GATE CS 2001 Q2.3 — Sets, Relations and Functions

To determine the correctness of statements $S_1$ and $S_2$, let us analyze them using the definition of the image of a set under a function $f: A \rightarrow B$.

Analysis of $S_1$: $f(E \cup F) = f(E) \cup f(F)$

By definition, an element $y \in B$ is in $f(E \cup F)$ if and only if there exists some $x \in E \cup F$ such that $f(x) = y$.

Since the logical equivalences hold in both directions, we have $f(E \cup F) = f(E) \cup f(F)$. Thus, $S_1$ is correct.

Analysis of $S_2$: $f(E \cap F) = f(E) \cap f(F)$

While it is always true that $f(E \cap F) \subseteq f(E) \cap f(F)$, the reverse inclusion does not necessarily hold unless $f$ is injective (one-to-one).

Counterexample:

• Let $A = \{1, 2\}$ and $B = \{3\}$.
• Define $f: A \rightarrow B$ as $f(1) = 3$ and $f(2) = 3$.
• Let $E = \{1\}$ and $F = \{2\}$ be subsets of $A$.

Now, let's compute both sides:

• $E \cap F = \emptyset \implies f(E \cap F) = \emptyset$
• $f(E) = \{3\}$ and $f(F) = \{3\} \implies f(E) \cap f(F) = \{3\}$

Since $\emptyset \neq \{3\}$, the statement $f(E \cap F) = f(E) \cap f(F)$ is false in general. Thus, $S_2$ is incorrect.

Conclusion

Only $S_1$ is correct. Hence, the correct option is A.