The hyperreals are "real-closed," which means that any first-order statement that is true of the reals is true of the hyperreals. However, they also satisfy a "transfer principle," and it seems the existence of this "transfer principle" is strictly stronger than a field being real-closed.
In particular, the hyperreals have the property that for any function $f: \Bbb R \to \Bbb R$, there is an extension $^*f: ^*\Bbb R \to ^*\Bbb R$ which satisfies all of the same first-order properties that $f$ does, which would seem to be stronger than just saying that the entire collection of hyperreals satisfies those first-order predicates about the reals.
My questions:
- Is this the correct idea, that this "extensibility" of real functions to nonstandard functions is strictly stronger than just being real-closed?
- Is there some way to explain what, in model theoretic terms, the transfer principle actually is? That is, it isn't just that the hyperreals are a nonstandard model of the first-order theory of the reals, but that the functions from $^*\Bbb R \to ^*\Bbb R$ is a model of the first-order theory of the functions from $\Bbb R \to \Bbb R$, or something like that?
- There are real-closed fields which are a strict subset of the reals, such as the real algebraic numbers. Are there fields, which satisfy this stronger property, which are also a strict subset of the reals? In general, what is the smallest field which satisfies this stronger property?
Best Answer
1.
Your "extensibility" condition is strictly stronger than being a real-closed field.
You already know that every structure which satisfies your extensibility condition is a real-closed field. I will now show that the converse does not hold.
Consider e.g. the theory of the structure $\mathfrak{R} = (\mathbb{R}, +, -,\cdot, \sin, 0, 1)$. Clearly $\mathfrak{R} \models \exists x. x \neq 0 \wedge \sin(x) = 0$. Moreover, by the transcendence of $\pi$ we have that for any polynomial $p$ with integer coefficients, $\mathfrak{R} \models \forall x. x \neq 0 \wedge \sin(x) = 0 \rightarrow p(x) \neq 0$. Since $p$ has integer coefficients, this is a first-order sentence in the language of fields.
But this means that the field $\overline{\mathbb{Q}} \cap \mathbb{R}$ of real algebraic numbers does not satisfy the extension property. If it did, $\sin: \mathbb{R} \rightarrow \mathbb{R}$ would have to "extend" to a function $~^\star\!\sin: \overline{\mathbb{Q}} \cap \mathbb{R} \rightarrow \overline{\mathbb{Q}} \cap \mathbb{R}$ that takes on a transcendental value. That's of course not possible.
2.
General transfer principles take more than just RCFs and Łoś ultraproduct theorem; we usually formulate them in terms of superstructures/universes. Most textbooks on Nonstandard Analysis will introduce the required notions (universes/superstructures), and state and prove the Transfer Principle. You could consult e.g.
Goldblatt's Lectures on the Hyperreals Part 3, or
Loeb's Nonstandard Analysis for the Working Mathematician Chapter 2.
3.
All fields satisfying your condition have cardinality at least continuum. This follows from the fact that every Dedekind cut $c \subseteq \mathbb{R}$ defines a function $f_c: \mathbb{R} \rightarrow \{0,1\}$, and you can distinguish two different cuts $c \subseteq d$ using the first-order theory of the structure $(\mathbb{R}, +, -, \cdot, f_c, f_d, 0, 1)$ simply by finding a rational $\frac{p}{q}$ that belongs to $d$ but not to $c$. This is possible because $\mathbb{Q}$ is dense in $\mathbb{R}$.
Since $\mathfrak{R} \models \exists! x_c. f_c(x_c) = 0 \wedge (\forall y. (\exists z. y - x_c = z^2) \rightarrow f_c(y) = 1)$, and $\mathfrak{R} \models x_c \neq x_d$ for any cut $d \neq c$ and corresponding unique $x_c, x_d$, a field satisfying your property will have at least as many elements as $\mathbb{R}$. Moreover, if you take a proper subset $S \subseteq \mathbb{R}$ containing $0,1$ and closed under all the usual operations $+, -, \cdot$, then you will not be able to "extend" $f_c: \mathbb{R} \rightarrow \mathbb{R}$ to $S$ for any $c \not\in S$. So $\mathbb{R}$ is minimal with this extension property.