Let f : A → B be a function and E, F are subsets of A. Show thatf (E ∪ F ) = f (E) ∪ f (F ) and f (E ∩ F ) ⊂ f (E) ∩ f (F )

Sure, let's go through this step by step.

1. f(E ∪ F) = f(E) ∪ f(F)

Proof:

Let's take an element y in f(E ∪ F). This means there exists an element x in E ∪ F such that f(x) = y. Now, x being in E ∪ F means that x is in E or x is in F. If x is in E, then y is in f(E). If x is in F, then y is in f(F). Therefore, y is in f(E) ∪ f(F). This shows that f(E ∪ F) is a subset of f(E) ∪ f(F).

Now, let's take an element y in f(E) ∪ f(F). This means that y is in f(E) or y is in f(F). If y is in f(E), then there exists an x in E such that f(x) = y. If y is in f(F), then there exists an x in F such that f(x) = y. In either case, there exists an x in E ∪ F such that f(x) = y. Therefore, y is in f(E ∪ F). This shows that f(E) ∪ f(F) is a subset of f(E ∪ F).

Since each set is a subset of the other, we can conclude that f(E ∪ F) = f(E) ∪ f(F).

2. f(E ∩ F) ⊂ f(E) ∩ f(F)

Proof:

Let's take an element y in f(E ∩ F). This means there exists an x in E ∩ F such that f(x) = y. Now, x being in E ∩ F means that x is in both E and F. Therefore, y is in both f(E) and f(F), which means y is in f(E) ∩ f(F). This shows that f(E ∩ F) is a subset of f(E) ∩ f(F).

However, the reverse inclusion does not necessarily hold. That is, we cannot say that f(E) ∩ f(F) is a subset of f(E ∩ F). This is because there might be different elements x1 in E and x2 in F such that f(x1) = f(x2) = y. In this case, y is in f(E) ∩ f(F), but there is no x in E ∩ F such that f(x) = y. Therefore, we can only say that f(E ∩ F) is a subset of, but not necessarily equal to, f(E) ∩ f(F).