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Can the existence of nonstandard hyperreal extensions be proved?

  1. Feb 23, 2012 #1
    I have begun to read about the hyperreals, and am wondering whether the natural extensions of real-valued functions to hyperreal-valued 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.)


  2. jcsd
  3. Feb 23, 2012 #2


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    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 non-standard 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 pre-empt 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.
  4. Feb 23, 2012 #3
    Thank you! Is non-standard analysis helpful in any way? Could I prove the theorems of calculus without having to define the hyperreal number line? Or are they simply used as a convenience? I don't want to expose myself to anything I don't need... but I do need my rigor.

  5. Feb 24, 2012 #4


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    some find it "more intuitive".


    if one is working with the hyperreals, certain statements in calculus become more "direct", for example:

    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 non-sensical 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 non-standard 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.
  6. Feb 24, 2012 #5


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    It allows you to stream-line a number of various arguments and concepts -- e.g. in an argument where one considers things that are "sufficiently small" or "sufficiently large", one can often replace them with "infinitesimal" and "transfinite" which are often easier to work with.

    Yes, and no. Any theorem about standard analysis that can be proven by non-standard methods can be proven using standard methods. But as a practical matter, we have this quote from mathworld:
    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.​

    This deserves a bit of nitpicking, since I see a lot of people get this wrong. If the derivative is defined (and dx is a nonzero infinitesimal), then the equation is true. However, the derivative is only defined if the right-hand side has the same value for all choices of dx.

    For the simplest stuff, the difference between standard and non-standard methods is usually quite superficial; they generally have the same issues and concerns and you can translate directly back and forth. The winnings of non-standard analysis don't really show up until you start putting things together to make more complex arguments or start treating more sophisticated notions.
  7. Feb 28, 2012 #6


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    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?
  8. Feb 28, 2012 #7
    Terence Tao wrote an absolutely clear and brilliant exposition of ultrafilters and nonstandard analysis that develops the subject from the ground up without much background required.

  9. Feb 28, 2012 #8


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    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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