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Nonstandard analysis is a way of doing calculus and analysis with infinitesimals. The historical approach of Leibniz, Euler, and others to infinitesimal calculus was gradually replaced by epsilon, delta techniques in the context of a real continuum, in the 19th century. It was not until the 1960s that Abraham Robinson developed a theory of a hyperreal continuum that allows for a development of analysis procedurally akin to that of its founders.

5 votes
Accepted

Star-transfer of powerset

$\cal{P}({}^\ast\mathbb{R})$ is the full standard power set of the nonstandard reals, the set of all subsets of ${}^\ast\mathbb{R}$. This power set includes the subsets consisting solely of infinites …
Joel David Hamkins's user avatar
9 votes
Accepted

Metrization of hyperreals

I am not sure whom you are addressing in your question, but some of your remarks relate to issues brought up at this MO question. If not, could you let us know to which post you were referring? It is …
Joel David Hamkins's user avatar
2 votes

Compactness and omega models

There is no property of $T$ alone that will ensure that $T+S$ always has an $\omega$ model in the circumstances you describe. In fact, there is no computably axiomatizable theory $T$ with the property …
Joel David Hamkins's user avatar
5 votes

Hyperreal finitely-additive measure on [0,1) assigning $b-a$ to $[a,b)$ or $(a,b]$ and infin...

I think this is a very interesting question. In response to your comment, let me argue that if 1 holds and the measure is additive, then the singleton values are all the same. This is the sense in w …
Joel David Hamkins's user avatar
9 votes
Accepted

Isomorphisms between non-standard reals.

If the continuum hypothesis holds, then both of these ultrapowers are saturated models of cardinality $\omega_1$, and one can see that they are isomorphic by a back-and-forth argument. When the CH f …
Joel David Hamkins's user avatar
4 votes

Hyperfinite set containing the reals, with specified upper bound on internal cardinality?

Andreas has pointed out that in any sufficiently saturated nonstandard model $\mathbb{R}^\ast$, the answer is yes. Meanwhile, let me point out that if, as is commonly done, one builds one's hyperreal …
Joel David Hamkins's user avatar
12 votes

Are hyperreal numbers isomorphic to formal power series?

I would like to point out that it is not true that every every hyperreal can be represented by a Laurent series in the way you describe. (Let me assume that by the term "hyperreals", you mean a nonst …
Joel David Hamkins's user avatar
4 votes

Turing degree of a turing machine with access to an (arbitrary) nonstandard integer

I shall give two different interpretations of the question. (The second interpretation using true arithmetic is modified in this update.) Using arbitrary nonstandard models of PA. Let us say that a T …
Joel David Hamkins's user avatar
31 votes
Accepted

Is non-existence of the hyperreals consistent with ZF?

The answer is yes, provided ZF itself is consistent. The reason is that the existence of the hyperreals, in a context with the transfer principle, implies that there is a nonprincipal ultrafilter on $ …
Joel David Hamkins's user avatar
56 votes

How helpful is non-standard analysis?

The other answers are excellent, but let me add a few points. First, with a historical perspective, all the early fundamental theorems of calculus were first proved via methods using infinitesimals, r …
Joel David Hamkins's user avatar
6 votes

tennenbaum phenomena for the reals?

This is a very interesting question! One way to interpret the question is like this: we have the structure $\langle\mathbb{R},{+},{\cdot},{\lt}\rangle$, which is a real-closed field, and Tarski prove …
Joel David Hamkins's user avatar
45 votes
Accepted

Does every series of hyperreal numbers converge to some hyperreal number?

The answer is strongly negative. Arbitrary extensions. The first thing to say is that whenever one extends $\newcommand\R{\mathbb{R}}\R$ to a larger ordered field $F$, one has immediately destroyed (e …
Joel David Hamkins's user avatar
2 votes

Isomorphism types or structure theory for nonstandard analysis

Let me offer one counterpoint to John's excellent answer. Under the Continuum Hypothesis, the ultrapower version of R* will be saturated in any countable language. That is, it will realize all finite …
Joel David Hamkins's user avatar
20 votes
5 answers
2k views

Isomorphism types or structure theory for nonstandard analysis

My question is about nonstandard analysis, and the diverse possibilities for the choice of the nonstandard model R*. Although one hears talk of the nonstandard reals R*, there are of course many non-i …
Joel David Hamkins's user avatar
31 votes

Why does CH imply that there is a unique ultrapower of $\mathbb{N}$?

The point is that the ultrapower of any structure $\mathcal{M}$ by a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ is countably saturated, that is, it realizes any finitely satisfiable $n$-type with …
Joel David Hamkins's user avatar

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