1.6 The Completeness Property 41Theorem 1.6.9 Any complete ordered field is Archimedean.Proof. Suppose F is a complete ordered field. For contradiction, suppose
F is not Archimedean. That is, it is not true that Vx > 0, 3 n E Np 3 n > x.
Equivalently, 3 x > 0 in F 3 Vn E Np, n ~ x.
This means that the set Np is a nonempty set in F with an upper bound,
x. Then, since Fis complete,
3xo = supNp.
Now, Vn E Np, n + 1 E Np, so n + 1 ~ x 0 , which implies that n ~ x 0 - l.
In summary,
Vn E Np, n ~ x o - 1.
This says that xo - 1 is an upper bound for Np; but x o is the least upper
bound for Np. Contradiction. Therefore, Fis Archimedean. •Theorem 1.6.10 If an ordered field Fis complete, then 3x E F 3 x^2 = 2.
*Proof. Suppose Fis a complete ordered field. Define the following subsets
of F:
A= {x > 0: x^2 < 2} and B = {x > 0: x^2 > 2}.
Then A and B are nonempty, since 1 E A and 2 E B. Also, A is bounded
above. Since Fis complete, we may let u =sup A. Then u 2: 1, since 1 EA.We first establish some preliminary results:f/ l. Suppose y satisfies 0 < y < u. Then y < supA, so 3a EA 3 y <a~ u,
and so y^2 < a^2 < 2. (See Theorem 1.2.8 (e).) Thus, y EA.,., 2. Suppose y satisfies y > u. Then let u < t < y. (Such a t exists; for
example, t = ~.) Since t > sup A , t tf-_ A; so t^2 2: 2. Then y^2 > t^2 2: 2.
Thus, y EB.;:,, 3. Summarizing (1) and (2), every positive number less than u lies in A , and
every number greater than u lies in B.We now prove that u^2 = 2. We shall use the forcing principle [Theorem 1.5.9
(d)]. Let c > 0. Choose o E F to be any element such that 0 < o <min { u, :u }·
c
Then O < o < u and 0 < o < - , so by (3) above, u - o EA and u + o EB.
4u
Thus,
(u-o)^2 <2<(u+o)^2.
Multiplying through by -1, we have
-(u + 6)^2 < -2 < -(u - 6)^2.