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Let $\mathbb{Q}$ be the set of rational numbers. Show that every member of ${}^{*}\mathbb{R}$ is infinitely close to some member of ${}^{*}\mathbb{Q}$.

This is an exercise on page 180, A Mathematical Introduction to Logic, Herbert B. Enderton（2ed)

${}^{*}\mathbb{R}$ is the universe of ${}^{*}\mathfrak{R}$, which is a nonstandard structure by using compactness theorem.

$\mathfrak{R}$ is the standard structure of the language, in which:

  • $\forall$ refers to all real numbers.
  • For each $r \in \mathbb{R}$, there's a constant $c_r$
  • For each relation $R$ and operation $F$ there is an predicate symbol $P_R$ and a function symbol $f_F$.

${}^{*}\mathfrak{R}$ is the structure satisfying $\operatorname{Th}\mathfrak{R} \cup \{c_r P_{<} v_1 | r \in \mathbb{R}\}$ ($\operatorname{Th}\mathfrak{R}$ is the set of all theorems of $\mathfrak{R}$ )

${}^{*}\mathbb{Q}$ is defined similarly.

Two elements are infinitely close, if their difference is infinitesimal.

It seems to me, each element of ${}^{*}\mathbb{Q}$ is either a sum of a rational number and an infinitesimal or simply an infinite element. If so, how can there be an element in ${}^{*}\mathbb{Q}$ that is infinitely close to an irrational number in ${}^{*}\mathbb{R}$ ?

zyx
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Metta World Peace
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  • Have you tried thinking in terms of ultraproducts? That will show you that you're mistaken about what $^{\ast}\mathbb{Q}$ looks like. – Qiaochu Yuan May 09 '13 at 05:25
  • @QiaochuYuan: No. In fact, this is the first time I hear of that concept. – Metta World Peace May 09 '13 at 05:27
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    Take a look here and here. An irrational $\alpha\in{}^\Bbb R$ is represented by a sequence $\langle x_n:n\in\Bbb N\rangle$ such that each $x_n$ is irrational. For $n\in\Bbb N$ let $q_n\in\Bbb Q$ with $|x_n-q_n|<2^{-n}$. Then $\langle q_n:n\in\Bbb N\rangle$ represents a rational in ${}^\Bbb Q$ that is infinitely close to $\alpha$. – Brian M. Scott May 09 '13 at 05:33
  • How did you construct your definition of real numbers in the first place? – Ethan Splaver May 09 '13 at 06:28
  • @Ethan: It is assumed as something given in Enderton's book. – Metta World Peace May 09 '13 at 06:42

1 Answers1

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A well-known construction of R starts with the hyperrationals, deletes the infinite hyperrationals, and factors the resulting ring by the ideal consisting of infinitesimals. This can be found for example in the book by Martin Davis. Thus every real is infinitely close to a suitable hyperrational.

To see that every hyperreal is infinitely close to a hyperrational, use the fact that the hyperreals are an elementary extension of the reals. Thus, a hyperreal $x$ also satisfies the formula that for every $\epsilon>0$, there is a hyperrational $r$ such that $|x-r|<\epsilon$. But over the hyperreals, we can choose $\epsilon$ to be a positive infinitesimal, showing that $x$ and $r$ are infinitely close.

Mikhail Katz
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