6 1. The Real Numbers

Inverse image behaves very well with respect to the set theory operations, as

the following theorem shows.

Theorem 1.1.6. If f : A → B is a function and E and F are subsets of B, then

(a) f

−1(E

∪ F) = f

−1(E)

∪ f

−1(F

);

(b) f −1(E ∩ F) = f −1(E) ∩ f −1(F ); and

(c) f

−1(E

\ F) = f

−1(E)

\ f

−1(F

) if F ⊂ E.

Proof. We will prove (a) and leave the other two parts to the exercises.

To prove (a), we will show that f −1(E ∪ F) and f −1(E) ∪ f −1(F ) have the

same elements. If x ∈ f −1(E ∪ F), then f(x) ∈ E ∪ F. This means that f(x) is in

E or in F. If it is in E, then x ∈ f −1(E). If it is in F, then x ∈ f −1(F ). In either

case, x ∈ f −1(E) ∪ f −1(F ). This proves that every element of f −1(E ∪ F) is an

element of f −1(E) ∪ f −1(F ).

On the other hand, if x ∈ f −1(E) ∪ f −1(F ), then x ∈ f −1(E), in which case

f(x) ∈ E, or x ∈ f −1(F ), in which case f(x) ∈ F. In either case, f(x) ∈ E ∪ F,

which implies x ∈ f −1(E ∪ F). This proves that every element of f −1(E) ∪ f −1(F )

is also an element of f −1(E ∪ F). Combined with the previous paragraph, this

proves that the two sets are equal.

Image does not behave as well as inverse image with respect the set operations.

The best we can say is the following:

Theorem 1.1.7. If f : A → B is a function and E and F are subsets of A, then

(a) f(E ∪ F) = f(E) ∪ f(F);

(b) f(E ∩ F) ⊂ f(E) ∩ f(F);

(c) f(E) \ f(F) ⊂ f(E \ F) if F ⊂ E.

Proof. We will prove (c) and leave the other parts to the exercises.

To prove (c), we must show that each element of f(E)\f(F) is also an element

of f(E \ F). If y ∈ f(E) \ f(F), then y = f(x) for some x ∈ E and y is not the

image of any element of F. In particular, x / ∈ F. This means that x ∈ E \ F and

so y ∈ f(E \ F). This completes the proof.

The above theorem cannot be improved. That is, it is not in general true that

f(E ∩ F) = f(E) ∩ f(F) or that f(E) \ f(F) = f(E \ F) if F ⊂ E. The first of

these facts is shown in the next example. The second is left to the exercises.

Example 1.1.8. Give an example of a function f : A → B for which there are

subsets E, F ⊂ A with f(E ∩ F) = f(E) ∩ f(F).

Solution: Let A and B both be in R and let f : A → B be defined by

f(x) =

x2.

If E = (0, ∞) and F = (−∞,0), then E ∩ F = ∅, and so f(E ∩ F) is also the empty

set. However, f(E) = f(F) = (0, ∞), and so f(E) ∩ f(F) = (0, ∞) as well. Clearly

f(E ∩ F) and f(E) ∩ f(F) are not the same in this case.