Call an integer ra>la 'far seeing integer' if xm > xn for all n>m. (The term 'far seeing' is invented for use in this particular proof and is not standard terminology.) There are two possibilities:

We assume the hypotheses of Theorem 3.17. Set [ao,&o] = [—К, К]. Show that we can find a sequence of pairs of points an and bn such that
%m £ [a-mbn] for infinitely many values of m, o>n-i < a,n < bn < bn-\, and bn- an = (6n_i - an_i)/2,
for all n > 1.
Show that an —> с as n —> со for some с 6 [UQ, bo]. Show further that we can find m(j) with m(j + 1) > m(j) and xmu^ G [a,j, bj] for each j > 1. Deduce the Bolzano-Weierstrass theorem.

If xn is a sequence of real numbers which is bounded above we write
limsupa;n = lim sup{xm : m>n}. n—>oo n >0°

Suppose that F is an ordered field for which the Bolzano-Weierstrass theorem holds (that is, every bounded sequence has a convergent subsequence). Suppose that an is an increasing sequence bounded above. Use the Bolzano- Weierstrass theorem to show that there exists an a 6 F and n(l) < n(2) < ... such that an(j) "~* a as 3 ~* °°- Show that an —> a as n —> oo and so F obeys the fundamental axiom.

Produce a version of the proof just given in which we do not assume that a and b take particular values.

In this exercise you may use the axiom of Archimedes and the fact that any non-empty bounded set of integers has a maximum.