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(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.

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(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.

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HallsofIvy

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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.

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.

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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.

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Hyperreals are not used in computation at all.the ordering is very computation heavy in practice.

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http://www.merriam-webster.com/dictionary/computing

I was saying that it seems like the process to determine ordering for two specific sequences can be a little clunky.

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