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.
Previous Page Next Page