Let X = { 1, 2, 3} and Y= {4, 5}. Find whether the following subsets of `X xx Y` are functions from X to Y or not.
(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} (ii) g = {(1, 4), (2, 4), (3, 4)}
(iii) h = {(1, 4), (2, 5), (3, 5) } (iv) k = {(1, 4), (2, 5)}

Explanation

To determine whether the given subsets of \( X \times Y \) are functions from \( X \) to \( Y \), we need to check if each element in the domain \( X \) is assigned to a unique element in the codomain \( Y \).

### Given Sets:
- \( X = \{ 1, 2, 3 \} \)
- \( Y = \{ 4, 5 \} \)

### Function Definitions:
A relation \( f \) from \( X \) to \( Y \) is a function if:
1. Every element in \( X \) is associated with an element in \( Y \).
2. Each element in \( X \) is associated with exactly one element in \( Y \).

### Checking Each Subset:

#### (i) \( f = \{(1, 4), (1, 5), (2, 4), (3, 5)\} \)
1. Check the images:
- \( f(1) = 4 \) and \( f(1) = 5 \) (two images for 1)
- \( f(2) = 4 \)
- \( f(3) = 5 \)
2. Since \( 1 \) has two images, \( f \) is **not a function**.

#### (ii) \( g = \{(1, 4), (2, 4), (3, 4)\} \)
1. Check the images:
- \( g(1) = 4 \)
- \( g(2) = 4 \)
- \( g(3) = 4 \)
2. Each element in \( X \) is assigned a unique image in \( Y \) (even though they map to the same value, each element has a defined image). Thus, \( g \) is a **function**.

#### (iii) \( h = \{(1, 4), (2, 5), (3, 5)\} \)
1. Check the images:
- \( h(1) = 4 \)
- \( h(2) = 5 \)
- \( h(3) = 5 \)
2. Each element in \( X \) has a defined image in \( Y \), and no element in \( X \) has more than one image. Thus, \( h \) is a **function**.

#### (iv) \( k = \{(1, 4), (2, 5)\} \)
1. Check the images:
- \( k(1) = 4 \)
- \( k(2) = 5 \)
- \( k(3) \) has no image.
2. Since \( 3 \) does not have an image, \( k \) is **not a function**.

### Summary of Results:
- \( f \) is **not a function**.
- \( g \) is a **function**.
- \( h \) is a **function**.
- \( k \) is **not a function**.