Let f : A → B be an injective (one-to-one) function. Define g : 2A → 2B as :…
2003
Let f : A → B be an injective (one-to-one) function.
Define g : 2A → 2B as :
g(C) = {f(x) | x ∈ C}, for all subsets C of A.
Define h : 2B → 2A as :
h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B. Which of the following statements is always true ?
- A.
g(h(D)) ⊆ D
- B.
g(h(D)) ⊇ D
- C.
g(h(D)) ∩ D = ф
- D.
g(h(D)) ∩ (B - D) ≠ ф
Attempted by 127 students.
Show answer & explanation
Correct answer: A
Answer: g(h(D)) ⊆ D is always true.
Proof. Take any element y in g(h(D)). Then y = f(x) for some x in h(D). By definition of h, x ∈ A and f(x) ∈ D. Therefore y = f(x) ∈ D. This shows every element of g(h(D)) lies in D, so g(h(D)) ⊆ D.
Equality remark: g(h(D)) = D ∩ f(A). Thus g(h(D)) equals D exactly when every element of D is in the image f(A).
Why the other statements fail (brief counterexamples):
The claim that g(h(D)) ⊇ D need not hold. Example: A = {1}, B = {a,b}, f(1)=a, and D={b}. Then h(D)=∅ and g(h(D))=∅, so g(h(D)) does not contain b.
The claim that g(h(D)) ∩ D = ∅ need not hold. Example: A={1}, B={a}, f(1)=a, D={a}. Then h(D)={1}, g(h(D))={a}, and the intersection is {a}.
The claim that g(h(D)) meets B \ D (i.e., g(h(D)) ∩ (B \ D) ≠ ∅) is impossible because g(h(D)) ⊆ D, so its intersection with B \ D is always empty.