The Real And Complex Number Systems

Without loss of generality, we assume that p a and q b. Claim that f is strictly
decreasing on a, b.
Suppose NOT, then there exist x, y a, b, with x y such that fx fy. (""
does not hold since f is 1-1.) Consider x, y and by above method, we know that f| x,y has
the maximum at y, andf| a,y has the minimum at y. Then it implies that there exists
By; x, y such that f is constant on By; x, y, which contradicts to 1-1. Hence,
for any x y a, b, wehavefx fy. ("" does not hold since f is 1-1.) So, we
have proved that f is strictly decreasing on a, b.
Reamrk: 1. Here is another proof by Exercise 4.61. It suffices to show that 1-1 and
continuity imply that f does not have a local maximum or a local minimum at any interior
point.
Proof: Suppose NOT, it means that f has a local extremum at some interior point x.
Without loss of generality, we assume that f has a local minimum at the interior point x.
Since x is an interior point of a, b, then there exists an open interval
x , x a, b such that fy fx for all y x , x . Note that f is 1-1, so
we have fy fx for all y x , x x. Choose y 1 x and
y 2 x, x , then we have fy 1 fx and fy 2 fx. And thus choose r so that
fy 1 r fx fp r, wherep y 1 , x by Intermediate Value Theorem,
fy 2 r fx fq r, whereq x, y 2 by Intermediate Value Theorem,
which contradicts to the hypothesis that f is 1-1. Hence, we have proved that 1-1 and
continuity imply that f does not have a local maximum or a local minimum at any interior
point.
2. Under the assumption of continuity on a compact interval, one-to-one is
equivalent to being strictly monotonic.
Proof: By the exercise, we know that an one-to-one and continuous function defined
on a compact interval implies that a strictly monotonic function. So, it remains to show that
a strictly monotonic function implies that an one-to-one function. Without loss of
generality, let f be increasing on a, b, then as fx fy, we must have x y since if
x y, then fx fy and if x y, then fx fy. So, we have proved that a strictly
monotonic function implies that an one-to-one function. Hence, we get that under the
assumption of continuity on a compact interval, one-to-one is equivalent to being strictly
monotonic.
4.63 Let f be an increasing function defined on a, b and let x 1 ,..,x n be n points in
the interior such that a x 1 x 2 ... x n b.
n
(a) Show that k1
fx k fx k fb fa .
Proof: Let a x 0 and b x n1 ;sincef is an increasing function defined on a, b, we
know that both fx k and fx k exist for 1 k n. Assume that y k x k , x k1 , then
we have fy k fx k and fx k1 fy k1 . Hence,
n
k1
fx k fx k fy k fy k1
n
k1
fy n fy 0
fb fa .
(b) Deduce from part (a) that the set of dicontinuities of f is countable.