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Questions about hyperreal numbers/functions

  1. Jun 15, 2009 #1
    Questions about hyperreal numbers, as used in non-standard analysis. Please be patient with this long post. This should probably go in linear & abstract algebra forum, but it has ideas from linear algebra, set theory, and calculus. doesn't fit into any one of the forums.

    1. From Wiki: "Unlike the reals, the hyperreals do not form a standard metric space..." Why not?
    From definition of metric space: d: SxS -> R(reals) is a metric on S iff: d(x,y) = 0 iff x=y (infinitely close numbers are the same); d(x,z) ≤ d(x,y) + d(z,y). Is it because the metric on *R is not |x-y|, but st(|x-y|), (st is standard part function; allowing some numbers to be infinitely close to each other)?

    2. What are the resp. cardinalities of the hypernaturals *N and the hyperintegers *Z? If *Z is countable, then it will establish countability of *Q (the hyperrationals; set of all a/b such that a, b are in *Z, and b is nonzero). It will mean that not all hyperreal numbers (*R has cardinality of reals = uncountable) can be expressed as *Q (cf. question 4). It will also mean that *Q will be "finer" (has smaller numbers/distances) than R but still has less numbers than R, which is very counterintuitive...

    3. Can digits go on beyond the infinite (digits in the infinite H-th place) digits, truly forever, in *R?

    4. Are there "hyper-hyperreals", "hyper-hyper-hyperreals", [...] "hyper^n-reals", etc? apologies for some informal abuse of "^n".

    5. usage the transfer priniple when defining hyperreal extensions of functions defined in terms of integrals, for example, erf and the gamma function. I think you will need to extend the epsilon-delta limit definition "for all real ε>0, there is a real δ>0..." to "for all HYPERreal ε>0, there is a HYPERreal δ>0...". I can't think of anything else besides introducing "hyper-hyperreals" to have the "standard hyperreal part", and i think the limit is the only way to go.
     
    Last edited: Jun 16, 2009
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  3. Jun 16, 2009 #2

    CRGreathouse

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    st(|x-y|) isn't a metric. Why?
     
  4. Jun 16, 2009 #3

    Hurkyl

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    You're refering to the external cardinality I assume? The usual nonstandard model, I believe, has |*N| = |*R| = |R|.

    (Note that *R is not hypercountable -- there does not exist an internal bijection between *N and *R)


    In a hyperdecimal number, there is one digit for each hyperinteger.

    Just like in a standard decimal number, there is one digit for each standard integer.

    There are lots of nonstandard models of real analysis, of arbitrarily large size. It would be fair to call the corresponding internal reals a "set of hyperreal numbers". However, "the hyperreals" refers to a specific nonstandard model. I'm not aware of other nonstandard models having specific names.

    I can't work out what you're thinking here; I think you're confused. I think you miss the point of the transfer principle.

    The three statements are, I believe, equivalent definitions of (internal) limits of hyperreals:

    1. The operator [itex]{}^\star\lim[/itex] is the transfer of the standard limit

    2.
    [tex]{}^\star \lim_{x \rightarrow a} f(x) = L[/tex]
    if and only if
    For all [itex]\epsilon \in {}^\star \mathbb{R}[/itex] with [itex]\epsilon > 0[/itex], there exists a [itex]\delta \in {}^\star \mathbb{R}[/itex] with [itex]
    \delta > 0[/itex] such that for all [itex]x \in {}^\star \mathbb{R}[/itex]: [itex]0 < x < \delta[/itex] implies [itex]|f(x) - L| < \epsilon[/itex]

    3.
    [tex]{}^\star \lim_{x \rightarrow a} f(x) = L[/tex]
    if and only if
    For all [itex]x \in {}^{\star \star} \mathbb{R}[/itex] with [itex]{}^\star st(x) = a[/itex] and [itex]x \neq a[/itex], we have [itex]{}^\star st(f(x)) = L[/itex].


    Which one is simpler? :wink: (I hope all of my notation was clear)


    The transfer principle is the whole reason why we care about nonstandard analysis; if you're not going to use it, then you shouldn't bother with hyperreals.

    Normally, if we were to invent some auxiliary number system to help us study real analysis, we would have make all sorts of definitions and prove all sorts of theorems about the auxiliary number system before we could really use it to help us study the real numbers.

    But the big idea behind nonstandard analysis is that every definition or theorem we prove about the real numbers automatically becomes a definition or theorem about the hyperreals, because of the transfer principle. If I want to use the hyperreal gamma function, I don't have to do any work at all -- I don't need to make any definitions, I don't need to prove any theorems, I just transfer the standard gamma function.
     
    Last edited: Jun 16, 2009
  5. Jun 16, 2009 #4
    @CRGreathouse: I meant, is st(|x-y|) the distance function on the set *R? is that why *R is not a metric space?

    @Hurkyl: *N is uncountable, then is *Q is uncountable, like the standard reals R? show me that there's no bijection between *N and N? also, if |*N|= |*R| they should have a bijection.

    also, every arithmetical truth about the real numbers is true for *R. you can just extend standard operators. then i don't think there's any point in nonstandard analysis.
     
    Last edited: Jun 16, 2009
  6. Jun 16, 2009 #5

    CRGreathouse

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    If you go through the definition of a metric, you'll see that st(|x-y|) fails. I was hoping you could figure out which part of the definition it fails on -- clearly, it;s related to the 'standard part' function.
     
  7. Jun 16, 2009 #6

    CRGreathouse

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    That's a silly statement. What about statements relating to infinitesimals, where the nonstandard numbers can make finer distinctions than the real numbers? What if the nonstandard numbers are easier to work with in some cases, so the fastest way to prove an arithmetical property is though *R?
     
  8. Jun 16, 2009 #7

    Hurkyl

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    d(x,y) = |x-y| is not an (external) metric because it isn't real valued.
    d(x,y) = st(|x-y|) is not an (external) metric, because it doesn't have the property "d(x,y)=0 implies x=y"


    I'm going to assume you're familiar with the construction of *N as an ultrapower of N.

    Suppose sn be a sequence of elements of *N.
    For each n, Let tn,m be a sequence of elements of N that represent sn.

    Define a sequence
    [tex]b_n = 1 + \max_{x, y \leq n} t_{x,y}[/tex]

    and let a be the hyperinteger that corresponds to the sequence bn.

    It shouldn't be too hard to show that a > sn for each n.

    Therefore, for any countable subset of *N, there exists a hyperineger not contained in that subset.


    (Every analytical truth as well) But there's one more ingredient -- the fact you can (externally) relate R with *R. e.g. [itex]\mathbb{R} \subseteq {}^\star \mathbb{R}[/itex].

    The fact we can relate R with *R gives us a way to make convenient statements about R. The transfer principle gives us a way to make convenient statements about *R. When put together, they are incredibly useful.
     
  9. Jun 16, 2009 #8
    only at CRGreathouse: so *R is a pseudometric space, in which indistinguishable (st(|x-y|=0) numbers do not have to be the same.

    I meant that the standard part definitions of the derivative and the integral are basically the same as the limits definition, because the epsilon-delta definition of limit is equivalent to the standard part definition, isn't it? both involve a kind of "rounding" process.

    Hurkyl, you missed this. if |*N|= |*R| they should have a bijection between each oterh.
     
  10. Jun 16, 2009 #9

    Hurkyl

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    This is one of my favorite proofs of all time:


    Theorem: Let f be a continuous, real-valued function on the interval [0,1]. Then, f has a maximum.

    Proof: Let H be a transfinite hyperinteger. The (internal) set [itex]\{ {}^\star f(x/H) \mid 0 \leq x \leq H \}[/itex] is hyperfinite, and thus has a maximum. Let a be a hyperinteger such that f(a/H) is that maximum.

    Let M = st *f(a/H)
    Because f is continuous, M = f( st(a/H) ).

    Let x be any real number in [0,1]. Let b = floor(xH). Observe:

    [tex]f(x) \approx {}^\star f(b / H) \leq {}^\star f(a/H) \approx M[/tex]

    therefore [itex]f(x) \leq M[/itex].

    We have shown M is an upper bound, and that f attains the value M. Therefore, M is the maximum of f on [0,1].



    The intuitive sketch:

    1. Cover [0,1] with a (hyperfinite) set of points, making sure to have at least one in the monad of each standard point.

    2. The value of the function at any standard point is infinitessimaly close to the value at one of our chosen points.

    3. Finite sets have a maximum.
     
    Last edited: Jun 16, 2009
  11. Jun 16, 2009 #10

    Hurkyl

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    That one's easy (at least, it is if we assume the continuum hypothesis.

    *N is uncountable, therefore [itex]|{}^\star \mathbb{N}| > |\mathbb{N}|[/itex].
    By the continuum hypothesis, this implies [itex]|{}^\star \mathbb{N}| \geq |\mathbb{R}|[/itex].
    Elements of *N are named by elements of NN. (That is, by sequences of natural numbers) Therefore, [itex]|{}^\star \mathbb{N}| \leq |\mathbb{N}|^{|\mathbb{N}|} = |\mathbb{R}|[/itex].
    Therefore [itex]|{}^\star \mathbb{N}| = |\mathbb{R}|[/itex].

    *R contains R, therefore [itex]|{}^\star \mathbb{R}| \geq |\mathbb{R}|[/itex].
    Elements of *R are named by elements of RN. (That is, by sequences of real numbers) Therefore, [itex]|{}^\star \mathbb{R}| \leq |\mathbb{R}|^{|\mathbb{N}|} = |\mathbb{R}|[/itex].
    Therefore [itex]|{}^\star \mathbb{R}| = |\mathbb{R}|[/itex].

    Therefore [itex]|{}^\star \mathbb{R}| = |{}^\star \mathbb{N}|[/itex].

    I believe this can be proven without assuming CH, but I don't see an easy way to do it.
     
  12. Jun 16, 2009 #11
    what is the diff between "external" bijections and "internal" bijections, such that there is an external bijection but no internal bijection between *N and *R?
     
  13. Jun 16, 2009 #12

    Hurkyl

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    If I had to give a short answer, I would say that:
    1. If you produce something by the transfer principle, it's internal
    2. If you study the nonstandard model with nonstandard analysis, it's internal
    3. If you study the nonstandard model with standard analysis, it's external

    (internal / external only applies to the nonstandard model)

    e.g. we should expect "the set of all finite hypernatural numbers" is an external set, because the definition of "finite" invokes the standard model.


    However, I really don't think one can reasonably give a comprehensive answer in this medium -- your best chance of understanding it is probably to study reference materials.
     
  14. Feb 7, 2011 #13
    Yes, there are as many "hypers" as you like. One of my students did a thesis on the extension to second order hyperreals, which can be constructed as sequences of hypereals, indexed by the hypernaturals and sorted into equivalence classes by a co-countable ultrafilter (no countable sets)

    The construction can be continued in a natural way.
     
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