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Results tagged with nonstandard-analysis
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user 1946
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 $ …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k