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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.
17
votes
2
answers
2k
views
Do the surreal numbers enjoy the transfer principle in ZFC?
The surreal field $\newcommand\No{№}\No$ is definable in ZFC, and it is easy to see that the surreal order is $\kappa$-saturated for every cardinal $\kappa$, precisely because we fill any specified ga …
8
votes
In hyperreal field, can ln(ε) and ln(ω) be expressed as infinite sums?
To help avoid any misunderstanding that may arise for readers of this question, let me say that when understood in the usual sense, there are no nontrivial convergent sequences or series at all in the …
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 …
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 …
47
votes
4
answers
4k
views
Which topological spaces admit a nonstandard metric?
My question is about the concept of nonstandard metric space that would arise from a use of the nonstandard reals R* in place of the usual R-valued metric.
That is, let us define that a topological sp …
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 …
70
votes
Accepted
A remark of Connes on non-standard analysis
...as soon as you have a non-standard number, you get a non-measurable set.
Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X …
22
votes
5
answers
1k
views
What is the spectrum of possible cofinality types for cuts in an ordered field? Or in a mode...
I am interested to know the full range of possibilities for the cofinality type of cuts in an ordered field and in other structures, such as nonstandard models of arithmetic.
Definitions. Specifical …
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 …
7
votes
1
answer
562
views
Are the definable hyper-reals, using quantifiers only over the standard reals and natural nu...
This question arose today at Yevgeny Gordon's talk, "Will nonstandard analysis be
the analysis of the future?" at the CUNY Logic
Workshop. Here is my way of asking it.
Consider the ordered real field …
10
votes
Automorphisms of the hyperreals over the rationals and nontrivial automorphism groups
As a general principle, every first-order theory with infinite models, such as the theory of real-closed fields, will have models with rich automorphism groups.
The general reason is that one can ta …
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 …
29
votes
Accepted
What are the advantages of the more abstract approaches to nonstandard analysis?
To my way of thinking, there are at least three distinct
perspectives one can naturally take on when undertaking work in
nonstandard analysis. In addition, each of these perspectives can
be varied on …
3
votes
"Lebesgue-measurable" cardinals and real-closed fields
You say that $\kappa$ admits a $[0,1]$-valued measure, but actually, it is a $\{0,1\}$-valued measure; every subset of $\kappa$ has measure $0$ or $1$. If I am understanding your suggestion, you inten …
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 …