Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 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!

share|cite|improve this answer

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.

share|cite|improve this answer

$\{1,2,3, \ldots \omega\}$ 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, $B = \mathcal P(\Bbb N)$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.