Week 1: Sets, Relations & Functions

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2026-09-12T00:00:00.000Z

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Number systems, sets, subsets, Cartesian products, equivalence relations, partial orders, and function classifications. # Mathematics I — Week 1: Sets, Relations & Functions ## TL;DR Week 1 is fundamentally about **objects and mappings between objects**: $$ \text{Numbers} \rightarrow \text{Sets} \rightarrow \text{Re...

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Mathematics I — Week 1: Sets, Relations & Functions

TL;DR

Week 1 is fundamentally about objects and mappings between objects:

\text{Numbers} \rightarrow \text{Sets} \rightarrow \text{Relations} \rightarrow \text{Functions}

The most exam-important ideas are:

set notation and operations
Cartesian product
relation properties:
- reflexive
- symmetric
- antisymmetric
- transitive
equivalence relations
functions:
- domain, codomain, range
- injective / one-one
- surjective / onto
- bijective

If these become intuitive, Week 1 becomes much easier.

1. Number systems

Before sets, IITM starts with the basic hierarchy of numbers.

1.1 Natural numbers

Usually:

\mathbb N=\{1,2,3,4,\dots\}

Some textbooks include

0

\mathbb N

, so always check the convention being used.

Think:

Numbers used for counting.

1.2 Whole numbers

\{0,1,2,3,\dots\}

Natural numbers plus zero.

1.3 Integers

\mathbb Z=\{\dots,-3,-2,-1,0,1,2,3,\dots\}

The symbol

\mathbb Z

comes from German Zahlen, meaning numbers.

Contains:

positive integers
negative integers
zero

2. Rational numbers

A number is rational if it can be written as

\frac pq

where

p,q\in\mathbb Z,\qquad q\neq0.

The set of rational numbers is:

\mathbb Q

Examples:

\frac23,\quad -\frac51,\quad 7,\quad 0.25

because

7=\frac71

and

0.25=\frac14.

Decimal characterization

A rational number has a decimal expansion that is either:

terminating, or
eventually repeating.

Examples:

\frac14=0.25

\frac13=0.3333\dots

3. Irrational numbers

Numbers that cannot be expressed as

\frac pq

for integers

p,q

with

q\neq0

Examples:

\sqrt2,\quad \sqrt3,\quad \pi,\quad e.

Their decimal representations are:

non-terminating
non-repeating

4. Real numbers

The real numbers contain both rational and irrational numbers:

\mathbb R = \mathbb Q \cup \{\text{irrational numbers}\}.

Hierarchy:

\mathbb N \subseteq \mathbb Z \subseteq \mathbb Q \subseteq \mathbb R.

This inclusion chain is worth memorizing.

5. What exactly is a set?

A set is a well-defined collection of distinct objects.

Example:

A=\{1,2,3\}

The objects

1,2,3

are called elements or members of

A

Notation:

2\in A

means:

$2$ belongs to $A$ .

Meanwhile,

5\notin A

means:

$5$ does not belong to $A$ .

6. Sets ignore order

These represent the same set:

\{1,2,3\}

\{3,1,2\}

Therefore:

\{1,2,3\}=\{3,1,2\}.

Unlike lists or sequences, order is irrelevant.

7. Sets ignore repetitions

\{1,1,2,2,3\} = \{1,2,3\}.

An element is either present or absent.

There isn't a meaningful concept of an element occurring "twice" in an ordinary set.

8. Describing a set

Roster form

Explicitly list elements:

A=\{2,4,6,8\}.

Set-builder form

Describe the property elements satisfy:

A=\{x\in\mathbb N:x\text{ is even and }x<10\}.

Read:

$A$ is the set of all natural numbers $x$ such that $x$ is even and less than $10$ .

The colon

:

means such that.

Sometimes a vertical bar is used:

\{x\in\mathbb N\mid x<10\}.

9. Empty set

A set containing no elements is the empty set:

\varnothing

\{\}.

Its cardinality is

|\varnothing|=0.

Important distinction:

\varnothing\neq\{\varnothing\}.

Why?

\varnothing

contains zero elements.

But

\{\varnothing\}

contains one element, namely the empty set.

Therefore:

|\{\varnothing\}|=1.

This is a classic conceptual trap.

10. Cardinality

The number of elements in a finite set

A

is written:

|A|

or sometimes:

n(A).

Example:

A=\{4,7,12\}

therefore

|A|=3.

11. Subsets

A

is a subset of

B

if every element of

A

also belongs to

B

Notation:

A\subseteq B.

Formally:

A\subseteq B \iff \forall x,\ x\in A\Rightarrow x\in B.

Example:

A=\{1,2\}

B=\{1,2,3\}

Then:

A\subseteq B.

12. Every set is a subset of itself

For every set

A

A\subseteq A.

Also:

\varnothing\subseteq A

for every set

A

Why?

To violate

\varnothing\subseteq A

we would need an element that belongs to

\varnothing

but not

A

But

\varnothing

has no elements.

Therefore no counterexample exists.

This idea—something being true because no counterexample can exist—is called vacuous truth.

You'll encounter it repeatedly in mathematics and logic.

13. Proper subset

A\subseteq B

and

A\neq B,

then

A

is a proper subset of

B

Common notation:

A\subset B.

Unfortunately, some textbooks use

\subset

merely to mean subset, so pay attention to the convention.

14. Equality of sets

Two sets are equal when they contain exactly the same elements.

A=B

iff

A\subseteq B

and

B\subseteq A.

This is a useful proof technique:

To prove two sets are equal, prove each is a subset of the other.

15. Power set

The power set of

A

is the set containing all subsets of $A$ .

Notation:

\mathcal P(A).

Example:

A=\{a,b\}.

Its subsets are:

\varnothing

\{a\}

\{b\}

\{a,b\}.

Therefore:

\mathcal P(A) = \{ \varnothing, \{a\}, \{b\}, \{a,b\} \}.

Power-set cardinality

|A|=n,

then

|\mathcal P(A)|=2^n.

Why?

For every element, you have two choices:

\text{include it} \quad\text{or}\quad \text{exclude it}.

For

n

elements:

\underbrace{2\times2\times\dots\times2}_{n\text{ times}} = 2^n.

Example:

|A|=5

implies

|\mathcal P(A)|=2^5=32.

16. Union

The union of

A

and

B

contains everything belonging to at least one of them.

A\cup B = \{x:x\in A\text{ or }x\in B\}.

Example:

A=\{1,2,3\}

B=\{3,4,5\}

then

A\cup B=\{1,2,3,4,5\}.

Note that

3

does not appear twice.

17. Intersection

The intersection contains elements belonging to both sets.

A\cap B = \{x:x\in A\text{ and }x\in B\}.

For

A=\{1,2,3\}, \qquad B=\{3,4,5\},

we get

A\cap B=\{3\}.

18. Disjoint sets

Two sets are disjoint when they share no elements:

A\cap B=\varnothing.

Example:

A=\{1,2\}

B=\{5,6\}.

19. Set difference

A-B

contains elements belonging to

A

but not

B

Formally:

A-B = \{x:x\in A\text{ and }x\notin B\}.

Example:

A=\{1,2,3,4\}

B=\{3,4,5\}

then

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

Notice generally:

A-B\neq B-A.

In the same example:

B-A=\{5\}.

So set difference is not commutative.

20. Complement

Suppose

U

is the universal set.

The complement of

A

contains everything in

U

that is not in

A

A^c=U-A.

Example:

U=\{1,2,3,4,5\}

A=\{1,3,5\}

then

A^c=\{2,4\}.

The universal set matters.

Without specifying the universe, a complement can be ambiguous.

21. Core set identities

These behave surprisingly similarly to algebra.

Commutative laws

A\cup B=B\cup A

A\cap B=B\cap A.

Associative laws

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

(A\cap B)\cap C=A\cap(B\cap C).

Distributive laws

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

and

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

Both directions exist.

22. Identity laws

A\cup\varnothing=A

A\cap U=A.

23. Domination laws

A\cup U=U

A\cap\varnothing=\varnothing.

24. Idempotent laws

A\cup A=A

A\cap A=A.

"Idempotent" means roughly:

applying an operation again produces no further change.

25. Complement laws

A\cup A^c=U

A\cap A^c=\varnothing.

Also:

(A^c)^c=A.

26. De Morgan's laws

Extremely important:

(A\cup B)^c=A^c\cap B^c

and

(A\cap B)^c=A^c\cup B^c.

A useful verbal interpretation:

Complement flips OR ↔ AND.

So:

NOT( $A$ or $B$ )

becomes

NOT $A$ and NOT $B$ .

27. Cartesian product

This concept is the bridge from sets → relations → functions.

For sets

A

and

B

A\times B = \{(a,b):a\in A,\ b\in B\}.

Example:

A=\{1,2\}

B=\{x,y\}.

Then:

A\times B = \{ (1,x),(1,y),(2,x),(2,y) \}.

Important:

(a,b)

is an ordered pair.

Therefore generally:

(a,b)\neq(b,a).

Cardinality

|A|=m

and

|B|=n,

then:

|A\times B|=mn.

Example:

|A|=3,\quad |B|=5

gives:

|A\times B|=15.

28. What is a relation?

A relation from

A

B

is simply a subset of the Cartesian product:

R\subseteq A\times B.

This definition is crucial.

Suppose:

A=\{1,2,3\}

and

B=\{a,b\}.

Then:

A\times B= \{(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)\}.

A possible relation is:

R=\{(1,a),(2,a),(3,b)\}.

Because:

R\subseteq A\times B.

29. Relation on a set

A relation on $A$ satisfies:

R\subseteq A\times A.

Example:

A=\{1,2,3\}.

Define:

R=\{(1,1),(1,2),(2,2),(3,3)\}.

This is a relation on

A

30. Reading relation notation

Instead of writing

(a,b)\in R,

we sometimes write:

aRb.

Both mean:

$a$ is related to $b$ under relation $R$ .

Example:

R

means "less than":

3R5

means:

3<5.

31. Reflexive relation

A relation

R

A

is reflexive if every element is related to itself.

Formally:

\forall a\in A,\quad (a,a)\in R.

Mental model

Every node has a self-loop.

Example:

A=\{1,2,3\}

and

R=\{(1,1),(2,2),(3,3),(1,2)\}.

Reflexive?

Yes.

Because all required pairs

(1,1),(2,2),(3,3)

exist.

Extra pairs do not hurt reflexivity.

32. Not reflexive

Suppose:

R=\{(1,1),(2,2)\}

A=\{1,2,3\}.

It is not reflexive, because:

(3,3)\notin R.

One missing self-pair is enough.

33. Symmetric relation

R

is symmetric when:

(a,b)\in R \Rightarrow (b,a)\in R.

Mental model:

Every arrow has a reverse arrow.

If:

(1,2)\in R,

symmetry requires:

(2,1)\in R.

Self-pairs don't cause problems

For:

(1,1),

its reverse is itself:

(1,1).

So self-pairs automatically satisfy symmetry.

34. Antisymmetric relation

This one causes a lot of confusion.

A relation is antisymmetric if:

(a,b)\in R \text{ and } (b,a)\in R \Rightarrow a=b.

Equivalent intuition:

Two different elements cannot have arrows in both directions.

This does not mean:

"not symmetric."

A relation can actually be both symmetric and antisymmetric.

Example:

R=\{(1,1),(2,2),(3,3)\}.

It is symmetric.

It is also antisymmetric.

Why?

There are no distinct

a,b

with both directions present.

35. Symmetric vs antisymmetric

This distinction is worth burning into memory:

Symmetric

For every arrow:

a\to b

you must have:

b\to a.

Antisymmetric

If both:

a\to b

and

b\to a

exist, then:

a=b.

So for distinct elements:

a\neq b,

antisymmetry forbids having both directions simultaneously.

36. Transitive relation

R

is transitive if:

(a,b)\in R

and

(b,c)\in R

implies:

(a,c)\in R.

Mental model:

a\to b\to c

requires shortcut:

a\to c.

Example:

If:

(1,2)\in R

and

(2,3)\in R,

then transitivity requires:

(1,3)\in R.

37. The important trap with transitivity

Transitivity only imposes a requirement when a valid two-step chain exists.

Suppose:

R=\{(1,2),(3,4)\}.

There is no chain:

aRb,\quad bRc.

Therefore nothing violates transitivity.

So the relation is vacuously transitive.

Again, absence of a counterexample matters.

38. Equivalence relation

An equivalence relation satisfies all three:

\boxed{ \text{reflexive + symmetric + transitive} }

Example:

On integers, define:

aRb

a

and

b

have the same remainder when divided by

3

For example:

1R4

because:

1\equiv4\pmod3.

This relation is:

reflexive
symmetric
transitive

therefore it is an equivalence relation.

39. Why "equivalence"?

An equivalence relation formalizes the idea:

Treat these objects as "the same" according to some property.

Example:

Under parity:

aRb \iff a\text{ and }b\text{ have the same parity}.

Then integers split into:

\{\dots,-4,-2,0,2,4,\dots\}

and

\{\dots,-3,-1,1,3,5,\dots\}.

These are equivalence classes.

40. Partial order

Another major relation type is a partial order.

A relation is a partial order when it is:

\boxed{ \text{reflexive + antisymmetric + transitive} }

Compare:

Equivalence relation

R+S+T

where

S=

symmetric.

Partial order

R+A+T

where

A=

antisymmetric.

A classic example:

\subseteq

on sets.

Why?

Reflexive

A\subseteq A.

Antisymmetric

A\subseteq B

and

B\subseteq A,

then:

A=B.

Transitive

A\subseteq B

and

B\subseteq C,

then:

A\subseteq C.

41. Relation-property cheat sheet

Property	Mathematical condition	Mental picture
Reflexive	$aRa$ for every $a$	every node loops to itself
Symmetric	$aRb\Rightarrow bRa$	every arrow reverses
Antisymmetric	$aRb\land bRa\Rightarrow a=b$	no two-way arrows between distinct nodes
Transitive	$aRb\land bRc\Rightarrow aRc$	every two-step path has shortcut

This table is one of the highest-value things to memorize in Week 1.

42. Functions

Now we narrow relations further.

A function

f:A\to B

assigns every element of $A$ exactly one element of $B$ .

Here:

$A$ = domain
$B$ = codomain

If:

f(a)=b,

then

b

is the image of

a

43. Function as a special relation

Remember:

R\subseteq A\times B.

A function is a relation satisfying two additional conditions:

Every domain element must be used

For every:

a\in A,

there must be some:

b\in B

such that:

(a,b)\in f.

Exactly one output per input

You cannot have:

(a,b_1)

and

(a,b_2)

with:

b_1\neq b_2.

This is allowed:

f(1)=5

f(2)=5.

So:


text
1 ─┐
   ├──> 5
2 ─┘

is perfectly valid.

What is forbidden is:


text
       ┌──> 5
1 ─────┤
       └──> 8

because one input gets two outputs.

45. Domain, codomain and range

For:

f:A\to B,

Domain

All allowed inputs:

A.

Codomain

The declared target set:

B.

Range

Outputs that are actually achieved:

f(A) = \{f(a):a\in A\}.

Therefore always:

\text{Range}\subseteq\text{Codomain}.

These are not necessarily equal.

46. Example

Suppose:

f:\{1,2,3\}\to\{a,b,c,d\}

with:

f(1)=a,\qquad f(2)=b,\qquad f(3)=b.

Then:

Domain

\{1,2,3\}

Codomain

\{a,b,c,d\}

Range

\{a,b\}.

Notice:

\{a,b\}\subsetneq\{a,b,c,d\}.

47. One-one / injective function

A function is injective or one-one when distinct inputs produce distinct outputs.

Formally:

f(a)=f(b) \Rightarrow a=b.

Equivalent form:

a\neq b \Rightarrow f(a)\neq f(b).

Mental model:

No output receives arrows from two different inputs.

48. Onto / surjective function

A function is surjective or onto if every element of the codomain gets hit.

Formally:

\forall b\in B,\quad \exists a\in A \text{ such that } f(a)=b.

Equivalent statement:

\text{Range}=\text{Codomain}.

Mental model:

Nothing in the codomain is left unused.

49. Bijective function

A function is bijective if it is both:

\boxed{\text{injective + surjective}}

So there is a perfect one-to-one correspondence between domain and codomain.

Example:

f:\{1,2,3\}\to\{a,b,c\}

with:

1\mapsto a,\qquad 2\mapsto b,\qquad 3\mapsto c.

This is bijective.

50. Why bijections matter

Bijective functions have inverses.

If:

f:A\to B

is bijective, then:

f^{-1}:B\to A

exists.

This becomes important later when Maths 1 reaches inverse functions in Week 5. The official syllabus places functions, exponential functions, composition and inverse functions there.

So Week 1 is already building foundations for Week 5.

51. Fast finite-set tests

Suppose:

f:A\to B.

If:

|A|>|B|,

then

f

cannot be injective.

Why?

More inputs than available outputs means at least two inputs must share an output.

This is the pigeonhole principle.

If:

|A|<|B|,

then

f

cannot be surjective.

Why?

There aren't enough inputs to hit every codomain element.

If:

|A|=|B|

for finite sets, then:

f\text{ injective} \iff f\text{ surjective} \iff f\text{ bijective}.

This shortcut is extremely useful.

52. Week 1 conceptual map


text
NUMBERS
│
├── Natural
├── Integer
├── Rational
└── Real
     │
     ▼
SETS
│
├── subset
├── power set
├── union/intersection
├── complement
└── Cartesian product
     │
     ▼
RELATIONS
│
├── reflexive
├── symmetric
├── antisymmetric
└── transitive
     │
     ├── equivalence relation
     └── partial order
          │
          ▼
FUNCTIONS
│
├── domain
├── codomain
├── range
├── injective
├── surjective
└── bijective

That is essentially the conceptual skeleton of Week 1.

53. Things you should be able to do after Week 1

You don't need to memorize every sentence above. You need to be able to perform these operations:

Convert between roster and set-builder notation.
Determine subsets and proper subsets.
Compute $A\cup B$ , $A\cap B$ , $A-B$ , and $A^c$ .
Compute power-set size using $2^n$ .
Construct $A\times B$ .
Recognize whether something is a relation.
Test a relation for reflexivity.
Test symmetry.
Distinguish symmetry from antisymmetry.
Test transitivity carefully.
Identify equivalence relations.
Identify partial orders.
Decide whether a relation qualifies as a function.
Find domain, codomain and range.
Decide whether a function is injective, surjective or bijective.

If you can do those without looking anything up, Week 1 is basically under control.

The one thing I'd memorize verbatim

\boxed{ \begin{aligned} \text{Reflexive: } & aRa \\ \text{Symmetric: } & aRb\Rightarrow bRa \\ \text{Antisymmetric: } & aRb\land bRa\Rightarrow a=b \\ \text{Transitive: } & aRb\land bRc\Rightarrow aRc \end{aligned} }

And:

\boxed{ \text{Equivalence} = \text{Reflexive + Symmetric + Transitive} }

\boxed{ \text{Partial Order} = \text{Reflexive + Antisymmetric + Transitive} }

Those definitions produce a large fraction of the reasoning questions in this part of the course.

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:

Question :: Answer.

Power Set cardinality of set A with $n$ elements :: $2^n$
Cartesian Product cardinality $|A \times B|$ :: $|A| \times |B|$
De Morgan's Law for $(A \cup B)'$ :: $A' \cap B'$
De Morgan's Law for $(A \cap B)'$ :: $A' \cup B'$
Equivalence Relation properties :: Reflexive, Symmetric, and Transitive
Partial Order properties :: Reflexive, Antisymmetric, and Transitive
Antisymmetric condition :: If $(a, b) \in R$ and $(b, a) \in R$ , then $a = b$
Injective (one-to-one) condition :: $f(x_1) = f(x_2) \implies x_1 = x_2$
Surjective (onto) condition :: Range of $f$ = Codomain of $f$
Bijective function :: Both Injective and Surjective (invertible)

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.

Question 1

Let

A = \{1, 2, 3, 4\}

. What is the cardinality of the power set

\mathcal{P}(A)

MCQ

3 Unit Assessment

A. 4

B. 8

C. 16

D. 32

Question 2

A relation

R

on set

S

is an equivalence relation if and only if it satisfies which three properties?

MCQ

3 Unit Assessment

A. Reflexive, Symmetric, and Antisymmetric

B. Reflexive, Symmetric, and Transitive

C. Reflexive, Antisymmetric, and Transitive

D. Symmetric, Antisymmetric, and Transitive

Question 3

Let

f: \mathbb{R} \to \mathbb{R}

be defined by

f(x) = 2x + 5

. Which of the following describes

f

MCQ

3 Unit Assessment

A. Injective but not surjective

B. Surjective but not injective

C. Bijective (both injective and surjective)

D. Neither injective nor surjective

What this note is really saying

Relevant recall pieces for this note

Mathematics I — Week 1: Sets, Relations & Functions

TL;DR

1. Number systems

1.1 Natural numbers

1.2 Whole numbers

1.3 Integers

2. Rational numbers

Decimal characterization

3. Irrational numbers

4. Real numbers

5. What exactly is a set?

6. Sets ignore order

7. Sets ignore repetitions

8. Describing a set

Roster form

Set-builder form

9. Empty set

10. Cardinality

11. Subsets

12. Every set is a subset of itself

13. Proper subset

14. Equality of sets

15. Power set

Power-set cardinality

Why?

16. Union

17. Intersection

18. Disjoint sets

19. Set difference

20. Complement

21. Core set identities

Commutative laws

Associative laws

Distributive laws

22. Identity laws

23. Domination laws

24. Idempotent laws

25. Complement laws

26. De Morgan's laws

27. Cartesian product

Cardinality

28. What is a relation?

29. Relation on a set

30. Reading relation notation

31. Reflexive relation

Mental model

32. Not reflexive

33. Symmetric relation

Self-pairs don't cause problems

34. Antisymmetric relation

35. Symmetric vs antisymmetric

Symmetric

Antisymmetric

36. Transitive relation

37. The important trap with transitivity

38. Equivalence relation

39. Why "equivalence"?

40. Partial order

Equivalence relation

Partial order

Reflexive

Antisymmetric

Transitive

41. Relation-property cheat sheet

42. Functions

43. Function as a special relation

Every domain element must be used

Exactly one output per input

44. Many inputs may share one output

45. Domain, codomain and range

Domain

Codomain

Range

46. Example

Domain

Codomain

Range

47. One-one / injective function