Bridge to Abstract Mathematics: Mathematical Proof and Structures

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9.3 COMPLETENESS IN AN ORDERED FIELD 311

less than n, no matter how close to n, is less than some number in the
sequence. Stated differently, .n is the smallest real number greater than
each number in the sequence.
Because of these two properties (to be formalized later in Definitions 1
and 3), we say that the real number n is the least upper bound in R of the
given infinite set of real numbers. Note, however, a third significant prop-
erty of the given sequence; each of its members is not only real, but ra-
tional as well. Now suppose we try to find a rational number that is the
"least upper bound in Q" of this infinite set of rational numbers. Your
intuition may suggest that no such rational number exists; indeed, none
does (see Example 3, for a rigorous treatment of a similar example).
The preceding example is at the heart of the theoretical difference between
R and Q. Let us begin now to take a more systematic approach.


DEFINITION 1
Let F be an ordered field. A subset S of F is said to be bounded above in F if and
only if there exists an element u E F such that x 5 u for all x E S. Any such field
element u is called an upper bound of S in F.

In the ordered field R the interval (0, 1) is bounded above by 1, as well
as by any real number larger than 1. The same is true of the interval [0, 11.
In the ordered field Q the subset S = {x E Qlx2 5 3) is bounded above by
rational numbers such as 1.8, 1.74, and 1.733. The real number fi is
not an upper bound of S in Q, because fi $ Q. Clearly fi is an upper
bound in R for the subset I = {x E R 1x2 5 3) of R.
The concepts bounded below and bounded in an ordered field are defined
in a manner analogous to Definition 1. Specifically, we have Definition 2.


DEFINITION 2
A subset S of an ordered field F is said to be bounded below in F if and only if there
exists an element IE F such that 15 x for all x E S. Any such field element I is
called a lower bound of S in F. S is said to be bounded in F if and only if there
exists b E F such that 1x1 I b for all x E S.
It is easy to show that a subset of an ordered field F is bounded in F if
and only if it is bounded both above and below in F (see Exercise 2). The
subset (- 5, GO) of R is bounded below but not above in R, whereas (-6,171
is a bounded subset of the ordered field R.
There. is, as you may have noticed, a similarity between Definition 1,
above, and Definition 2 of Article 7.4. The relationship is that every ordered
field is a partially ordered set, and indeed, a totally ordered set (recall
Theorem 3, Article 9.2), so that the theory developed in Article 7.4 applies
to ordered fields. The following definition is analogous to, and consistent
with, Definition 3, Article 7.4. It is, however, designed to take more direct
advantage of the additional knowledge we gain about a totally ordered set
from the fact that it is also an ordered field.

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