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Joel David Hamkins
Joel David Hamkins
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Meanwhile, if you think about numbers like $e$ and $\pi$, it is not clear how to define them in $\R^*$ without quantifying over all the (possibly nonstandard) natural numbers. For example, $e$ is the limit of $(1+\frac 1n)^n$, and so $e^*$ is the unique $x$ in $\R^*$ such that $$\forall \epsilon>0\ \exists N\ \forall n\geq N \ |(1+\frac 1n)^n-x|<\epsilon.$$ This definition can be undertaken in $\R$, using the predicate for $\mathbb{N}$, which allows one to define exponentiation and so on. But in $\R^*$, these quantifiers are not only over the standard numbers; we have to quantify also over the nonstandard numbers. If you restrict to standard $\epsilon$ and standard $N$ and $n$ only, then there will be an entire interval of hyper-reals that are that close to those numbers---anything infinitesimally close to $e$ will do. So the restricted-scope version of the definition will not succeed as a definition.

Meanwhile, if you think about numbers like $e$ and $\pi$, it is not clear how to define them in $\R^*$ without quantifying over all the (possibly nonstandard) natural numbers. For example, $e$ is the limit of $(1+\frac 1n)^n$, and so $e^*$ is the unique $x$ in $\R^*$ such that $$\forall \epsilon>0\ \exists N\ \forall n\geq N \ |(1+\frac 1n)^n-x|<\epsilon.$$ But these quantifiers are not only over the standard numbers; we have to quantify also over the nonstandard numbers. If you restrict to standard $\epsilon$ and standard $N$ and $n$ only, then there will be an entire interval of hyper-reals that are that close to those numbers---anything infinitesimally close to $e$ will do. So the restricted-scope version of the definition will not succeed as a definition.

Meanwhile, if you think about numbers like $e$ and $\pi$, it is not clear how to define them in $\R^*$ without quantifying over all the (possibly nonstandard) natural numbers. For example, $e$ is the limit of $(1+\frac 1n)^n$, and so $e^*$ is the unique $x$ in $\R^*$ such that $$\forall \epsilon>0\ \exists N\ \forall n\geq N \ |(1+\frac 1n)^n-x|<\epsilon.$$ This definition can be undertaken in $\R$, using the predicate for $\mathbb{N}$, which allows one to define exponentiation and so on. But in $\R^*$, these quantifiers are not only over the standard numbers; we have to quantify also over the nonstandard numbers. If you restrict to standard $\epsilon$ and standard $N$ and $n$ only, then there will be an entire interval of hyper-reals that are that close to those numbers---anything infinitesimally close to $e$ will do. So the restricted-scope version of the definition will not succeed as a definition.

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Joel David Hamkins
Joel David Hamkins
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Are the definable hyper-reals, using quantifiers only over the standard reals and natural numbers, the same as the algebraic numbers?

This question arose today at Yevgeny Gordon's talk, "Will nonstandard analysis be the analysis of the future?" at the CUNY Logic Workshop. Here is my way of asking it.

Consider the ordered real field $\newcommand\R{\mathbb{R}}\R$ with a predicate for the natural numbers, and a nonstandard version of it $\R^*$, the hyper-reals, which an ordered field extending $\R$ and having the transfer property, a map $a\mapsto a^*$, which preserves the truth of any statement in the language of ordered fields, allowing also the predicate for the natural numbers. The hyper-real numbers of the form $a^*$ are referred to as the standard elements of $\R^*$.

Question. Suppose that a hyper-real number $a\in\R^*$ is definable in $\R^*$ by a formula $\varphi$ in the language of ordered fields with a predicate for natural numbers, but where the scope of the quantifiers is only over the standard elements. Must $a$ be algebraic?

It is easy to see that every algebraic hyper-real number is definable in this way. For example, the hyper-real $\sqrt{2}$ is definable in $\R^*$ as the unique $x$ for which $x^2=2$ and $0<x$. Similarly, any algebraic hyper-real (algebraic over the standard integers) is the unique solution in $\R^*$ in a certain standard rational interval of a polynomial equation over the standard integers. So in fact, every algebraic hyper-real is quantifier-free definable in $\R^*$, even without the predicate for the natural numbers. So the question is equivalent to asking:

Question. If a hyper-real is definable by a formula whose quantifiers have scope restricted to the standard reals (allowing a predicate for the natural numbers), then is it quantifier-free definable?

Meanwhile, if you think about numbers like $e$ and $\pi$, it is not clear how to define them in $\R^*$ without quantifying over all the (possibly nonstandard) natural numbers. For example, $e$ is the limit of $(1+\frac 1n)^n$, and so $e^*$ is the unique $x$ in $\R^*$ such that $$\forall \epsilon>0\ \exists N\ \forall n\geq N \ |(1+\frac 1n)^n-x|<\epsilon.$$ But these quantifiers are not only over the standard numbers; we have to quantify also over the nonstandard numbers. If you restrict to standard $\epsilon$ and standard $N$ and $n$ only, then there will be an entire interval of hyper-reals that are that close to those numbers---anything infinitesimally close to $e$ will do. So the restricted-scope version of the definition will not succeed as a definition.

Similarly, it is not clear how to define $\pi$ or indeed any other transcendental hyper-real number while quantifying only over standard numbers.

I believe that Tarski's theorem on real-closed fields will prove the positive result for the special case of the question, where $\varphi$ does not use the predicate for the natural numbers (but still has the scope of all quantifiers restricted to the standard hyper-reals). My reason for this expectation is that I believe we can apply Tarski's elimination of quantifiers procedure to such a $\varphi$ and thereby prove that $\varphi(x)$ is equivalent to a quantifier-free assertion in the language of ordered fields. Then, using the fact the algebraic numbers form an elementary substructure of the reals, as ordered fields, it follows that the existence of a solution in $\R$ is equivalent to the existence of a solution in the algebraic numbers. And so the given number must be algebraic.

But I am a little fuzzy on the details of how the scope-restriction affects this argument, and so if you can affirm or refute it, I would be grateful.