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One definition of a finite set is that it can be injected into an initial segment of $ \Bbb N$, thus any $n$ in $\Bbb N$ is finite.

Accordingly, if it's legitmate to define every element in $^* \Bbb N$ as its own intial segment, then every element $^* \Bbb N$ is finite.

What about Dedekind-infinity? Can we define an Dedekind-infinite element in $^* \Bbb N$ by constructing an injection into its proper subset?

Added: I'm not sure whether the construction of $^* \Bbb N$ is compatible with $\bf ZFC$ in the same way as $\Bbb N$ is, which is probably not, I guess. But how?

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  • $\begingroup$ how exactly are you considering an element in $^*\mathbb N$ to be a set? $\endgroup$ – Ittay Weiss May 17 '13 at 7:56
  • $\begingroup$ @IttayWeiss: Perhaps I got it wrong. Can't we define an element in $^{\ast} \Bbb N$ as its own initial segment? $\endgroup$ – Metta World Peace May 17 '13 at 8:00
  • $\begingroup$ I'm not sure how you intend to do that. $\endgroup$ – Ittay Weiss May 17 '13 at 8:03
  • $\begingroup$ You say that something is finite, and then ask whether it can be Dedekind infinite. But "finite" implies "Dedekind-finite" in ZF. $\endgroup$ – Trevor Wilson May 17 '13 at 15:52
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There is a problem with this idea because we use $\in$ to construct $\omega$ which is a surrogate for $\Bbb N$ in $\sf ZFC$. This is fine because both $\in$ and $\Bbb N$ are well-founded.

But $^*\Bbb N$ is not well-founded. The non-standard integers must have a decreasing sequence. So we cannot use $\in$ to model them in a model of $\sf ZFC$. Of course, we can have non-standard models of $\sf ZFC$ whose $\omega$ is actually $^*\Bbb N$, but that's besides the point. Inside these models, the set which they think is $\omega$ is well-founded (internally!).

You could perhaps use $\subseteq$ to define the order $^*\Bbb N$ somehow and use $\omega$ for the initial segment of the standard integers. In that case it shouldn't be hard to note that the standard integers correspond to Dedekind-finiteness, and the non-standard integers are Dedekind-infinite sets.

Why? Because any non-standard integer would have to have infinitely many smaller integers, which correspond to infinitely many subsets, which means that it is a Dedekind-infinite set (remember we assume $\sf AC$).

To be fair, though, I don't see what good comes from that definition.

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