8

This is a follow-up question of a previous one.

Is it true ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$ ?

It seems to me it's true because for any $x \in{}^{*}\mathbb{R}$, there is a $y \in {}^{*}\mathbb{Q}$ such that $y < x$ and vice versa, and because for any $x' \in \mathbb{R}$, there is a $y' \in {}^{*}\mathbb{Q}$ such that $y' = x'$.

Community
  • 1
Metta World Peace
  • 6,403
  • 2
    Your second fact is not true: no irrational element of $\mathbb{R}$ is in ${}^\star \mathbb{Q}$. The best you can do is to find a hyperrational that is infinitesimally distant. –  May 12 '13 at 17:38
  • (... but you can always do that much) –  May 12 '13 at 17:46

4 Answers4

10

No. The theory of $\Bbb R$ is not the same first-order theory of $\Bbb Q$. If you define $^*\Bbb Q$ and $^*\Bbb R$ as ultrapowers then they have the same first-order theories as their standard counterparts.

In particular, $^*\Bbb Q$ satisfies $\forall x(x\cdot x\neq(1+1))$; whereas $^*\Bbb R$ satisfies $\exists x(x\cdot x=1+1)$, and that $x$ is the equivalence class of $f(n)=\sqrt2$ in the ultrapower.

You can do a bit better, e.g. since in $\Bbb R$ it is true that every odd-degree polynomial has a root, so we can find many algebraic numbers in $^\ast\Bbb R$ which do not exist in $^*\Bbb Q$.

One more thing on the use of the transfer principle. The transfer principle is really just Los theorem, that the ultrapower of a structure is elementary equivalent to it, i.e. a sentence is true in the ultrapower if and only if it is true in the structure.

Structures, and models come with a language. If your language is the language of ordered-fields then $\Bbb Q$ is not definable. This is because definable sets of real numbers are finite unions of intervals (possibly one-point intervals). The rationals are clearly not such set.

If, on the other hand, you are free to add whatever predicate that you choose to add and whatever function, constants or otherwise, then clearly you can add an unary predicate symbol $Q$ which is interpreted as true for $x$ if and only if $x\in\Bbb Q$. Since $\Bbb Q\models\forall x.Q(x)$ we have that $^*\Bbb Q$ must satisfy the same sentence, and by a similar reasoning we have that $Q^{^*\Bbb R}=\, ^*\Bbb Q$. So it's quite obvious now that $\exists x.\lnot Q(x)$ is true in $\Bbb R$ and therefore in $^*\Bbb R$, and so we have $^*\Bbb R\neq\, ^*\Bbb Q$.

But usually when we talk about $^*\Bbb R$ we talk about it as an ordered field, in which case there is no $Q$ predicate like that, and the rationals are not definable there. So we have that there is no formula which states $\Bbb Q\neq\Bbb R$.

Asaf Karagila
  • 393,674
  • Thank you, Asaf. Your argument is, of course, quite right as always. But it looks like a black box to me. Is it possible to find an element in the symmetic difference ${}^{\ast}\mathbb{R} \triangle {}^{\ast}\mathbb{Q}$? – Metta World Peace May 12 '13 at 16:30
  • Metta, is that better? :-) – Asaf Karagila May 12 '13 at 16:35
  • What about a sequence of rational numbers that converge to $\sqrt{2}$? Is the equivalence class of such sequence equals $\sqrt{2}$? – Metta World Peace May 12 '13 at 16:43
  • 1
    Metta, recall that $^*\Bbb R\models\varphi([f]_U)$ if and only if ${n\in\Bbb N\mid\Bbb R\models\varphi(f(n))}\in U$. That should answer your question. Remember the ultrapower is a transfer of the field structure, rather than the topological/metric structure. – Asaf Karagila May 12 '13 at 16:45
  • So, the equivalence class of a sequence of rational numbers that converge to $\sqrt{2}$ only equals a hyperreal whose standard part is $\sqrt{2}$ instead of $\sqrt{2}$ in $\mathbb{R}$. – Metta World Peace May 12 '13 at 16:49
  • Probably. Because you can write the formula $\varphi(x,y,z)=|x-y|<z$, and for every $\varepsilon>0$ we have that $\varphi(\sqrt2,f(i),\varepsilon)$ holds for large enough $i$, where $f(i)$ is a sequence approaching $\sqrt2$. – Asaf Karagila May 12 '13 at 16:52
  • Got it. Tons of thanks :-) – Metta World Peace May 12 '13 at 16:59
  • 1
    Note that by the same argument one can easily show that the result is not $\sqrt2$, because we can find a sequence of $\varepsilon_i$ which goes to $0$ faster than $f(i)$ approaches $\sqrt2$, so there is some infinitesimal distance between the two. – Asaf Karagila May 12 '13 at 17:01
5

No, ${}^*\Bbb Q\ne{}^*\Bbb R$. It’s not true that $\Bbb R\subseteq{}^*\Bbb Q$. In the ultrapower construction this is especially easy to see: in ${}^*\Bbb R$, $\Bbb R$ is identified with the set of equivalence classes of constant sequences of reals, and the class of a constant sequence with irrational value is easily seen to be different from every element of ${}^*\Bbb Q$, though it’s also easy to see that there are elements of ${}^*\Bbb Q$ infinitely close to each such class.

Brian M. Scott
  • 616,228
  • 1
    To be more explicit, note that every element of $\Bbb Q$ is algebraic, and as in my answer, algebraicity is preserved (as any other first-order property), so the formula defining each algebraic number must interpret the same algebraic number in the canonical embedding, so all the irrationals must not appear in $^*\Bbb R$. – Asaf Karagila May 12 '13 at 16:47
3

By transfer, every element in $^*\mathbb Q$ is a quotient of two elements from $^*\mathbb Z$. By transfer, not every element in $^*\mathbb R$ is the quotient of two elements from $^*\mathbb Z$. So, the two enlargements are not equal.

What you are alluding to (I think) in the question is that both enlargements give rise the same completion of $\mathbb Q$, namely $\mathbb R$.

Ittay Weiss
  • 79,840
  • 7
  • 141
  • 236
0

The simplest argument is to transfer the formula $\mathbb{Q}\not=\mathbb{R}$.

In more detail, the symbols "Q" and "R" in the language are interpreted as respectively the rationals $\Bbb Q$ and the reals $\Bbb R$ in the standard model, and as $^\star\Bbb Q$ and $^\star\Bbb R$ in the non-standard model. The fact that $\Bbb Q$ is PROPERLY contained in $\Bbb R$, and therefore distinct from it, is encoded by the formula Q$\not=$R. By the transfer principle, the same formula remains true when interpreted over the nonstandard model. Hence $^\star\Bbb Q\not= {}^\star\Bbb R$.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
  • 2
    And what formula would that be? – Asaf Karagila May 13 '13 at 08:14
  • The transfer principle does not merely assert that the hyperreals are an elementary extension of the reals. It asserts more generally that every symbol you have in the theory is suitably interpreted in the non-standard extension, and has the same properties. In particular, we have a symbol Q which in the standard theory is interpreted as $\mathbb{Q}$ and the non-standard theory is interpreted as $\star\mathbb{Q}$. Similarly, this lets you conclude that $\sin\pi n=0$ even when $n$ is non-standard, where "sin" is interpreted in the non-standard theory, etc. – Mikhail Katz May 13 '13 at 08:19
  • And what first-order formula defines $\Bbb Q$ in $\Bbb R$? Please, enlighten me. – Asaf Karagila May 13 '13 at 08:22
  • And what first order formula defines the sine function? Yet we have what Keisler calls its "natural extension". I am not sure what you are getting at. In the language we have a symbol for whatever we choose to have a symbol for, for example all real sets and functions, including Q, R, and sin. You seem not to be distinguishing sufficiently between "elementary extension" and "an extension satisfying transfer". – Mikhail Katz May 13 '13 at 09:58
  • Thank you very much for your answer. Perhaps you want to @AsafKaragila to make him aware your update and comment. Here's a link about how it works. http://meta.math.stackexchange.com/questions/2063/ping-only-works-for-the-first – Metta World Peace May 13 '13 at 13:29
  • So... just add a predicate for $\Bbb Q$. Hey, why not? Let's add a predicate for every subset of $\Bbb R$ while we're at it. – Asaf Karagila May 13 '13 at 14:56
  • I am not sure what you are getting at. It is standard procedure to include "natural extensions" (as Keisler calls them) of real sets and functions in a non-standard framework. If you want to do calculus for example you are going to need at least as much. – Mikhail Katz May 17 '13 at 09:53