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

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.

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.

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

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

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

**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.

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

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

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 .