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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 confused, because to model an infinite stream of coin toss, we use $$ X=(2^\omega,\sigma(2^\omega),\text{ counting measure}). $$ So if $A$ is external, $X$ is an external probability space, and I cannot construct it's associated Loeb space.

Is $A$ internal?
Or, should I use something else than $X$ to model coin tosses?

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3 Answers 3

up vote 2 down vote accepted

Consider the family $$\mathcal{R} := \{A \in 2^\mathbb{N} \: | \: \exists\, n \in \mathbb{N} \: : \: (\forall\, k \leq n \: : \: k \in A) \: \wedge \: (\forall\, m > n \: : \: m \notin A)\}.$$ The transfer of this family contains your set, so it is internal. On the other hand, the set of all sets $\{1, \ldots, \omega\}$ with $\omega$ infinite is external!

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Yes, the description $1 \le x \le k$ is a first-order description of a set, regardless of what element $k$ is. This has nothing to do with "standard part" at all.

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$\{1,2,3, \ldots \omega\}$ with $\omega \in {}^{\ast}\Bbb {N} \setminus \Bbb N$ is an example of non-standard but internal set, though every standard entity is internal.

$\{1,2,3, \ldots \omega\}$ is not standard, since there's no $A \in \Bbb N$ such that ${}^{\ast}A = \{1,2,3, \ldots \omega\}$. Suppose otherwise: If $A$ is finite, then $A = {}^{\ast}A $, and hence ${}^{\ast}A$ is finite. If $A$ is infinite, and hence unbounded, then ${}^{\ast}A$ is unbounded in ${}^{\ast} \Bbb N$.

By definition, $\{1,2,3, \ldots \omega\}$ is internal, only when $\{1,2,3, \ldots \omega\} \in {}^{\ast}B$ for some $B \in V(\Bbb N)$, i.e. $\{1,2,3, \ldots \omega\}$ . In this case, $B = \mathcal P(\Bbb N)$.

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