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Let $ \mathscr{S}$ be a statement in a superstructure $\hat{S}$.
Let $^*$ denote the transfer of an element of $\hat{S}$ via the transfer principle.

The transfer principle says that $\mathscr{S}$ is true if and only if $ ^*\mathscr{S}$ is true.

Let our statement $\mathscr{S}$ be: $\,\,a\in V$ with $V\in \hat{S}$, then it feels clear that $^*\mathscr{S}$ is $\,\,^*a\in {^*V}$, and that both $^*a$ and $^*V$ exist.
If we start though with $\,\,b\in {^*V}$, how do I know whether there is an element $a\in V$ so that $^*a = b$?

For example, if we take $\infty \in\, ^*\mathbb{N}$, where $\infty$ be some infinite element of $\,^*\mathbb{N}$, then
$\infty \in\, ^*\mathbb{N}$ is a true statement. However, there is no element that transfers to $\infty$, is there?

So, doesn't that mean that there are statements in the non-standard-realm that can't be transferred from the standard-realm?

Sudix
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    The transfer principle only works if every constant in the statement is standard. $\infty$ isn't (at least, not in the sense you're using it). –  Aug 29 '17 at 19:35

2 Answers2

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There are statements in nonstandard-land which don't transfer. But such statements can't be first-order expressible, for instance. There's no way to express the property "$n$ is infinite" in a statement to which the transfer principle applies; similarly, the statement "$x$ is not a standard real" doesn't transfer. All statements about internal sets do transfer, if I recall correctly; but you need to be careful to justify that the sets under consideration are internal. $\{1\}$ is internal because $1$ can be defined in a first-order way (it's the unique real such that $1x = x$ for all $x$); the set of all standard reals is not internal.

You should be careful to find an exact statement of the transfer principle so that you know what restrictions need to be placed on the statements you're considering.

Patrick Stevens
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  • $\exists x\in\mathbb{R}\forall a\in \mathbb{R} : x>a$ would be the statement for an infinite number x in $\mathbb{R}$. As it is false, wouldn't that mean the internal set of hyperreals doesn't contain infinities? – Sudix Aug 29 '17 at 19:47
  • That statement is false in $\mathbb{R}$, so its transfer is false in $^\mathbb{R}$. The transferred statement is $\exists x \in ,^\mathbb{R} \forall a \in ,^*\mathbb{R} : x > a$. – Patrick Stevens Aug 29 '17 at 19:52
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    Oh, I'm a dumbus - but it finally made click for me – Sudix Aug 29 '17 at 19:54
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For a statement to be tranferable it needs to be expressed in a language common to both models. Since the symbol $\infty$ is not part of such a common language, the statement cannot be transfered.

More specifically, you typically have three objects: a symbol in the language, its interpretation in one model, and its interpretation in the other model. Thus, you may have a symbol $R$ in the language, its interpretation $\mathbb R$ in the standard model, and its interpretation ${}^\ast\mathbb R$ in the nonstandard model.

A nonstandard element of ${}^\ast\mathbb R$ does not have a counterpart in the language.

Mikhail Katz
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  • I should mention that I am using the term "language" loosely in my answer. A better choice would probably be "theory" or more precisely "symbols available in a theory". – Mikhail Katz Sep 01 '17 at 07:37