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Suppose, we want to add positive elements that are smaller than all positive elements in ${}^{\ast}{\Bbb R}$. One way to do this, as shown in this master’s thesis, is to construct sequences of elements in ${}^{\ast}{\Bbb R}$ that range over ${}^{\ast}{\Bbb N}$ and find an ultrafilter on ${}^{\ast}{\Bbb N}$, ${}^{\ast}{\mathcal U}$.

Can we find such ultrafilter ${}^{\ast}{\mathcal U}$? Do we need to require all sets in ${}^{\ast}{\mathcal U}$ to be internal sets? Or we have to live with a cheap version, ${}^{\ast}\mathcal {P}{(\Bbb N)} \cap {}^{\ast}{\mathcal U}$?

The author want to eliminate the possibility of constructing an positive element in ${}^{\ast}\Bbb R$ that is strictly smaller than all positive elements in ${}^{\ast\ast}\Bbb R$ by excluding $\Bbb N$ from ${}^{\ast} \mathcal U$. But it seems to me it's far from enough. ${}^{\ast} \mathcal U$ shouldn't contain any element with the cardinality of $\Bbb N$. I think what he need is a uniform ultrafilter in which all elements have the cardinality of $2^{\aleph_0}$. Is it right?

Metta World Peace
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  • I don't understand your question fully. Can't we just let $\mathcal F = {A \subseteq {}^\mathbb N \mid \text{ there is a finite set $E$ such that } {}^\mathbb N - A = \mathbb N \cup E }$. Then $F$ is a filter, take an ultrafilter ${}^\mathcal U$ containing it. Then ${}^\mathcal U$ is an ultrafilter on ${}^*\mathbb N$ not containing $\mathbb N$ as wished. – martini May 22 '13 at 15:11
  • @martini:Both $\Bbb N$ and ${}^{\ast} \Bbb {N} \setminus\Bbb N$ are external. If we exclude $\Bbb N$, then ${}^{\ast}\mathcal U$ must contain ${}^{\ast} \Bbb {N} \setminus\Bbb N$, right? – Metta World Peace May 22 '13 at 15:33
  • Yes, but the author of the thesis you linked remarks that we need to exclude $\mathbb N$ to get what we want. – martini May 22 '13 at 18:20
  • @martini: Thank you for your comment. I feel less puzzled now.The author want to eliminate the possibility of constructing an positive element in ${}^{\ast}\Bbb R$ that is strictly smaller than all positive elements in ${}^{\ast\ast}\Bbb R$ by excluding $\Bbb N$ from ${}^{\ast} \mathcal U$. But it seems to me it's far from enough. ${}^{\ast} \mathcal U$ shouldn't contain any element with the cardinality of $\Bbb N$. I think what he need is an uniform ultrafilter in which all elements have the cardinality of $2^{\aleph_0}$. – Metta World Peace May 22 '13 at 19:18
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    I didn't dive into the thesis this deep, but this can be achieved, I think. Let $\mathcal F = {A \subseteq {}^\mathbb N \mid |{}^\mathbb N - A| < |{}^\mathbb N|}$. This is a filter, take an ultrafilter ${}^\mathcal U$ containing it. ${}^\mathcal U$ cannot contain a set of cardinality less then $|{}^\mathbb N|$. – martini May 22 '13 at 19:47
  • @martini: Thanks for the information. – Metta World Peace May 22 '13 at 19:51
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    I don’t think that he needs a uniform u.f.: he just needs to be sure that no member of $\mathscr{U}$ is countable. Any u.f. extending the co-countable filter on ${}^*\Bbb N$ should do nicely. – Brian M. Scott May 23 '13 at 07:43
  • @BrianM.Scott: I see. Uniform ultrafilter is too strong, unless we assume continuum hypothesis. – Metta World Peace May 23 '13 at 08:13

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Why would you need such a thing as **R? To define a *Riemann integral over infinitesimal intervals. And why would one want to do that? Because that is one way to integrate over fractals, where, for instance, the Cantor set is 2^omega intervals of length 2^-omega. Now integrate your function over each of these intervals, and add them all up.

user110946
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$^\star\mathbb{R}$ only satisfies countable saturation, so to defeat it it suffices to take an index set of cardinality the continuum in the ultraproduct construction, so using $^\star\mathbb{N}$ is fine. It is not entirely clear to me why one needs such a $^{\star\star}\mathbb{R}$.

Mikhail Katz
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    What is you are doing calculus, except that your functions are already defined on the hyperreals. Then you would need numbers smaller than any hyperreal. Ergo hyperhyperreal. – Christopher King May 30 '13 at 00:02
  • I understand that, but practically speaking this can be done within a single $^\star\mathbb R$ suitably defined. I don't think one needs the iterated construction. I recall seeing a discussion of this in one of Terry Tao's blogs. If you are interested I can try to look it up. – Mikhail Katz May 30 '13 at 07:06
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    Well, sure a single $\mathbb{R}**$ could be used practically. But that would be boring. – Christopher King May 30 '13 at 16:39
  • Infinitesimals are never boring. See http://link.springer.com/article/10.1007%2FBF03023265?LI=true# :-) – Mikhail Katz May 31 '13 at 08:42
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    I have had occasion to make use of the iterative construction -- I had to iterate something like four times to make the intuitive argument I wanted to make. (although technically it was semi-algebraic, so I just needed real closed fields rather than hyperreals, but the same idea) –  Feb 10 '14 at 14:03
  • @Hurkyl, that's interesting. Can you elaborate? – Mikhail Katz Feb 10 '14 at 16:06
  • It's been long enough I don't recall the details, but it was the sort of thing where I had to make a sufficiently small perturbation to a function to change a double root into two single roots, so I make an infinitesimal one. But now I have two roots that are infinitesimally separated, and I need to pick a third point that is sufficiently near one but not the other... and other similar things. At each step, it was far easier to just add a new infinitesimal, rather than try to argue how "close" was "close enough". –  Feb 10 '14 at 19:47