Talk:Hyperfinite set

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This is really only a rough start, since I have neither the sourcing nor the knowledge to provide a more detailed coverage (non-standard analysis is itself new to me). All expansion/correction is welcome. —Anonymous Dissident^Talk 11:58, 6 December 2009 (UTC)Reply

The ultrapower construction of the hyperreals provides the context where hyperfinite sets can be described explicitly. An arbitrary hyperfinite set ${\displaystyle [A_{n}]}$ in *R is defined by a sequence ${\displaystyle \langle A_{n}\rangle }$ of finite sets ${\displaystyle A_{n}\subset \mathbb {R} ,n=1,2,\ldots }$, see Goldblatt (1998).

Goldblatt, R. (1998) Lectures on the hyperreals. An introduction to nonstandard analysis. Graduate Texts in Mathematics, 188. Springer-Verlag, New York. —Preceding unsigned comment added by Tkuvho (talk • contribs) 15:06, 6 December 2009 (UTC)Reply

Perhaps "most other" should be changed to "finite". Tkuvho (talk) 12:40, 7 December 2009 (UTC)Reply

True. Fixed. —Anonymous Dissident^Talk 13:37, 7 December 2009 (UTC)Reply

The current version states that "Hyperfinite sets may be used to approximate other sets, where approximation in this context is considered to be a minimisation of symmetric difference". I am not sure what this is referring to. Tkuvho (talk) 12:43, 7 December 2009 (UTC)Reply

I've removed it. Could you provide a better definition of "approximate", since I wasn't entirely sure how to express it. —Anonymous Dissident^Talk 13:13, 7 December 2009 (UTC)Reply

I think the point is that every real set can be approximated, in the larger *R domain, by a hyperfinite set in such a way that every point of the real set is infinitely close to a suitable member of the hyperfinite set. Typically one may work with partitions of a compact interval into infinitesimally short subintervals of equal length. Tkuvho (talk) 13:17, 7 December 2009 (UTC)Reply

Yes, and the best way I could think of to express that was through symmetric difference, which is intuitively but not technically clear. I'll keep thinking about the phraseology. —Anonymous Dissident^Talk 13:24, 7 December 2009 (UTC)Reply

Perhaps one could say that they "look" the same under any finite microscope, no matter how powerful. To tell the difference, you need an infinite-magnification microscope. Tkuvho (talk) 13:47, 7 December 2009 (UTC)Reply

That's probably too informal an analogy. What do you think of the current phrasing? Maybe "or, roughly speaking, be "almost equal" to". —Anonymous Dissident^Talk 13:50, 7 December 2009 (UTC)Reply

I don't see that the parenthetical remark adds any new information. What one could say is that their "halos" are the same (more precisely the union of the halos of all points). Tkuvho (talk) 13:52, 7 December 2009 (UTC)Reply

Hmm. I think I'll just remove it – it does seem redundant. —Anonymous Dissident^Talk 13:57, 7 December 2009 (UTC)Reply

"subsets of hyperfinite sets are not hyperfinite, often because they do not contain the extreme elements of the parent set." The reference to Ambrosio's text is a little puzzling. Did he have anything to say about hyperfinite sets? Tkuvho (talk) 12:45, 7 December 2009 (UTC)Reply

Yes, he said something very similar to that. —Anonymous Dissident^Talk 13:41, 7 December 2009 (UTC)Reply

Still, this statement is a little too obscure than is necessary. Thus, the simplest example of a hyperfinite set is an initial string {1,2,3,...;...,H} of *N, where H is an infinite hypernatural. This set contains the set of natural numbers N as every element of N is less than H. Now N is not hyperfinite since it is not an internal set. I think every internal subset of a hyperfinite set will also be hyperfinite, but I don't have a reference right now. If this is true, then the sentence needs to be simplified. It could be that in some setting more general than *R, something more complicated could happen. Do Ambrosio et al give any indication of what's involved? Tkuvho (talk) 14:32, 12 July 2010 (UTC)Reply