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# 5.2 Completeness

5.21   Definition (Completeness axiom.) Let be an ordered field. We say that is complete if it satisfies the condition:

Every binary search sequence in converges to a unique point in .

5.22   Example. The field is not complete, since in example 5.16 we found a binary search sequence in that does not converge.

5.23   Definition (Real field, .) A real field is a complete ordered field. We will use the name to denote a real field.

5.24   Remark. It is not at all clear that any real fields exist. If real fields do exist, there is a question of uniqueness; i.e., is it the case that any two real fields are essentially the same"? I don't want to worry about what mathematical existence means, so let me formulate the questions: Are the axioms for a real field consistent; i.e., is it the case that no contradictions can be derived from them? Note that we are not entirely free to throw axioms together. If I were to make a definition that a 3-field is an ordered field in which , I would immediately get a contradiction: and . All I can say about consistency is that no contradictions have been found to follow from the real field axioms. There exist proofs that any two real fields are essentially the same, cf. [35, page 129]. (This source uses a different statement for the completeness axiom than we have used, but the axiom system is equivalent to ours.) There also exist constructions of pairs of very different real fields, cf. [41].

In what follows, I am going to assume that there is a real field (which I'll call the real numbers). Any theorems proved will be valid in all real fields.

5.25   Theorem (Archimedean property 1.) Let be a real field, and let . Then there is an integer such that .

Proof: Let , and suppose (in order to get a contradiction) that there is no with . Then for all . Now is a binary search sequence in . Since , I have for all . We see that , but clearly . Since completeness of implies that a binary search sequence has a unique limit, this yields a contradiction, and proves the theorem.

5.26   Corollary (Archimedean property 2.) Let . Then there is some such that .

Proof: By the theorem, there is some with . Then .

5.27   Corollary (Archimedean property 3.) Let be a real number, and let be a positive real number. Suppose
 (5.28)

Then .

Proof: Let , and let satisfy
 (5.29)

Suppose, in order to get a contradiction, that . Then by Archimedian property 2 there is some such that , i.e. . This contradicts (5.29).

5.30   Theorem. If , then there is an integer and a number such that .

In order to prove this theorem, I will use the following lemma.

5.31   Lemma. If , then the interval contains an integer.

Proof:
Case 1.
: Suppose and does not contain an integer. I will show that for all . This contradicts the Archimedean property, so no such can exist. For each , let ". Then is true, since I assumed that . Let be a number such that is true; i.e., . If were , we'd have , and this cannot happen, since contains no integers. Hence,

and by induction, for all . This gives the desired contradiction.
Case 2.
: If , then by Case 1 there is an integer with

Then

If , then contains . If , then contains .

Proof of theorem 5.30. Let . By the lemma, there is an integer with . Then

and gives the desired decomposition.

5.32   Theorem. There is a number such that .

Proof: Let be the binary search sequence constructed in example 5.16. We know there is a unique such that for all . Then , and by our construction for all , so

 (5.33)

for all .

By Archimedean property 3, we conclude that , i.e., .

5.34   Theorem. Let . Then there is a binary search sequence in such that and for all , and such that .

Proof: I will suppose . The case where is left to you. By the Archimedean property of , there is an integer such that , so . Now build a binary search sequence as follows:

From the construction, we have and . A simple induction argument shows that and for all , and an induction proof similar to the one in example 5.16 shows that
for all so .

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