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I know that there are an infinite number of hyperreals. But is it true that there are only two hyperreals with standard part equal to $0$ (the "finite" infinitesimal one and the "infinite" hyperreal)?

Put differently, is it wrong to view the hyperreals as a field "generated" by $\mathbb{R} \cup \{\infty, 1/\infty\}$ whereby every real number $r \in \mathbb{R}$ is associated with its hyperreal shadow $s = r + 1/\infty$ with $s \approx r$ uniquely?

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    $\begingroup$ You can also see Hyperreals from the perspective of the compactness and Lowenheim-Skolem theorems in logic: once you have a model , you can find models of any infinite cardinality; the Hyperreals are an uncountable model for the structure of the Reals. $\endgroup$ Oct 23 '13 at 4:26
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Yes, it’s completely wrong. For example, if $\epsilon$ is any positive infinitesimal (i.e., a positive hyperreal whose standard part is $0$), then so is $\epsilon^2$, and of course $0<\epsilon^2<\epsilon$, so $\epsilon^2\ne\epsilon$, and $\epsilon^2$ is therefore another hyperreal whose standard part is $0$. For that matter, $\epsilon x$ is a positive infinitesimal for each positive standard real number $x$, and no two of these infinitesimals are equal.

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  • $\begingroup$ Can real numbers be infinitely close to multiple different hyperreals? I know that each finite hyper real is associated with a unique real, but what about in the reverse direction? $\endgroup$ Oct 23 '13 at 4:12
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    $\begingroup$ @user1770201: Yes, each real is infinitely close to infinitely many different hyperreals. There are infinitely many infinitesimals, and if $x\in\Bbb R$, then $x+\epsilon$ is a hyperreal infinitely close to $x$ whenever $\epsilon$ is an infinitesimal. $\endgroup$ Oct 23 '13 at 4:14
  • $\begingroup$ If @Brian is correct ("Yes, each real is infinitely close to infinitely many different hyperreals. There are infinitely many infinitesimals, and if x∈R, then x+ϵ is a hyperreal infinitely close to x whenever ϵ is an infinitesimal.") it would seem to me that the Hyperreal numbers (since they are so abundant) deserve a different cardinality greater than that of the real numbers. However from a related question (Cardinality of the set of hyperreal numbers), the cardinality of the Hyperreal numbers is 2 to the aleph not. $\endgroup$
    – user120660
    Jan 11 '14 at 19:36
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    $\begingroup$ Bart, you should post this as a different question. You can link to this one. But before you do, let me tell you why your argument is very insufficient. If you order $\Bbb R^2$ lexicographically, you can think about it as $(x,y)$ where $x$ is a real number and $y$ is an infinitesimal (this is not quite true, of course, but you can think about this for a moment). Near every real number there are continuum many "infinitesimally close" numbers. But the cardinality of $\Bbb R^2$ is the same as that of $\Bbb R$. The fact that "every point has many points near by" do not necessarily increase size. $\endgroup$
    – Asaf Karagila
    Jan 11 '14 at 19:48
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The wording of your question suggests that you are assuming that the standard part of an infinite hyperreal is equal to $0$. This is not correct. The standard part function is only defined on the subring of ${}^\ast\mathbb{R}$ given by limited (finite) hyperreals. Thus, the standard part of an infinite (unlimited) hyperreal is undefined.

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  • $\begingroup$ It is fair, incidentally, to extend the standard part function so that $\operatorname{st}(x) = +\infty$ for every positive unlimited extended hyperreal. (indeed, the positive unlimited extended hyperreals are precisely the monad at $+\infty$) $\endgroup$
    – user14972
    Feb 15 '14 at 10:05

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