# Sequences in nonstandard analysis(basic question)

1. Aug 18, 2009

### Bourbaki1123

Given two sequences (hyperreal numbers):

(2,1,2,1....) and (1,3,1,3...) how can I order these? They are not compliments and don't seem to fit into any of the possible orderings. It seems that intuitively the second is larger than the first.

2. Aug 18, 2009

### HallsofIvy

What makes you think that they can be ordered? What definition of "order" are you using? If this is a homework or schoolwork question, it should be posted in the "homework" section.

3. Aug 18, 2009

### Bourbaki1123

No, I am self studying from Lectures on the Hyperreals by Goldblatt. The hyperreals are an ordered field, so there must be some way to order the equivalence classes for those two sequences right? The orderings r<s ,r>s and r=s of equivalence classes of sequences of real numbers should be determined by taking a representative from each class and determining if {n in N: s(n) = r(n)} is in a nonprincipal ultrafilter defined on N, or equivalently if it is cofinite with N.

Since the if the union of disjoint sets is containted in an ultrafilter F, so are one of the sets in the union; it must be true that one of [r<s], [r>s] or [r=s] (where [r>s] is the set of natural numbers used for indicies of the elements of representative sequences chosen that agree) is in F, but it seems as though none of these could be in F or that more than one must be. Whichever is contained in the nonprincipal ultrafilter is supposed to be indicative of the ordering.

I'm certain I missed something here, if you are familiar with the Hyperreals could you try to help me out?

Edit: This question came to me when I saw a problem in the book; If A (a subset of R) is finite, show that A* (a corresponding subset of the hyperreals) has no nonstandard elements. A* is defined thusly: [r] is an element of A iff {n in N: r(n) is in A} is cofinite with N. The problem is that [1,2,1,2...] seems to be an element of A*, but it is nonstandard as far as I can tell. From this I wondered how you could even order elements like this. It seems like it must be the case that these are somehow excluded form the hyperreals, but I am not sure how or where in the book it could have said something indicating that.

Last edited: Aug 18, 2009
4. Aug 19, 2009

### g_edgar

This is what your ultrafilter is for. Either the set A of even numbers is in the ultrafilter or the set of odd numbers. Which is it? I don't care, but one of these two possibilities happens. If A is in it, then (2,1,2,1,...) represents the same hyperreal as (1,1,1,1,...) and (1,3,1,3,...) represents the same hyperreal as (3,3,3,3,...). They are easy to compare.

In general to compare two hyperreals, given sequences (a_n) and (b_n) representing them, look at three sets of natural numbers: indices where a_n < b_n; indices where a_n = b_n; indices where a_n > b_n. Exacly one of these three belongs to the ultrafilter. So exactly one of the three order relations <, =, > holds.

5. Aug 19, 2009

### Bourbaki1123

I had sort of figured out that first part, but what still confuses me is if say you have n numbers in A and then arrange them in a sequence where you have the whole set repeating. Then you have for each value, either the set of every nth number is in F or its compliment, and this is true for each n in A. It seems as though the ordering is very computation heavy in practice. I suppose if you check each of them, either you will have one in F or one will happen to only be missing a finite number of elements so the sequence will be equal to one of the [n].

6. Aug 20, 2009

### g_edgar

Hyperreals are not used in computation at all.

7. Aug 20, 2009