|-|b|\big|$ , and $|x|<k\iff-k<x<k$ , $|x|>k\iff x<-k$ or $x>k$ ; a quadratic condition has real solutions only when its discriminant is $\ge0$ , so $\{x\in\mathbb R: x^2+x+1=0\}=\emptyset$ .

Difference and complement identities

A - B = A \cap B' \qquad (A')' = A

Worked example

E

is the universal set, simplify

E - (E - (E - A))

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2024 · Q24Moderate

Example 2 · Sets & Relations · Set Operations, Identities, and Cartesian Products of Sets

For any three non-empty sets

A,B,C

, what is

(A\cup B)-\{(A-B)\cup(B-A)\cup(A\cap B)\}

equal to?

$A - B$ is not symmetric

A - B

(in A, not B) is different from

B - A

(in B, not A). Only their UNION,

(A-B)\cup(B-A)

, is symmetric — that is the symmetric difference. Don't write

A-B = B-A

The three disjoint pieces rebuild the union

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

— the only-A, both, and only-B regions tile the whole union. Recognising this collapses many 'simplify the expression' questions to

\emptyset

A

Drill 9 more on union, intersection, complement and difference

Concept 3 of 4: The laws of set algebra

Set operations obey a fixed list of laws — the same shape as the laws of logic. The NDA's favourite question gives you four 'identities' and asks which is NOT correct; the wrong one is always a real law with one operation swapped. Knowing the genuine laws makes it a spot-the-impostor exercise.

Definition

The laws you must recognise on sight (A, B, C are any sets):

Distributive: $A \cup (B \cap C) = (A\cup B)\cap(A\cup C)$ and $A \cap (B \cup C) = (A\cap B)\cup(A\cap C)$ .
De Morgan: $(A\cup B)' = A' \cap B'$ and $(A\cap B)' = A' \cup B'$ .
Absorption: $A \cup (A\cap B) = A$ and $A \cap (A\cup B) = A$ .
Idempotent / Identity / Complement: $A\cup A = A$ ; $A\cup\emptyset=A$ ; $A\cup A' = E$ , $A\cap A' = \emptyset$ .

Law	Statement	The impostor to watch for
Distributive	$A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$	Swapping $\cup$ / $\cap$ on one side breaks it
De Morgan	$(A\cup B)'=A'\cap B'$	$(A\cup B)' = A'\cup B'$ is WRONG — the operation flips In predicate form: $x\notin(A\cup B)\Rightarrow x\notin A$ AND $x\notin B$ (not OR).
Absorption	$A\cup(A\cap B)=A$	$A\cup(A\cap B)=A\cup B$ is WRONG — it collapses to just $A$
Subset test 'for all B'	$(A\cap B)\subseteq(C\cap B)\ \forall B \Rightarrow A\subseteq C$	Test such claims by choosing $B=\emptyset$ or $B=E$

Distractors are genuine laws with one operation flipped. Verify a suspect identity on a tiny example or a Venn diagram.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q3Moderate

Example 3 · Sets & Relations · Set Operations, Identities, and Cartesian Products of Sets

A

B

and

C

are subsets of a given set, then which one of the following relations is

\textbf{\text{not}}

correct ?

The wrong option is a real law with a flipped operation

These questions never use nonsense — every distractor is a genuine law with

\cup

\cap

swapped or absorption over-simplified. If unsure, plug in

A=\{1\}, B=\{2\}

and compute both sides.

Drill 6 more on the laws of set algebra

Concept 4 of 4: Symmetric difference and set-equality conditions

The symmetric difference

A \triangle B

collects elements in exactly one of the two sets — the union minus the overlap. It also gives a clean test for equality: two sets are equal exactly when their symmetric difference is empty. A related trap is that you CANNOT cancel sets like numbers.

Definition

Equality tools:

Symmetric difference: $A \triangle B = (A-B)\cup(B-A) = (A\cup B)-(A\cap B) = (A\cap B')\cup(A'\cap B)$ .
$A \triangle B = \emptyset \iff A = B$ ; also $A\cup B = A\cap B \iff A = B$ .
Cancellation fails: $A\cap B = A\cap C$ does NOT force $B=C$ ; $A\cup C = B\cup C$ forces $A=B$ only when C is disjoint from both.

Symmetric difference

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

Worked example

Sets satisfy

A\cap C = B\cap C

with

C

disjoint from both

A

and

B

. Also

A\cup C = B\cup C

. What can you conclude?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Apr 2018 · Q6Moderate

Example 4 · Sets & Relations · Set Operations, Identities, and Cartesian Products of Sets

Let A and B be subsets of X and

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

, where A' and B' are complements of A and B in X. What is C equal to?

You cannot cancel sets like numbers

A\cap B = A\cap C

does NOT give

B=C

, and

A\cup B = A\cup C

does NOT give

B=C

either. Cancellation only works when the cancelled set is disjoint from the rest. Always look for a small counterexample.

Drill 3 more on symmetric difference and set-equality conditions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Union, intersection, complement and difference
Difference and complement identities
$A - B = A \cap B' \qquad (A')' = A$
Symmetric difference and set-equality conditions
Symmetric difference
$A \triangle B = (A - B) \cup (B - A) = (A \cup B) - (A \cap B)$

Reference tables (1)

The laws of set algebra4 rows

Law	Statement	The impostor to watch for
Distributive	$A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$	Swapping $\cup$ / $\cap$ on one side breaks it
De Morgan	$(A\cup B)'=A'\cap B'$	$(A\cup B)' = A'\cup B'$ is WRONG — the operation flips In predicate form: $x\notin(A\cup B)\Rightarrow x\notin A$ AND $x\notin B$ (not OR).
Absorption	$A\cup(A\cap B)=A$	$A\cup(A\cap B)=A\cup B$ is WRONG — it collapses to just $A$
Subset test 'for all B'	$(A\cap B)\subseteq(C\cap B)\ \forall B \Rightarrow A\subseteq C$	Test such claims by choosing $B=\emptyset$ or $B=E$

Distractors are genuine laws with one operation flipped. Verify a suspect identity on a tiny example or a Venn diagram.

Watch out for (5)

Equal vs equivalent
→ Sets — notation, empty set, and equal vs equivalent
$A - B$ is not symmetric
→ Union, intersection, complement and difference
The three disjoint pieces rebuild the union
→ Union, intersection, complement and difference
The wrong option is a real law with a flipped operation
→ The laws of set algebra
You cannot cancel sets like numbers
→ Symmetric difference and set-equality conditions

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