Question

Let X and Y be finite sets and f: X → Y be a function. Which one of the following statements in TRUE?
1. For any subsets A and B of X, |f (A ∪ B)| = |f(A)| + |F(B)|
2. For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B)
3. For any subsets A and B of X, |g(A ∩ B)| = min{|f(A)|, |f(B)|}
4. For any subset S and T of Y, F^-1(S∪ T) = f^-1(S)∩ f^-1(T)

Answer 1

Correct Answer - Option 4 : For any subset S and T of Y, F^-1(S∪ T) = f^-1(S)∩ f^-1(T)

Concept:

First three statements can be proved invalid by taking a counter example. For example,

Suppose X = {p, q, r} and Y = (1, 2)

A function f maps each element of set X to exactly one element of set Y. The one element from Y is 1 to which each element of X is mapped.

This means,

f(p) = 1, f(q) = 1, f(r) = 1

Now, we take subsets of X as A = (p, q) and B = (q, r)

Explanation:

Option_1 - False

LHS = |f (A ∪ B)| = |f ({p, q, r})| = 3

RHS = |f(A)| + |F(B)| = 2 + 2 = 4

LHS ≠ RHS

Option_2 - False

LHS = f (A ∩ B) = f ({q}) = {1}

RHS = f(A) ∩ f(B) = {1, 1} ∩ {1, 1} = {1, 1}

LHS ≠ RHS

Option_3 - False

LHS = |f (A ∩ B)| = |f ({q})| = |{1}| = 1

RHS = min{|f(A)|, |f(B)|} = min(2, 2) = 2

LHS ≠ RHS

Option_4 - True

In a function, one value from domain set X can be mapped to exactly one value from range set Y.

This option assumes that the inverse of function f exists which is true.