1.8. COMPLETENESS OF R 19

16. De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents,not just integers.

1 = 1(1/4) = (cos 2π + i sin 2π)1/4

= cos (π/2) + i sin (π/2) = i.

Therefore, squaring both sides it follows 1 = −1 as in the previous problem. Whatdoes this tell you about De Moivre’s theorem? Is there a profound difference betweenraising numbers to integer powers and raising numbers to non integer powers?

17. Show that C cannot be considered an ordered field. Hint: Consider i2 = −1. Recallthat 1 > 0 by Proposition 1.4.2.

18. Say a + ib < x + iy if a < x or if a = x, then b < y. This is called the lexicographicorder. Show that any two different complex numbers can be compared with this order.What goes wrong in terms of the other requirements for an ordered field.

19. With the order of Problem 18, consider for n ∈ N the complex number 1 − 1n . Show

that with the lexicographic order just described, each of 1− in is an upper bound toall these numbers. Therefore, this is a set which is “bounded above” but has no leastupper bound with respect to the lexicographic order on C.

1.8 Completeness of RRecall the following important definition from calculus, completeness of R.

Definition 1.8.1 A non empty set, S ⊆ R is bounded above (below) if there exists x ∈ Rsuch that x ≥ (≤) s for all s ∈ S. If S is a nonempty set in R which is bounded above,then a number, l which has the property that l is an upper bound and that every other upperbound is no smaller than l is called a least upper bound, l.u.b. (S) or often sup (S) . If S is anonempty set bounded below, define the greatest lower bound, g.l.b. (S) or inf (S) similarly.Thus g is the g.l.b. (S) means g is a lower bound for S and it is the largest of all lowerbounds. If S is a nonempty subset of R which is not bounded above, this information isexpressed by saying sup (S) = +∞ and if S is not bounded below, inf (S) = −∞.

Every existence theorem in calculus depends on some form of the completeness axiom.

Axiom 1.8.2 (completeness) Every nonempty set of real numbers which is bounded abovehas a least upper bound and every nonempty set of real numbers which is bounded below hasa greatest lower bound.

It is this axiom which distinguishes Calculus from Algebra. A fundamental result aboutsup and inf is the following.

Proposition 1.8.3 Let S be a nonempty set and suppose sup (S) exists. Then for everyδ > 0,

S ∩ (sup (S)− δ, sup (S)] ̸= ∅.

If inf (S) exists, then for every δ > 0,

S ∩ [inf (S) , inf (S) + δ) ̸= ∅.

Proof: Consider the first claim. If the indicated set equals ∅, then sup (S) − δ is anupper bound for S which is smaller than sup (S) , contrary to the definition of sup (S) asthe least upper bound. In the second claim, if the indicated set equals ∅, then inf (S) + δwould be a lower bound which is larger than inf (S) contrary to the definition of inf (S). ■