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The second-order theory of real numbers is what you get when you take the axioms for ordered fields and add one more axiom, the least upper bond property, also known as Dedekind completeness: that every set that has an upper bound has a least upper bounds. This is a theory in the language of second-order logic, and it's categorical, meaning it only has one model. But if we make it into a first-order theory, taking only the first-order consequences of the completeness axiom, then the resulting theory is no longer categorical: in addition to the standard model, the real number systems, we get all sorts of non-standard models, known as real closed fields.

One example of a real closed field is the set of algebraic real numbers. That means that the proofs that pi and e are transcendental numbers (and possibly even the definition of pi and e) cannot be carried out in the theory of real closed fields. In other words, the proof of each must rely on some second-order consequence of the completeness axiom. My question is, what is the second-order consequence that is used in the proof of each, and where in the proof is it used?

Any help would be greatly appreciated.

Thank You in Advance.

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  • $\begingroup$ You could look at a transcendence proof and maybe see for yourself. Some intro Number Theory texts will have proofs for $e$. $\endgroup$
    – Gerry Myerson
    Sep 15, 2013 at 4:45
  • $\begingroup$ I've seen such proofs, but the relevant step doesn't jump out at me. $\endgroup$
    – Keshav Srinivasan
    Sep 15, 2013 at 4:51
  • $\begingroup$ How do you propose to even define $\pi$ or $e$ in $(\mathbb{R}; +,\times)$ using first-order logic? (You won't be able to.) $\endgroup$
    – Noah Schweber
    Aug 2, 2018 at 5:06

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If you look at the [wikipedia entry] (http://en.wikipedia.org/wiki/Real_closed_field) on real closed fields, you see that one characterization of such ordered fields is that the intermediate value theorem holds for polynomials over the field; related to this is the fact that odd degree polynomials always have a root. This gives a way of constructing certain numbers in $F$, say as the root of a certain odd degree polynomial.

Since $\pi$ and $e$ are transcendental, we know they can't be constructed this way (as roots of polynomials over the field of real algebraic numbers). And the other usual definitions of $\pi$ and $e$ are in terms of infinite series, or integrals, or other limiting processes which don't apply in a general real closed field.

So the basic second order property that is being used is the existence of certain limits to define $\pi$ and $e$. The ability to define numbers as limits is something that one has available in the reals that is not available in a general real ordered field.

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  • $\begingroup$ Why would you need second-order logic to define limits? The epsilon-delta definition Is first-order. $\endgroup$
    – Keshav Srinivasan
    Sep 15, 2013 at 4:31
  • $\begingroup$ @KeshavSrinivasan: Dear Keshav, The statement "any increasing bounded sequence of numbers $a_n$ has a limit" is not a first order axiom; among other things it quantifies over a countable number of variables of the field. Regards, $\endgroup$
    – Matt E
    Sep 15, 2013 at 4:37
  • $\begingroup$ P.S. The point is that this is the statement which produces $e$ and $\pi$ in $\mathbb R$. It doesn't hold in a general real closed field. $\endgroup$
    – Matt E
    Sep 15, 2013 at 4:38
  • $\begingroup$ It is true that that general statement is second-order, but isn't it any first-order consequence of it provable in the theory of real closed fields? Specifically, for any sequence $a_{n}$ that's first-order definable, you can prove that $a_{n}$ satisfies that statement. $\endgroup$
    – Keshav Srinivasan
    Sep 15, 2013 at 4:49
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    $\begingroup$ @KeshavSrinivasan: Dear Keshav, It's still not a first order statement. The definition of the limit of a sequence $a_n$ says (among other things) "For all $\epsilon > 0$, there exists $N$ such that for all $n > N$ \ldots'', and so quantifies over the natural numbers $N$ and $n$. This is not something you can do in the first order language of ordered fields. Regards, $\endgroup$
    – Matt E
    Sep 15, 2013 at 5:24

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