Question

Show that if f:A→B and G,H are subsets of B, then f⁻¹(G ∪ H) = f⁻¹(G) ∪ f⁻¹(H) and f⁻¹(G ∩ H) = f⁻¹ (G)∩ f⁻¹(H)

Answer 1

The question involves proving that the pre-images of the union and intersection of two subsets under a function equal the union and intersection of their pre-images respectively. This is shown by applying the definitions of pre-image and set operations union and intersection.

Step-by-step explanation:

The question pertains to the pre-image of sets under a function f from A to B. To show that f⁻¹(G ∪ H) = f⁻¹(G) ∪ f⁻¹(H) and f⁻¹(G ∩ H) = f⁻¹(G) ∩ f⁻¹(H), consider the definitions of pre-image and set operations.

• f⁻¹(G ∪ H) includes all elements x from A such that f(x) is in G ∪ H. This, by definition, is the union of elements that map into G and H separately, which means f⁻¹(G ∪ H) = f⁻¹(G) ∪ f⁻¹(H).
• Similarly, f⁻¹(G ∩ H) includes all elements x from A such that f(x) is in both G and H. This intersection means f⁻¹(G ∩ H) = f⁻¹(G) ∩ f⁻¹(H).