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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:

  1. Is this the correct idea, that this "extensibility" of real functions to nonstandard functions is strictly stronger than just being real-closed?
  2. 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?
  3. 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?
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2 Answers 2

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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.

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I'd explain the non standard reals as follows.

The language $L$ contains literally all functions ${\mathbb R}^n\to\mathbb R$ and all subsets of ${\mathbb R}^n$. (Recall: any object can be used as symbol.)

Consider ${\mathbb R}$ as an $L$-structure. Each symbol in $L$ is interpreted in itself.

Now let $^*{\mathbb R}$ be a proper elementary extension of ${\mathbb R}$.

This makes elementarity and tranfer principle the same thing.

P.S. In the context of NSA (non standard analysis) the real closedness of ${\mathbb R}$ is an infinitesimal detail (excuse the pun). Real closedness is concerned only with the order-algebraic part of ${\mathbb R}$. This is just a tiny fraction of the language you use in NSA.

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  • $\begingroup$ Thanks @PrimoPetri - what do you mean when you say "symbols can be anything?" $\endgroup$ Commented Mar 26, 2022 at 23:12
  • $\begingroup$ That sentence is reversed. I correct it in: anything can be a symbol. I.e. you can use functions and subsets of ${\mathbb R}$ as symbols. $\endgroup$ Commented Mar 27, 2022 at 7:37

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