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Questions tagged [nonstandard-analysis]

Non-standard ordered fields are fields which have infinitesimals, that is, positive numbers which are smaller than any positive real number. Non-standard analysis is analysis done over such fields (e.g. hyperreal fields). Please specify the exact framework for non-standard analysis you are using in your question (e.g., what definition of "hyperreal number" you are using).

A non-zero element $\varepsilon$ of an ordered field is infinitesimal if $|\varepsilon| < \frac{1}{n}$ for all $n \in \mathbb{N}$. Nonstandard analysis is analysis done in fields with infinitesimals.

There are many ordered fields which contain infinitesimals, but the most common is the hyperreal field. Denoted by ${}^*\mathbb{R}$, the hyperreal field has a subfield isomorphic to $\mathbb{R}$ and is therefore the perfect setting for formalising the arguments of Leibniz and Newton (without the need for limits). This came to fruition in the 1960's thanks to the work of Abraham Robinson.

See also .

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What are the advantages/disadvantages of non-standard analysis?

I'm not interested in an in-depth answer. Here are some specific questions for which I couldn't find an answer: With non-standard analysis, can we solve problems that can't be solved using standard analysis? Do we have some results that differ from…
user16538
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Is the empty set internal?

Is the empty set internal or not? And is there a proof (either way), or is it just a convention? If it's just a convention, why was that particular convention chosen?
Pablo Herrera
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A basic, basic question about the definition of a superstructure

A superstructure $V(X)$ over a set $X$ is defined as: $V_0(X) =X$ $V_{i+1}(X) = V_i(X) \cup P(V_i(X))$ $V(X) = ⋃_{i=0}^{\infty}V_i(X)$ My question is in regard the line item 2, where the set $V_{i+1}(X)$ is defined as the union of the set $V_i(X)$…
user153768
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Nonstandard analysis without transfer principle and mathematical logic

I noticed that a big part of nonstandard analysis aimed to work with ordinary real valued functions. It is a bit strange for me because when we expand rational numbers to reals, we do not formulate theorems for functions $\mathbb Q \to \mathbb Q$ -…
TheWildPalms
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How is it possible that $dx$ contains $dx^2$ $N$ and $N+1$ times simultaneously?

Let's assume that $N$ is even infinite hyperinteger and $N=1/dx$, $N=dx/dx^2$. Let's assume that we have expression $dx+dx^2$. We can add up $dx^2$ to $dx$ and $dx$ is obtained: $dx+dx^2=dx$ (algebraic property of infinitisemals). So, $dx$ contains…
Mike_bb
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When is the transfer a subset not its inclusion Into Ultra-power?

Prompted by a really cool proof of the Hahn–Banach theorem that relies only on nonstandard-analysis and the ultrafilter lemma, I have decided to take some time to learn more about the non-standard analysis. This quest has lead me to this paper and…
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A positive number smaller than any positive infinitesimals

I've been searching everywhere for an answer, but to no avail. I know that the hyperreals are a quotient of the space $\mathbb{R}^\mathbb{N}$ of sequences of real numbers modulo a fixed free ultrafilter on $\mathbb{N}$. Can this method be…
phst
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How to show that $f : {}^\ast \Bbb R \to {}^\ast \Bbb R$ is bounded, if it obtains a limited value everywhere?

Let $f : {}^\ast \Bbb R \to {}^\ast \Bbb R$. For all $x \in {}^\ast \Bbb R $, there exists $y \in \Bbb R $ such that $f(x) \leq y$. How to show that there exists $r \in \Bbb R$, such that for all $z \in {}^\ast \Bbb R$ we have $f(z) \leq r$? I…
Metta World Peace
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Can a set containing non-standard elements only, have a legal set formation?

Is it possible to form a set that contains only non-standard elements legally? A legal set formation means that only internal formulas can be used to form the set. Here's what I thought: A set containing non-standard elements must be a non-standard…
Metta World Peace
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Lectures on Non-Standard Analysis Book

I'm undergrad Applied Math Student and I'm writing my thesis about Non-Standard Analysis. I find a Spriger Book named Lectures On Non-Standard Analysis which are based on short courses of lectures delivered by Machover (first part) and the Master of…
Sergio Ivan
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Is the set $A=\{1,2,\ldots,\omega\}$ internal, when $\omega$ is infinite

Is the set $$ A=\{1,2,\ldots,\omega\}, $$ internal, when $$ \omega \in ^*\!\!\mathbb N, $$ is an infinite natural. I think the answer is no, because $$ ^\circ(\omega), $$ the standard part of $\omega$, is not in $\mathbb N$. But then, I'm…
user14108
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Hyperreal Numbers (Sequence Definition)

I am trying to understand the definition of a limit (for a sequence) regarding hyperreal numbers converging to $L$. The definition (see link here) states a real sequence of numbers converges to $L$ if every infinite hypernatural $H$, $x_H$ is…
W. G.
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Nonstandard part of a limited hyperreal

Let $b$ be a limited hyperreal and $x$ be its standard part, i.e. the unique real number infinitely close to $b$. Is it true that one can find an infinitesimal $\varepsilon$ such that $$b = \frac{x}{1+\varepsilon}$$
user34870
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Ordering the hyperreals and infinitesimals

I'm just getting into the hyperreals and infinitesimals and I would like to understand how one can determine when e is <, = or > g (where e and g are elements of the the infinitesimals). How does one order this field?
Spanki
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Let $f:\mathbb{^*R}\to \mathbb{^*R}$ be an external function with $(x,y\in\mathbb{^*R},x\approx y , x\le y ) \implies f(x)\le f(y)$. Is $f$ monotonic?

My main problem is that there's two methods yielding two different results: 1.We can count through $\mathbb{^*R}$ using nothing but infinitesimal steps. For example, we can partition the interval $^*[0,1]$ into $h\in\mathbb{^*N-N}$ steps, creating…
Sudix
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