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Apr 10, 2017 at 1:39 history edited Todd Trimble CC BY-SA 3.0
added a word to ensure countability
Jan 25, 2015 at 2:48 comment added Noah Schweber Again, "$No$" is from the surreals, not the hyperreals.
Jan 25, 2015 at 2:30 comment added Anixx @Noah S I think in $No(\omega_1)$ there is. But I still would like a clarification.
Jan 24, 2015 at 16:34 comment added Noah Schweber Anixx, based on your most recent comment I think you might be confusing the hyperreals with the surreals, which are two very different things. In particular, there's no "special" element of 'the' hyperreals which we deem "$\omega$."
Jan 24, 2015 at 14:50 comment added Todd Trimble @JoelDavidHamkins I was confused it seems; thanks for setting me straight! I'm going to use my mod powers to edit my comment now.
Jan 24, 2015 at 14:47 comment added Joel David Hamkins Todd, one may refer to $\sin(\omega)$ or $e^\omega$ for any hyperreal $\omega$, including infinite elements, simply by appealing to the transfer principle---every function on the reals has a nonstandard analogue, which satisfies all the same first-order expressible properties in the hyperreals that the original function did in the standard reals.
Jan 24, 2015 at 14:15 comment added Anixx Actually to be more precise I was referring to Conway's $No$ ohio.edu/people/ehrlich/Unification.pdf
Jan 24, 2015 at 13:41 comment added Todd Trimble Anixx, I think a problem here is exactly how you are defining the order if you are considering $x$ to be infinite and considering power series expressions such as $\sin x$. Is this going to be positive or negative (or zero)? I think you have to be careful here!
Jan 24, 2015 at 13:31 comment added Anixx So if we take $\omega$ as infinite element, there are formal power series (like $\sin \omega$) which are not hyperreals apparently, so the power series seems to be a bigger set than hyperreals?
Jan 24, 2015 at 13:24 history edited Anixx CC BY-SA 3.0
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Jan 24, 2015 at 13:19 comment added Todd Trimble Oh, I see (and sorry I was confusing). Right, the way that Puiseux series are usually set up, the $x$ will be infinitesimal [I edited a mistake here in my answer; I often get confused which way it should go], and there, there is no such series for that would express an infinite element such as $e^y$ where $y = x^{-1}$.
Jan 24, 2015 at 13:02 history edited Todd Trimble CC BY-SA 3.0
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Jan 24, 2015 at 13:01 comment added Anixx So $\sin (1/\omega)=\sin \varepsilon$ is definitely a hyperreal and infinitesimal, but what about $\sin \omega$ then?
Jan 24, 2015 at 12:56 comment added Anixx Oh, it seems you misunderstood me! In my question I meant $\omega$ to be infinite, as usually in non-standard analysis it is postulated to be equal to the first ordinal $\omega$.
Jan 24, 2015 at 12:55 comment added Todd Trimble Yes, $\sin \omega$ will be infinitesimal.
Jan 24, 2015 at 12:54 history edited Todd Trimble CC BY-SA 3.0
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Jan 24, 2015 at 12:54 comment added Anixx "No, you can form a hypperreal $\sin ω$". - hmm, I am curious, will it be an infinitesimal?
Jan 24, 2015 at 12:51 comment added Anixx I think the last comment is very valuable, can you please add this to the answer?
Jan 24, 2015 at 12:50 vote accept Anixx
Jan 24, 2015 at 12:48 comment added Todd Trimble (1) No, you can form a hypperreal $\sin \omega$. (2) Yes, one can regard the field of formal power series as an ordered subfield of the hyperreals.
Jan 24, 2015 at 12:45 comment added Anixx If the hyperreals are bigger, can we say that the formal power series (or converging power series or analytic functions) are isomorphic to a subset of hyperreals?
Jan 24, 2015 at 12:37 comment added Anixx By the way, $\sin \omega$ does belong to the formal power series but does not to hyperreals, is this wrong or correct?
Jan 24, 2015 at 12:31 history answered Todd Trimble CC BY-SA 3.0