GATE CS 2014 Set 3 Q2 — Sets, Relations and Functions

To find the correct statement, let us analyze each option:

• Option D: For any $x \in X$, we have: $$\begin{aligned} x \in f^{-1}(S \cap T) &\iff f(x) \in S \cap T \\ &\iff f(x) \in S \text{ and } f(x) \in T \\ &\iff x \in f^{-1}(S) \text{ and } x \in f^{-1}(T) \\ &\iff x \in f^{-1}(S) \cap f^{-1}(T) \end{aligned}$$ Since the logical equivalences hold in both directions, $f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$ is always TRUE.

We can disprove the other options using a simple counterexample:

Let $X = \{1, 2\}$, $Y = \{a\}$, and $f(1) = f(2) = a$. Let $A = \{1\}$ and $B = \{2\}$.

• Option A: $f(A \cup B) = f(\{1, 2\}) = \{a\} \implies |f(A \cup B)| = 1$. However, $|f(A)| + |f(B)| = 1 + 1 = 2$. Thus, $|f(A \cup B)| \neq |f(A)| + |f(B)|$.
• Option B: $A \cap B = \emptyset \implies f(A \cap B) = \emptyset$. However, $f(A) \cap f(B) = \{a\} \cap \{a\} = \{a\}$. Thus, $f(A \cap B) \neq f(A) \cap f(B)$.
• Option C: $|f(A \cap B)| = |f(\emptyset)| = 0$. However, $\min\{|f(A)|, |f(B)|\} = \min\{1, 1\} = 1$. Thus, $|f(A \cap B)| \neq \min\{|f(A)|, |f(B)|\}$.