• Let $X$ be any set and let $R$ be a relation from $X$ to $X$. Let $R$ be an equivalence relation.
• Let $x \in X$. Then equivalence class of $x$ will be,
  $[x]_R = \{y∈X : (x,y) ∈R\}$.

Let $x,y ∈X$ with $x\neq y$.
Suppose that $[x]_R \neq [y] _R$.
Claim: $[x]_R ∩ [y]_R = \phi$.

Suppose not, i.e,there exists $z ∈ [x]_R ∩ [y]_R$.

• $z∈[x]_R \ \& \ z∈[y]_R$
• $\Rightarrow (z, x)∈R \ \& \ (z, y)∈R$
• $\Rightarrow (x, z)∈R \ \& \ (z, y)∈R \$ (∵ $R$ is an equivalence relation)
• $\Rightarrow (x,y)∈R$ (∵ $R$ is transitive)
• $\Rightarrow y∈[x]_R$
• $∴ [y]_R ⊆ [x]_R$
• By symmetry of $x \ \& \ y$ we have $[y]_R =[x]_R$.
• Thus $[x]_R = [y]_R$
• Or $[x]_R ∩ [y]_R =\phi$

$\Z$ = Set of all integers.
$R=\{(m,n)∈ \Z \times \Z | m≡n (\text{mod} \ 9)\}$. Write down all the equivalence classes for the given relation.

$1∈ \Z$
$[1]_R = \{m|m-1 ≡ 0 \ (\text{mod} \ 9)\}$
$m-1≡0 (\text{mod} \ 9)$
Equivalently, $m-1$ is divisible by $9$.

$10$ is one such number.
$\{1,10,19,28,….\} = \{9n+1|n∈ \Z \}$

$[1]_R = \{1,10,19,28,…\}$

$[0]_R = \{0,9,18,27,…\}$

Note that there are exactly 9 equivalence classes.
$\{[0]_R, [1]_R, [2]_R, [3]_R, [4]_R, [5]_R, [6]_R, [7]_R, [8]_R \}$.

$R=\biggl\{((x_1, y_1),(x_2, y_2))∈ \R^2$\ $\{(0,0)\}\times \R^2$\ $\{(0,0)\} \mid \text{there exists a non zero scalar}| α ∈ \R \ \text{such that} \ (x_1, y_1) = α (x_2, y_2)\biggl \}$.
Write down the equivalence class for the given relation.

Suppose $(x_1, y_1)∈$ $\R^2$ \ $\{(0,0)\}$.

$((x_2, y_2), (x_1, y_1))∈ R$

$\Rightarrow (x_2, y_2) = α (x_1, y_1) \ \text{for some} \ α∈ \R \ \text{with} \ α≠0$.

$\Rightarrow (x_2, y_2) \ \& \ (x_1, y_1) \ \text{lie on the same line passing through the origin}.$

$\text{Thus} \ [(x_1, y_1)]_R \ \text{is the straight line passing through the origin}$
$\text{with the origin removed}.$

Definition A function from a nonempty set $A$ to a nonempty set $B$ is a relation from $A$ to $B$ with the following properties:
i) Domain $(f) = A$.

ii) Whenever $(a,b_1)∈f \ \text{and} \ (a, b_2)∈f$
$\Rightarrow b_1 = b_2$

$A= \{-5,-3,-2,0,1,3,5\}$
$B= \{ 1,2,4,9,16,24,30\}$

$R =\{(b,a)∈ B\times A \mid b=a^2 \}$.
Check if $R$ is a function or not.

Solution:

• Some elements of $B$ are not related to $A$. e.g. $2,16,24$ and $30$
• $(9,-3)∈R \ \& \ (9, 3)∈R$
• $∴ R$ is not a function.

$A=\{1,2,3,4,5,6\} \ \text{and} \ R = \{(a,b)∈ AxA | b= a+1 \}$
Check if $R$ is a function or not.

Solution:

• Domain $(R) = \{1,2,3,4,5\}≠A$
• $∴ R$ is not a function.

$f:\R → \R$

$f = \{(x,y) | y=x^2\}$. Check if $f$ is a function or not.

Solution:

• Domain$(f) =\R$.
• Suppose $(x, y_1)= (x, y_2)∈f$
• $(x, y_1) ∈f \Rightarrow y_1 = x^2$
• $(x, y_2) ∈f \Rightarrow y_2 = x^2$
• $\Rightarrow y_1 = y_2$.
• $∴$ f is a function.

• $f:A \rightarrow B$ is function
  i.e, the relation $‘f’$ is a function from a non empty set $A$ to the nonempty set $B.$
• If the pair $(a,b) ∈ f$ this will be denoted as $b=f(a).$

• Suppose $f:A \rightarrow B$ is a function.
• Range $(f) = {b∈B: b= f(a)}.$

$f:\R \rightarrow \R$
$f(x)=x^2$
Find the range of the function.

Solution:

• Let $y∈ \R$ with $y \ge 0.$
• Let us assume that $y≠0 \Rightarrow y>0$.
• $\Rightarrow$ there are exactly two square roots of $y$, namely $+\sqrt{y} \ \ \& \ -\sqrt{y}.$
• Let $x=+\sqrt{y} \ \Rightarrow \ y=x^2$
• Thus, Range$(f) = \{y∈\R | y \ge 0 \}.$

$f:\Z \rightarrow \Z$
$f(n) =n+1$
Find the range of the function.

Solution:

• Range $(f) = \Z$.
• Let $m ∈ \Z$
• Let $n=m-1 \Rightarrow f(n) =n+1 =(m+1) +1 =m.$
• Thus Range $(f) =\Z.$

• $f:\R \rightarrow \R$
  $f(x) =x$
• Domain $(f) = \R$
• Range $(f) = \R$

1. $f(x) =x^2$

• Domain $(f) = \R$
• Range $(f) = \{y \mid y \ge 0\}$

2. $f(x) = \sin x$

• Domain $(f) = \R$
• Range $(f) = [-1,1]$

3. $f(x) =|x|$

• Domain $(f) =\R$
• Range $(f) = \{ y∈ \R | y≥0\}.$

4. $f(x) =x^3$

• Domain $(f) =\R$
• Range $(f) = \R$.

$f:X \rightarrow Y$
Suppose $A,B ⊆ X$.
Find the relation between $f(A∩B)$ and $f(A)∩f(B)$.

Solution:

$f:\R → \R \ \ \text{and} \ f(x) =x^2.$

• $A=[0,4] ; B=[-4, 0]$
• $A∩B = {0}.$
• $f(A) = [0,16] ;f(B) = [0,16] ;f(A∩B) ={0}.$
• $∴ f(A∩B) ⊆ f(A)∩f(B).$

• Claim: For any $A,B ⊆ X,$
  $f(A∩B) ⊆ f(A)∩f(B)$
• Let $y ∈ f(A∩B)$
  $\Rightarrow$ there exists $x∈ A∩B$ such that $f(x)=y.$
• $x∈ A∩B \Rightarrow x∈A \ \& \ x∈B$
• $\Rightarrow f(x) ∈ f(A) \ \& \ f(x) ∈ f(B)$
• $\Rightarrow f(x) ∈ f(A) ∩ f(B).$
• $\Rightarrow y ∈ f(A) ∩ f(B).$
• Thus $f(A∩B) ⊆ f(A) ∩ f(B).$

$f:X \rightarrow Y$ Suppose $A,B ⊆ x.$
Find the relation between $f(AUB)$ and $(A) U f(B)$.

Solution:

• Claim: $f(AUB) = f(A) U f(B).$
• Let $y ∈ f(AUB)$
  $\Rightarrow$ there exists $x∈AUB$ such that $y=f(x).$
• $x∈AUB \ \Rightarrow \ x∈A \ \text{or} \ x∈B$
• $\Rightarrow f(x)∈f(A) \ \text{or} \ f(x)∈f(B)$
• $\Rightarrow f(x) ∈ f(A) U f(B)$ $\Rightarrow y ∈ f(A) U f(B)$
• $∴ f(AUB) ⊆ f(A) U f(B).$

• Let $y ∈ f(A) U f(B)$
• $\Rightarrow y ∈ f(A) \ \text{or} \ y ∈ f(B)$
• $y ∈ f(A) \Rightarrow \ \text{there exists} \ x∈A \ \text{such that} \ y=f(x)$
• $y ∈ f(B) \Rightarrow \ \text{there exists} \ x’∈B \ \text{such that} y=f(x’).$
• $x∈A \Rightarrow x∈ AUB \Rightarrow f(x) ∈ f (AUB)$
• $x’∈B \Rightarrow x’∈ AUB \Rightarrow f(x’) ∈ f (AUB)$
• $∴ y = f(x) = f(x’) ∈ f (AUB)$
• $∴ f(A) U f(B) ⊆ f (AUB)$
• $∴ f(A) U f(B) = f (AUB)$

Thank you