Final answer:
In mathematics, any subset A within a set X is always a subset of the preimage of its image through function f, A⊆f⁻¹(f(A)). Furthermore, if the function f is one-to-one, then the original set A and the preimage of its image through f are equal, A=f⁻¹(f(A)).
Step-by-step explanation:
The student is inquiring about set theory and functions in the context of mathematics, specifically the relationship between a set A and its image through a function f, as well as the preimage of that image when f is a one-to-one function. We need to show that for each subset A ⊆ X, it is true that A ⊆ f−1(f(A)) and that if f is one-to-one, then A = f−1(f(A)). To prove this, we first consider any element a in set A. By definition of the function, f(a) will be an element of f(A). Now, because f−1 is the preimage function, any element that maps to f(A) through f must originate from A; thus, a must be in f−1(f(A)), proving that A ⊆ f−1(f(A)).
For the second part, we consider f being one-to-one, or injective, which means no two distinct elements in the domain X map to the same element in the codomain. This implies that if f(a) is in f(A), the only corresponding preimage in X must be a itself. Therefore, for every element in f(A), there is one and only one preimage in A, which leads us to conclude that A = f−1(f(A)) when f is one-to-one.