Let f : A -> B be a function, and let E and F be subsets of A. Consider the following statements about images. S1: f(E ∩ F) = f(E) ∩ f(F) S2: f(E ∪ F) = f(E) ∪ f(F) Which of the following is true about S1 and S2? A. Only S2 is correct B. Only S1 is correct C. Both S1 and S2 are correct D. None of S1 and S2 is correct

GATE 2001 · Discrete Mathematics · Functions · medium

Answer: Only S2 is correct.

1. Prove S2: f(E ∪ F) = f(E) ∪ f(F): b in f(E ∪ F) iff some a in E or a in F has f(a)=b, iff b in f(E) or b in f(F). So f(E ∪ F) = f(E) ∪ f(F). S2 is always true.
2. Disprove S1 with a counterexample: Let A = {1, 2}, B = {a}, f(1) = f(2) = a. Set E = {1}, F = {2}. Then E ∩ F = ∅, so f(E ∩ F) = ∅. But f(E) = {a} and f(F) = {a}, so f(E) ∩ f(F) = {a} ≠ ∅. S1 fails.