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Can the existence of nonstandard hyperreal extensions be proved? 
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#1
Feb2312, 08:12 PM

P: 783

I have begun to read about the hyperreals, and am wondering whether the natural extensions of realvalued functions to hyperrealvalued functions is simply a definition of the hyperreals, or can it be proved? Or is it accepted as an axiom?
For example, if f(x) = sin(x), then is the existence of f*(x) = sin*(x) provable, or is it assumed? (f*(x) and sin*(x) represent the natural extensions.) Thanks! BiP 


#2
Feb2312, 08:34 PM

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PF Gold
P: 16,091

Well, there are always a variety of ways to about defining things. Which theorems are axioms^{*} and which are not isn't really important, since you can always shuffle the presentation of the theory around to change things.
General, nonstandard models of real analysis are usually defined by elementary equivalence: a superstructure^{**} a nonstandard model if and only if the transfer principle works. (more or less, anyways) So, [itex]{}^\star\!\!\sin[/itex] exists because of that. The elements of the most commonly used hyperreal field can be explicitly named by sequences of real numbers. Then, [itex]{}^\star\!\!\sin[/itex] has an explicit formula: you just take the sine of each term in the sequence. (aside: to preempt any misunderstanding, while the names are explicit, the same hyperreal number has many different such names. You need an ultrafilter to tell whether two names denote the same hyperreal number. The ultrafilter can't be explicitly constructed in ZF  its existence is usually derived from the axiom of choice) *: If P is an axiom, then P is also a theorem, by the trivial proof: "because P is true, P is true". **: Don't worry about the technical meaning of this word, it's not really important. 


#3
Feb2312, 10:48 PM

P: 783

BiP 


#4
Feb2412, 12:41 AM

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P: 906

Can the existence of nonstandard hyperreal extensions be proved?
f'(x) = st((f(x+dx)f(x))/dx) and there is no mention of limits (and thus no epsilons or deltas to quantify over). it is (for some) easier to understand "dx" as a very small (hyperreal) number (an infinitesimal, close to 0, but not equal to it), then to deal with the limiting process of something which appears to approach 0/0, a nonsensical number. because no formal system of hyperreals existed when calculus was invented, concerns about the rigorousness of calculus persisted until the 19th century, when formal definitions of the real numbers, limits and continuity were given that alleviated these concerns. it was even later (around 1966 i think) that the original methodolgies of Newton and Liebniz were shown to be logically consistent (that is, a suitable definition of infinitesimal given) by Abraham Robinson. at this point in time, both standard, and nonstandard analysis are considered "rigorous". it does not appear that either approach can prove something the other cannot. that said, most courses offered in advanced analysis will nevertheless take the "standard" approach. 


#5
Feb2412, 06:37 AM

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PF Gold
P: 16,091

Crucially, however, the angle at which the nonstandard analyst looks at the axioms of analysis provides for an average case reduction in complexity that provides shorter proofs of various results, and will one day lead to the proof of a result which is not accessible to classical mathematics without nonstandard methods, precisely because its classical proof is too long to write down in the length of time humans will reside on Earth. For the simplest stuff, the difference between standard and nonstandard methods is usually quite superficial; they generally have the same issues and concerns and you can translate directly back and forth. The winnings of nonstandard analysis don't really show up until you start putting things together to make more complex arguments or start treating more sophisticated notions. 


#6
Feb2812, 11:49 AM

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P: 1,172

Just curious:
What kind of ultrafilter do we use to mod out the ultraproduct? I guess we want to identify some elements that are close to each other, but I am not sure how. Also: is the transfer principle the same as elementary equivalence? 


#7
Feb2812, 02:12 PM

P: 800

http://terrytao.wordpress.com/2007/0...onmanagement/ 


#8
Feb2812, 03:32 PM

Sci Advisor
P: 1,172

Never mind my first question; I just found out that we only need an ultrafilter over the natural, and that the independence of the choice of U is equivalent to AC.



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