314 Functions, graphs, and numbers
course in calculus to explore these matters thoroughly. In order to
understand the proofs, it is necessary to sketch and study illustrations of
various kinds, and progress is very slow. Until considerable experience
has been obtained, it is not easy to reproduce the proofs even after they
have been completely understood. Students who peek at the proofs
can be compared with children of jewelers who peek at the innards of
watches. They start accumulation of knowledge of reasons why things
tick, and the overwhelming importance of getting started must be
recognized by everyone who knows that we toddle and walk before we
run. So far as this course is concerned, it is of primary importance to
understand the axiom and theorems of this section and to cultivate the
habit of formulating and using precise mathematical statements.
We begin a campaign to learn something about continuous functions
and differentiable functions and their graphs by proving the following
theorem.
Theorem 5.45 If f is continuous over an interval a < x < b, then f is
bounded over the interval, that is, there is a constant M such that
If(x)l < M (a<x<=b).
Our proof will use the Dedekind axiom. Let xi be put in A if xi 5 a.
Moreover, let xi be put in -4 if a < xi 5 b and there is a constant Ml
such that I f(x)1 5 Mi when a 5 x < xi. Let B contain all other num-
bers. This determines a Dedekind partition, and we can let be the
partition number. It is easy to see that a 5 t <_ b, but the remainder of
the proof is more delicate. Since f has right-hand continuity at a, we
can let e = 1 and choose a positive number S such thatf(a) - 1 <--_ f(x)
f (a) + 1 and hence If(x) l < Jf (a) l+ 1 whenever a x 5 a + S.
Hence f is bounded over the interval a < x <-- a + S, so a + S belongs to
A and i;? a + S > a. Our next step is to show that = b. If E < b,
then we have a < t < b as in Figure 5.451. Since f is continuous at ,
A 0 B
b x2
Figure 5.451
a
we can let .E = 1 and choose a positive number 5 such that a < t - S <
+ 5 < b as in Figure 5.451 and Jf (x) l < l f (i;) J + 1 when - 5 <
x S + S. But - 5 belongs to A, so there must be a constant MI
such that Jf(x)J S MI when a S x < t - S. If we let M2 be the greater
of MI and If(t) + 1, then I f(x) S M2 when a < x 5 + S. Therefore,
+ S must belong to A and we have a contradiction of the fact that the
partition number t must be either the greatest number in -4 or the least
number in B. All this shows that i = b, and we are almost finished.
Let e = I and choose a positive number 6 such that a < b- 8 < b and
