Timeline for Are there results unique to non-standard analysis or surreal numbers that have not been reconciled with classical analysis?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26 at 15:44 | vote | accept | Sergey Grigoryants | ||
| May 8 at 12:52 | vote | accept | Sergey Grigoryants | ||
| May 8 at 12:52 | |||||
| May 5 at 7:12 | comment | added | Mikhail Katz | @TimothyChow: moreover, they are not that relevant to analysis, either. | |
| May 5 at 7:12 | answer | added | Mikhail Katz | timeline score: 5 | |
| Apr 17 at 18:37 | comment | added | Timothy Chow | I don't think that nonstandard analysis and surreal numbers should be mentioned in the same breath here, as if somehow, surreal numbers are highly relevant to nonstandard analysis. As explained elsewhere on MO, the surreal numbers are not particularly relevant to nonstandard analysis. | |
| Apr 17 at 16:54 | comment | added | Michael Hardy | One might think that if two assignments of probabilities to $0,1,2,3,\ldots$ summing to $1$ differ by a nonzero infinitesimal for every such integer, then their moments likewise differ by nonzero infinitesimals. But it can happen that their $n\text{th}$ moments for every $n \in \{1,\ldots, N\}$ for some infinite integer $N$ agree exactly, and for $n>N$ differ by more than an infinitesimal. That is what happens with the $\operatorname{Poisson}(1)$ distribution and the distribution of the number of fixed points of a uniformly distributed random permutation of a set of cardinality $N. \qquad$ | |
| Apr 17 at 16:48 | comment | added | Michael Hardy | For example $e^{a+\varepsilon} - e^a$ is infinitesimal if $\varepsilon$ is infinitesimal and $a$ is real. But if $a$ is an infinitely large nonstandard real, then for some infinitesimals $\varepsilon,$ that quantity is not infinitesimal. Similarly, if $a\ne0$ is infinitesimal then for some such $\varepsilon,$ $\sin(1/(a+\varepsilon)) - \sin(1/a)$ is not infinitesimal. | |
| Apr 17 at 16:45 | comment | added | Michael Hardy | Here is a point of view on uniform continuity that seems to belong only to nonstandard analysis. It is not something that cannot be "reconciled" with standard methods, so I'm putting it in a comment rather than an answer. "Everybody knows" that a standard function $f$ is continuous if for every real number $a,$ the nonstandard counterpart of $f$ satisfies $f(a+\varepsilon)-f(a)$ is infinitesimal whenever $\varepsilon$ is infinitesimal. But if the same is true of every nonstandard real number $a$ rather than only of every real number $a,$ then $f$ is uniformly continuous. And conversely. | |
| Apr 17 at 16:26 | history | asked | Sergey Grigoryants | CC BY-SA 4.0 |