Let f:A->B where A and B are nonempty. Prove that f(S1 U S2) = f(S1) U f(S2)
To prove that f(S1 U S2) = f(S1) U f(S2), where f: A -> B and A and B are nonempty sets, we need to prove that any element in the set on the left side of the equation is also in the set on the right side, and vice versa.
First, let's prove that f(S1 U S2) is a subset of f(S1) U f(S2).
Take an arbitrary element y in f(S1 U S2). By definition, this means that there exists an element x in S1 U S2 such that f(x) = y.
Since x is in S1 U S2, it must either be in S1 or S2.
Case 1: x is in S1.
In this case, we have x ∈ S1 and f(x) = y. Since x is in S1, it follows that y = f(x) is also in f(S1).
Case 2: x is in S2.
In this case, we have x ∈ S2 and f(x) = y. Since x is in S2, it follows that y = f(x) is also in f(S2).
Therefore, in both cases, we have shown that y = f(x) is in either f(S1) or f(S2), meaning y is in f(S1) U f(S2).
This establishes that f(S1 U S2) is a subset of f(S1) U f(S2).
Now, let's prove that f(S1) U f(S2) is a subset of f(S1 U S2).
Take an arbitrary element z in f(S1) U f(S2). By definition, z is in either f(S1) or f(S2).
Case 1: z is in f(S1).
In this case, there exists an element a in S1 such that f(a) = z. Since a is in S1, it follows that a is also in S1 U S2. Therefore, z = f(a) is in f(S1 U S2).
Case 2: z is in f(S2).
In this case, there exists an element b in S2 such that f(b) = z. Since b is in S2, it follows that b is also in S1 U S2. Therefore, z = f(b) is in f(S1 U S2).
Therefore, in both cases, we have shown that z is in f(S1 U S2).
This establishes that f(S1) U f(S2) is a subset of f(S1 U S2).
Since we have proven that f(S1 U S2) is a subset of f(S1) U f(S2) and f(S1) U f(S2) is a subset of f(S1 U S2), we can conclude that f(S1 U S2) = f(S1) U f(S2).