Other Versions of the Fundamental Axiom

The supremum

Since all of analysis depends on the fundamental axiom, it is not surprising that mathematicians have developed a number of different methods of proof to exploit it. We have already seen the method of 'lion hunting'. In this chapter we see two more: the 'supremum method' and the 'Bolzano-Weierstrass method'.

Lemma 3.4.

Proof. The only problem is setting out the matter in the right way.
Suppose a and a1 least upper bounds for a non-empty set A of real numbers. Then
(i) a > a for all a e A.
(ii) If (3 > a for all a e A, then /3 > a.
(ii') a1 > a for all a e A.
(ii') If /3 > a for all a e A, then /3 > a'.

Supremum principle

If A is a non-empty set of real numbers which is bounded above (that is, there exists а К such that a < К for all a 6 A), then A has a supremum.
Note that the result is false for the rationale.

Exercise 3.10.

(i) If A satisfies the hypotheses of Theorem 3.7, explain carefully why we can find an integer r(j) such that
r(j)2~J > a for all a & A but
there exists an a(j) 6 A with a(j) > (r(j) — Y)2~3.
[Hint. Recall that every non-empty set of the integers bounded below has a minimum.]

Theorem 3.12.

We work in Ж. If f : [a, b] —»• M is continuous and /(a) > 0 > f(b), then there exists а с & [a, b] such that /(c) = 0.

We proceed as follows

If a < с < b, we proceed as follows. Since с = sup E we can find XQ G E such that 0 < c — xo < 5 and so \f(xo) — /(c) | < 6. Since xo G £7, /(жо) > О and so /(c) > —6. On the other hand, choosing yo = c + min(6 — c, 5)/2 we know that 0 < yo — с < 5 and so |/(жо) — /(c) | < e. Since yo > с it follows that yo ^ E so f(yo) < 0 and /(c) < 6. We have shown that |/(c)| < e.

Exercise 3.13.

In both the 'lion hunting' and the 'supremum argument' proofs we end up with a point с where /(c) = 0, that is a 'zero of f'. Give an example of a function f : [a,b] —> Ж satisfying the hypotheses of the intermediate value theorem for which 'lion hunting' and the 'supremum argument' give different zeros.

Lemma 3.14.

Let U be the open interval (a,(3) on the real line R. Suppose that a, b G U and b > a. If f : U —>• R is differentiable with f'(t) < К for all t 6 U and e > 0 then f(b)-f(a)<(b-a)(K + e).

Theorem 3.15.

Let (F, +, x, >) be an ordered field. Then the following two statements are equivalent.
(i) Every increasing sequence bounded above has a limit.
(ii) Every non-empty set bounded above has a supremum.
The next exercise sketches a much more direct proof that the supremum principle implies the fundamental axiom.