Calculus in ordered fields

Question

Is there any ordered field smaller that the set of real numbers in which we can do calculus, also with many restrictions ?

If not why ?

Answer 1

I remember reading in Körner's book, "A Companion to Analysis", he discusses problems with calculus over $\mathbb{Q}$, and ordered fields it seems. I think there was a problem with continuity over $\mathbb{Q}$ and of course the fact it's not complete. It might be worth a look if you've access to it.

Section 1.3 and Chapter 1 anyway it think he deals with ordered fields:

http://books.google.ie/books?id=H3zGTvmtp74C&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false

Answer 2

Some amount of calculus can be done over any field: see "algebraic geometry". The purely algebraic context does have a lot of deficiencies and technical obstacles, though, that make it a rather complicated subject.

You can do a lot more calculus over any real closed field. All such things are "elementarily equivalent": any semi-algebraic statement -- i.e. using just $+, \times, <$ and (firest-order) logic -- is true in one real closed field if and only if it is true in all real closed fields.

Since the real numbers are a real closed field, this lets you transfer semi-algebraic facts about the real numbers to all real closed fields.

The smallest real closed field, incidentally, is the field of all real algebraic numbers. There are also real closed fields with transfinite and infinitesimal elements.

The main thing to be careful of is that in standard calculus, you have the full power of set theory at your disposal, giving you access to things like the recursion and induction over the integers, sequences, Taylor series, Riemann integrals, and the like.

You don't have those in real-closed fields: you only get what can be defined semi-algebraically. e.g. while you can't define the transcendental function $\exp(x)$, you can define the function $\sqrt{x}$ on nonnegative numbers by the conditions

• $(\sqrt{x})^2 = x$
• $\sqrt{x} \geq 0$

It might be easier to instead think in terms of defining the graph of the function $\sqrt{x}$ in the $x-y$ plane; i.e. the graph of the square-root function is defined by

• $y^2 = x$
• $y \geq 0$.

For the purposes of this post, I will call functions whose graph can be defined semi-algebraically "semi-algebraic functions". A neat fact is that while we allow $\forall$ and $\exists$ when defining semi-algebraic functions, it turns out that it is always possible to give another definition that doesn't use them.

An example of an idea from calculus that does carry over to real closed fields is that of limits. If $f$ is a semi-algebraic function, then if you look at the definition of limits, you'll see that

$$ \lim_{x \to a} f(x) = L $$

can be defined entirely in the language of real closed fields. Similarly, things like "continuous function" or "derivative" can be too. By the elementary equivalence I mentioned above, the fact we can prove theorems about the reals immediately gives us theorems about real closed fields: e.g.

Intermediate Value Theorem: If $f$ is a continuous semi-algebraic function on $[a,b]$, $f(a) < 0$, $f(b) > 0$, then there exists $c \in (a,b)$ such that $f(c) = 0$.

Mean Value Theorem: if $f$ is a differentiable semi-algebraic function on $[a,b]$, then there exists $c \in (a,b)$ such that $f(b) - f(a) = f'(c) (b-a)$.

Theorem: If $f$ is a continuous semi-algebraic function on $[a,b]$, then there exists a value $c \in [a,b]$ such that $f(c) \leq f(x)$ for all $x \in [a,b]$

and so forth.

(aside: these are actually "theorem schema" -- first-order logic doesn't let you have variables representing functions. However, for each particular function $f$, you can substitute in the definition of $f$ and state it in first-order logic. Any constants appearing in the definition of $f$ can be eliminated by proving the theorem with the constants converted into variables)