Calculus: Analytic Geometry and Calculus, with Vectors

5.4 Theorems about continuous and differentiable functions 315

If(x) I < f(b) + 1 when b - S < x < b. Since b - & belongs to 4, there
must be a constant M3 such that Jf(x)j < M3 when a S x < b - S. If
we let M be the greater of M3 and l f(b)l + 1, then I f(x)l <- M when
a < x < b. This completes the proof of Theorem 5.45.
To strengthen Theorem 5.45, and for other purposes, we need informa-
tion about upper and lower bounds. A set S of numbers is said to have an
upper bound M1 if x 5 M1 whenever x is inS and is said to have a least upper
bound (l.u.b.) or supremum (sup) M if M - CI--I C7--I
is an upper bound and there is no upper ml m x M M3 x
bound M1 for which M1 < M. Analo- Figure 5.453
gously, S is said to have a lower bound ml
if x > ml whenever x is in S and is said to have a greatest lower bound
(g.l.b.) or infimum (inf) m if m is a lower bound and there is no lower
bound ml for which m, > m. The inequality

(5.452) ml<m_<x<M<M1

shows how these numbers must be related when x is in S, and Figure
5.453 shows a way in which they are sometimes related.
Theorem 5.46 If a nonempty set S of numbers has an upper bound Mi,
then it has a least upper bound. Similarly, if a nonempty set S of numbers
has a lower bound ml, then it has a greatest lower
bound. A-T-F I B
As Figure 5.461 indicates, we make a partition x0Mi
of numbers by putting a number in B if it is an Figure 5.461
upper bound of S and putting a number in !4 if it is
not an upper bound of S. The set B contains M1, and if xo is a number in
S, then A contains the number xo - 1. Let be the partition number.
Let x be a number in S. Then, for each positive number e, x 5 + e.
Hence x <-- and it follows that t is an upper bound of S. If x' < ,then
x' is in .4 and hence x' is not an upper bound of S. Therefore, is the
least upper bound of S. This completes the proof of the first part of the
theorem. The second part is proved similarly.
Theorem 5.47 If f is continuous over an interval a < x <-- b, thenf(x)
attains minimum and maximum values over the interval at points of the
interval, that is, there exist numbers in, M, xi, and x2 such that a S x1 5
b, a5X2Sb,and

m = f(x1) S f(x) S f(x2) = M

whenever a 5 x 5 b.

To prove the part of the theorem involving M, we use Theorem 5.45 to

conclude that f must have an upper bound M1. It follows from Theorem

5.46 that the set of numbers f(x) for which a S x _<_ b has a least upper

bound which we now denote by M. Then f (x) < M when a <= x <_ b,

and it remains to be shown that there is a number X2 for which f(x2) = M.