1549901369-Elements_of_Real_Analysis__Denlinger_

1.5 The Archimedean Property 33

DENSE SETS IN ORDERED FIELDS

Definition 1.5.6 A set S is dense in an ordered field F if

Va< bin F , 3 x ES 3 a< x < b.

Theorem 1.5.7 (a) (Denseness of the Rationals) The rational numbers

form a dense set in any Archimedean ordered field.

(b) (Denseness of the Irrationals) In any Archimedean ordered field

with at least one irrational element, the irrational elements form a dense subset.

Proof. Suppose F is an Archimedean ordered field, and a < b in F.

(a) Since b-a > 0, Theorem 1.5.2 (b) tells us that 3n EN 3 n(b-a) > 1.

Then nb - na > 1, so by Corollary 1.5.5, 3 m E Z3

Hence,

na < m < nb.

m

a< - < b.

n

Thus, the rational numbers form a dense subset of F.

(b) Exercise 6. •

Theorem 1.5.8 If S is a dense set in an ordered field F, then between any

two elements of F there are infinitely many elements of S.

Proof. Exercise 8. •
As a consequence of Exercise 1.4.5, an Archimedean ordered field F with
one irrational element must contain at least as many irrational as rational
elements. Moreover, Theorem 1.5.7 says that they must be densely scattered
in F. A more surprising result is that in the real number system, the irrational
numbers far outnumber the rational numbers. The sense in which this is true,
and a proof of that result, is found in Section 2.8.

The following theorem continues in the vein of Theorem 1.5.2. It establishes

a method used frequently in analysis to prove both inequalities and equalities.

Theorem 1.5.9 (Forcing Principle) Suppose Fis an Archimedean ordered

field, and x, a, b E F.

(a) If Ve;> 0, x::::; c:, then x::::; 0.

(b) If Ve;> 0, x::::; a+ c:, then x::::; a.

(c) If Ve;> 0, JxJ::::; c:, then x = 0.

(d) IfVc: > 0, Ja - bl::::; c:, then a= b.