Talk:Hyperreal number

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Connection to P-point filter

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

(\forall \epsilon >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:

(\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

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]

Properties of infinite numbers missing.

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]

"The derivative of a function y(x) is defined not as dy/dx but as the standard part of dy/dx"

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

identical behavior.

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, $\omega$ 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]

comment

The problem may be in my poor english, but from

A hyperreal number x is said to be finite if, and only if, |x|<n for some integer n. x is said to be infinitesimal if, and only if, |x|<1/n for all positive integers .

I understand that the infinitesimal are a subset of the finite. If is true, I would state this explicitly. 176.206.13.217 (talk) 18:17, 6 June 2025 (UTC)[reply]