Let f:A->B, where A and B are nonempty

Prove that f(S1 - f(S2) is a proper subset of f(S1 -S2) fo all subsets S1 and S2 of A.

Give an example where there are subsets S1 and S2 of A such that
f(S1) - f(S2) does not equal f(S1-S2)

To prove that f(S1 - f(S2)) is a proper subset of f(S1 - S2) for all subsets S1 and S2 of A, we need to show that every element in f(S1 - f(S2)) is also in f(S1 - S2), and that there exists at least one element in f(S1 - S2) that is not in f(S1 - f(S2)).

Let's break down the proof into two parts:

1. Every element in f(S1 - f(S2)) is in f(S1 - S2):
- Assume y is an arbitrary element in f(S1 - f(S2)). This implies that there exists x in S1 - f(S2) such that f(x) = y.
- Since x is in S1 - f(S2), it means x is in S1 and x is not in f(S2).
- Now, consider S1 - S2. Since x is in S1 and x is not in S2, it follows that x is also in S1 - S2.
- Therefore, f(x) = y is in f(S1 - S2), which implies that every element in f(S1 - f(S2)) is in f(S1 - S2).

2. There exists an element in f(S1 - S2) that is not in f(S1 - f(S2)):
- Consider the scenario where S1 = {1, 2} and S2 = {1}.
- Let f be a function such that f(1) = 1 and f(2) = 2.
- Now, let's calculate f(S1 - f(S2)) and f(S1 - S2):
- f(S1 - f(S2)) = f({2}) = {2}
- f(S1 - S2) = f({2}) = {2}
- In this example, f(S1 - f(S2)) = f(S1 - S2) = {2}. Therefore, there is no element in f(S1 - S2) that is not in f(S1 - f(S2)).

Hence, we have proven that f(S1 - f(S2)) is a proper subset of f(S1 - S2) for all subsets S1 and S2 of A.