Talk:Hyperreal number

This article is rated B-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics Mid‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles Mid This article has been rated as Mid-priority on the project's priority scale.

Archives

Index 1

This page has archives. Sections older than 90 days may be automatically archived by ClueBot III when more than 5 sections are present.

in the subsection on the "intuitive approach", the following claim is made: "the true infinitesimals are the classes of sequences that contain a sequence converging to zero." This is an oversimplification that will not always be true. Depending on foundational assumptions, certain infinitesimal classes may not be representable by sequences tending to zero. Tkuvho (talk) 13:01, 23 June 2010 (UTC)[reply]

The sentence "The infinitesimals can be represented by the non-vanishing sequences converging to zero in the usual sense, that is with respect to the Fréchet filter" is just plain wrong. Tkuvho (talk) 13:08, 23 June 2010 (UTC)[reply]

The phrase "However, there may be infinitesimals not represented by null sequences; see P-point" was deleted in a recent edit. Why was it deleted? Tkuvho (talk) 11:51, 7 May 2013 (UTC)[reply]

What does P-point have to do with it. An infinitesimal in an ultra-power of R or Q (or any separable space) must, considered as a sequence, have zero as a limit point. It doesn't necessarily have to be "null" (converge to 0). On the other hand, a sequence converging to 0 does have to be infinitesimal (or 0). — Arthur Rubin (talk) 00:11, 8 May 2013 (UTC)[reply]

I see you said something else above in 2010. Could you explain? If U is a non-principal ultrafilter over N, then an infinitesimal in R^U must have (in R^N) a subsequence (with coordinate set in U) converging to 0; conversely if a sequence in R^N converges to 0, and has no zero components, then it is an infintesimal in R^U. — Arthur Rubin (talk) 00:17, 8 May 2013 (UTC)[reply]

This is assuming classical logic; I don't know how ultrafilters work in intuitionistic logic. — Arthur Rubin (talk) 00:19, 8 May 2013 (UTC)[reply]

The background logic is classical. The question is whether each infinitesimal is representable by a null sequence. In other words, whether a subsequence can be chosen which is supported on a member of the ultrafilter. For this to be true requires a special type of ultrafilter namely P-point (whose existence cannot be proved in ZFC). Tkuvho (talk) 11:19, 8 May 2013 (UTC)[reply]

Never mind. You're right. The sequence (a_n) corresponds to an infinitesimal iff

${\displaystyle (\forall \epsilon >0)(\left\{n\mid ||a_{n}|<\epsilon \right\}\in U).}$0)(\left\{n\mid ||a_{n}|<\epsilon \right\}\in U).}"/>

It does follow that (a_n) has a null subsequence, but it doesn't follow that the subsequence is in U. I don't see that the topological definition of P-point corresponds to the necessary property of an ultrafilter so that all infinitesimals correspond to a null sequence; It appears to be that:

${\displaystyle (\forall n)(U_{n}\in U)\rightarrow (\exists V\in U)(\forall n)((V\smallsetminus U_{n}){\text{ is finite}}).}$

— Arthur Rubin (talk) 20:16, 8 May 2013 (UTC)[reply]

This would mean that U, in ${\displaystyle 2^{\mathbf {N} }/{\text{finite}}}$, is closed under countable intersections, which might correspond to a P-point filter, although not exactly a P-point. — Arthur Rubin (talk) 20:21, 8 May 2013 (UTC)[reply]

Found it. If we use [[Ultrafilter#Ultrafilters on ω|P-point]], rather than the current redirect at [[Glossary of topology#P|P-point]], then the statement as you wrote it makes sense, although could use a a more detailed argument and a source. — Arthur Rubin (talk) 21:58, 8 May 2013 (UTC)[reply]

The source is Cutland, Nigel; Kessler, Christoph; Kopp, Ekkehard; Ross, David, On Cauchy's notion of infinitesimal. British J. Philos. Sci. 39 (1988), no. 3, 375–378. Tkuvho (talk) 12:01, 9 May 2013 (UTC)[reply]

The section headed "Properties of infinitesimals and infinite numbers" does not mention any properties of infinite numbers. Shame, because that's what I wanted to know about. Tesspub (talk) 10:28, 29 August 2014 (UTC)[reply]

This is incorrect; using Keisler's treatment ${\displaystyle \mathrm {d} x}$ and ${\displaystyle \mathrm {d} y}$ are infinitesimal increments along the tangent line while ${\displaystyle \Delta x}$ and ${\displaystyle \Delta y}$ are infinitesimal increments along the curve. So ${\displaystyle y'(x)={\frac {\mathrm {d} y}{\mathrm {d} x}}=\mathrm {st} {\frac {\Delta y}{\Delta x}}}$. 58.169.240.244 (talk) 15:17, 4 May 2015 (UTC)[reply]

This sentence "The transfer principle, however, doesn't mean that R and *R have identical behavior" is misleading. R and *R do have identical behavior as long as you don't write down statements that involve both standard and non-standard numbers. In the example, ${\displaystyle \omega }$ is a non-standard real whereas the dots ... are interpreted in the standard way (with the set of standard integers) (in other words, with a set that is undefinable in *R). Mixing standard with non-standard is really the only way that "non-identical" behavior can occur. If you stick with "all standard" or "all non-standard" then the behavior is identical. MvH (talk) 21:58, 7 February 2020 (UTC)[reply]

Well, that's if you restrict yourself to first-order logic. The models don't have all the same properties in higher-order logic. I'm not sure whether "behavior" is the right word to explain this; maybe you can offer an improvement? --Trovatore (talk) 22:02, 7 February 2020 (UTC)[reply]

There is no test that could distinguish (N, R) from (*N, *R). Everything proved for (N, R) holds for (*N, *R) and vice versa. Only if you compare (N, R) with say (N, *R) will there be different properties. Higher-order logic hides, but doesn't solve, the issue by moving it to first order set theory. The point is, the behavior is identical, as long as whenever you replace R by *R you also replace N by *N, Z by *Z, Q by *Q etc. MvH (talk) 15:46, 21 February 2020 (UTC)[reply]