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 there exists at least one element in f(S1 - S2) that is not in f(S1 - f(S2)).
First, let's prove that every element in f(S1 - f(S2)) is also in f(S1 - S2).
1. Take an arbitrary element x in f(S1 - f(S2)). This means that there exists an element y in S1 - f(S2) such that f(y) = x.
2. Since y is in S1 - f(S2), it implies that y is in S1 but not in f(S2).
3. Now, let's consider the set S1 - S2. Since y is in S1 but not in S2, it follows that y is in S1 - S2.
4. Therefore, f(y) = x is an element in f(S1 - S2), proving that every element in f(S1 - f(S2)) is also in f(S1 - S2).
Next, let's show that there exists at least one element in f(S1 - S2) that is not in f(S1 - f(S2)).
Suppose we have a function f that maps A to B, with nonempty sets A and B. Let's consider a specific example to illustrate this.
Example:
Let A = {1, 2, 3} and B = {4, 5, 6}.
Define f as follows:
f(1) = 4
f(2) = 5
f(3) = 6
Let's choose S1 = {1, 2} and S2 = {1, 3}.
Then, f(S1) - f(S2) would be {4, 5} - {4, 6} = {5}.
Similarly, S1 - S2 would be {2}, and f(S1 - S2) would be {f(2)} = {5}.
In this example, f(S1) - f(S2) is {5} and f(S1 - S2) is also {5}. Thus, f(S1) - f(S2) = f(S1 - S2).
Therefore, this example demonstrates that there are subsets S1 and S2 of A such that f(S1) - f(S2) does not equal f(S1 - S2).
In conclusion, we have proven that f(S1 - f(S2)) is a proper subset of f(S1 - S2) for all subsets S1 and S2 of A, and also provided an example where f(S1) - f(S2) does not equal f(S1 - S2).