40 Chapter 1 • The Real Number System
- For each of the sets listed in Exercise 1, tell whether or not the given set
is bounded below. For those that are, give three different lower bounds
and find the greatest lower bound. - Prove Theorem 1.6.2 (a).
- Prove Theorem 1.6.2 (b), Part 2.
- Finish proving Theorem 1.6.2 (c), that S has a minimum element.
- Complete the proof of Theorem 1.6.4 by proving Part 2.
- Prove T heorem 1.6.5 (b).
- Prove Theorem 1.6.5 (c).
- Prove Theorem 1.6.5 (d).
- Prove that N has no maximum element. Does it have a supremum? A
minimum? An infimum? - Prove Theorem 1.6.7. [Note: The "forcing principle" (Theorem 1.5.9)
must be "massaged" a little to make it apply here. See Exercise 1.5.12.]
12. Prove that in an ordered field F, u is a lower bound of a set A {=? -u
is an upper bound of the set -A = {-a : a E A}. [Thus, A is bounded
below iff -A is bounded above.]13. Let F be an Archimedean ordered field, A ~ F, and u E F. Prove the
modified c--criterion for supremum: if u rf. A, u = sup A {=? Ve > 0,
(a) 'Vx E A, x < u + c, and
(b) 3infinitely many x EA 3 x > u - c.14. State and prove a modified c--criterion for v
Exercise 13.]inf A if v rf. A. [SeeTHE COMPLETENESS AXIOMDefinition 1.6.8 An ordered field F is complete if it satisfies the
Completeness property (C): every nonempty subset of F that has an
upper bound in F has a least upper bound in F. (See also Theorem 1.6. 12
below.).
Caution: recall that the word "has" here does not mean "contains." The
supremum of A need not be a member of A.
As we shall shortly see, there are common fields that are not co mplete. But
first, we develop a few properties of complete ordered fields.