What type of real sequences correspond to infinitesimals in non-standard analysis?

Question

If it is necessary to answer the following two questions using nonprincipal ultrafilters, then so be it, but if not, I would prefer simple yes/no plus handwaving answers, since I have not really immersed myself in the technicalities of constructing the hyperreals yet.

Questions:

1. Does every non-zero infinitesimal correspond to a real sequence for which (1) only finitely many of its terms are zero, i.e. "most" or almost all of them are non-zero, and (2) the limit exists and is $0$?

2. Given any sequence $(f(n))$ for which the function $f: \mathbb{N} \to \mathbb{R}$ is $o(g(n))$ for another function $g: \mathbb{N} \to \mathbb{R}$, does one automatically have that the sequence $\left(\frac{f(n)}{g(n)}\right)$ corresponds to a non-zero infinitesimal? And if we have another sequence which decays to $0$ even more quickly, does this new sequence correspond to an even smaller non-zero infnitesimal than the previous one?

Clarifying example of 2: Given a function $h: \mathbb{N} \to \mathbb{R}$ such that $g(n) \in o(h(n))$, do we have that $\left(\frac{f(n)}{h(n)} \right)$ corresponds to an infinitesimal which is even smaller than $\left(\frac{f(n)}{g(n)}\right)$?

This is the impression I have from the examples on p. 913 of Keisler's Calculus: An Infinitesimal Approach, where the author lists several real sequences which are claimed to correspond to non-zero infinitesimals, each being smaller than the previous one:

$$\begin{array}{rcl} \left( \displaystyle\frac{1}{n} \right)_{n\in \mathbb{N}} \\ \left( \displaystyle\frac{1}{n^2} \right)_{n\in \mathbb{N}} \\ \left( \displaystyle\frac{1}{2^n} \right)_{n\in \mathbb{N}} \end{array}$$

Answer 1

1. No.

2. Yes and Yes. Any sequence convergent to $0$ also represents an infinitesimal in the sequence models of nonstandard analysis.

Answer 2

The OP asked: "Does every non-zero infinitesimal correspond to a real sequence for which (1) only finitely many of its terms are zero, i.e. 'most' or almost all of them are non-zero, and (2) the limit exists and is 0?"

The answer to (1) is evident from the example the OP mentioned from Keisler: for the sequence defined by $\frac{1}{n}$ , none of the terms are zero.

The answer to (2) involves, as the OP suspected, a discussion of nonprincipal ultrafilters. It turns out that the answer is "it depends". Namely, the answer depends on the nature of the ultrafilter $\mathcal{F}$ chosen in the construction of the hyperreals via the ultrapower $\mathbb{R^N}/\mathcal{F}$. If the ultrafilter is of a type known as a "P-point" then the answer to (2) is affirmative; namely every infinitesimal is representable by a sequence tending to zero. Otherwise the answer is negative: there exist "exotic" infinitesimals that cannot be represented by a sequence tending to zero relative to the ultrapower construction. Nonetheless one can always choose a subsequence that tends to zero; the problem is that the sequence may not be supported on a set belonging to the (non-P-point) ultrafilter.

For additional details, see this article.

Answer 3

I believe (but am not completely sure) that you can internalize this approach to foundations.

Let $H$ be the specific infinite hyperreal corresponding to the sequence $h_n = n$.

Then, every function $f : \mathbb{N} \to \mathbb{R}$ has a corresponding transfer to a function ${}^\star f : {}^\star \mathbb{N} \to {}^\star \mathbb{R}$, and the sequence $(f(n))$ corresponds to the specific hyperreal number ${}^\star f(H)$.

(I will now dispense with the stars)

We can use this to answer (2).

I assume $g$ is eventually nonzero since you are dividing by it. Recall that the definition of $f \in o(g)$ then becomes $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 0$.

The infinitesimal corresponding to $\frac{f(n)}{g(n)}$ is just $\frac{f(H)}{g(H)}$; by the characterization of limits we have $\frac{f(H)}{g(H)} \approx 0$, so it is infinitesimal as you predicted.

Then if $g \in o(h)$, we have $\frac{g(H)}{h(H)} \approx 0$ as well, and $$ \frac{(f/h)(H)}{(f/g)(H)} = \frac{g(H)}{h(H)} \approx 0$$

so that the hyperreal corresponding to $(f/h)$ is indeed infinitesimally smaller than the one corresponding to $(f/g)$.

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We can use this to answer some things related to (1) as well. For example, if $a_H \neq 0$, we can construct a new sequence $b_n$ whose terms are all nonzero, by

$$ b_n = \begin{cases} 1 & n = 0 \\ b_{n-1} & a_n = 0, n > 0 \\ a_n & a_n \neq 0, n > 0 \end{cases} $$

$b_n$ is a standard sequence, every term is nonzero, and $b_H = a_H$, so $a$ and $b$ represent the same number.

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Note that if we fix $H$, we cannot internally argue that, for each hyperreal $x$, there is a standard sequence $a$ with $a_H = x$ — in fact there exist models of nonstandard analysis where the cardinality of ${}^\star \mathbb{R}$ strictly exceeds that of $\mathbb{R}^\mathbb{N}$. This property is a peculiar feature of the specific model $\mathbb{R}^\mathbb{N} // \mathcal{U}$.