# Does an internal set with countably many infinitesimals contain hyperfinitely many infinitesimals?

Let $\mathscr{A}$ be an internal subset of $^*\mathbb{R}$ that contains a countable family, say $\{a_n\}_{n\in\mathbb{N}}$, of infinitesimal members, that is $a_n\approx 0$ for all $n\in\mathbb{N}$.

We can extend the sequence $a\colon\mathbb{N}\to\mathscr{A}$ to an internal hypersequence with the same name $a\colon{^*\mathbb{N}}\to\mathscr{A}$. It follows that the set $$\{n\in{^*\mathbb{N}}\mid |a_n|<\frac{1}{n}\}$$ is internal by the Internal Definition Principle and includes $\mathbb{N}$ by our assumption. Hence, by the Overflow Principle, there exists $H\in{^*\mathbb{N}}\setminus\mathbb{N}$ such that $a_N\approx 0$ for all $N\in{^*\mathbb{N}}$ with $N\leq H$.

Since the extended hypersequence $a$ is internal, $\{a_0,\dots,a_H\}$ is an infinite hyperfinite set of infinitesimal elements of $\mathscr{A}$.

Is this procedure correct? Is there a less sophisticated way to prove the statement?

Perhaps a slighly shorter argument exploits saturation. By countable saturation you can find an infinitesimal $\alpha>0$ such that each of the countably many infinitesimals is smaller than $\alpha$ in absolute value. Then the subset $\{x\in\mathcal{A}\colon |x|<\alpha\}$ is internal, infinite, and therefore is infinite hyperfinite.

The existence of such an upper bound $\alpha$ can be established as follows. Suppose not; then wlog we can assume that the sequence $\langle\epsilon_n\colon n\in \mathbb N\rangle$ of positive infinitesimals is nondecreasing. Consider the hyperreal interval $S_n=[\epsilon_n,\frac{1}{n}]$. Then $\langle S_n\colon n\in\mathbb N\rangle$ is a nested sequence and by saturation it must have a common element $c$. But $c$ can be neither infinitesimal nor appreciable. The contradiction proves the existence of $\alpha$.

Translating the question into the language of Edward Nelson's internal set theory (IST) reduces it to triviality: "hyperfinitely many" in Robinsonian NSA translates to "finitely many" in IST, so the question asks whether there is a finite subset of an infinite set.

The answer is yes.

(Please read the discussion in the comments for further details.)

• pash, "finitely many" could be "standardly finitely many" or "nonstandardly finitely many", so this does not answer the question. Commented Jun 17, 2017 at 18:46
• @MikhailKatz, in Robinsonian NSA nonstandard natural numbers are in the set $^*\mathbb N \supseteq \mathbb N$, but in IST they are in the ordinary set $\mathbb N$. This makes all the difference: in IST a natural number is finite, in the classical sense, no matter whether it is standard or nonstandard. The Robinsonian statement "n is hyperfinite" translates into the analogous statement of IST that "n is nonstandard finite"—but the less precise statement "n is finite" follows immediately. So in IST the OP's question simplifies to the trivial: "Does an infinite set contain a finite subset?"
– pash
Commented Jul 11, 2017 at 16:11
• Oh, @Mihkail Katz, if you and the OP are asking further whether the largest such finite subset has standard or nonstandard cardinality, then the answer to that too is obvious In IST: there is an initial subsequence of $\mathbb N$ of nonstandard cardinality, so, yes, there is a nonstandard finite subset of a countable set.
– pash
Commented Jul 11, 2017 at 16:19
• pash, the OP is asking a question about internal sets. Obviously he is working in Robinson's framework rather than Nelson's, since what corresponds to an internal set in Nelson's framework is simply a set. Commented Jul 12, 2017 at 6:53
• @MikhailKatz, right. And the question in the OP reduces to one expressible as an internal formula once you translate into the language of IST. That's rather the point. There is no statement of nonstandard analysis that is innately "Robinsonian" or "Nelsonian", and in this case (as is often the case, in my experience), translating into IST yields a simpler problem.
– pash
Commented Jul 20, 2017 at 2:49