2
$\begingroup$

I am more or less familiar with constructions of the real number system starting from the Peano axioms for $\mathbb{N}$ and proceeding to $\mathbb{Z}$, $\mathbb{Q}$, and $\mathbb{R}$.

Apostol's Calculus, however, starts with the field axioms for $\mathbb{R}$ (including completeness), and then defines $\mathbb{N}$ as the smallest inductive set in $\mathbb{R}$, as follows:

  • A set $\mathcal{S}\subseteq \mathbb{R}$ is said to be inductive if $1\in\mathcal{S}$ and for all $x\in\mathcal{S}$, $x+1\in\mathcal{S}$.
  • A real number is called a positive integer if it belongs to every inductive set.

Later on he says:

In a thorough treatment of the real-number system, it would be necessary at this statge to prove certain theorems about integers. For example, the sum, difference, or product of two integers is an integer, but the quotient of two integers need not be an integer. However, we shall not enter into the details of such proofs.

My questions are:

  1. How common is it to define the $\mathbb{N}$ in this way, rather than defining $\mathbb{R}$ constructively from $\mathbb{N}$?

  2. I would be interested to see how the proofs of the above statements would look like, as well as something like "if $n$ and $n+1$ are positive integers, there is no positive integer $m$ such that $n<m<n+1$".

$\endgroup$
  • 1
    $\begingroup$ Once you show that $\Bbb N$ satisfies Peano Axioms, everything follows. $\endgroup$ – Asaf Karagila Jun 23 '17 at 8:50
1
$\begingroup$

As you guessed Apostol's treatment of $\mathbb N$ is pretty unusual. Perhaps the reason he wanted to do it this way is so as to emphasize the fact that $\mathbb N$ is literally a subset of $\mathbb R$ rather than being isomorphic or even naturally isomorphic to a subset of $\mathbb R$. The problem is that in the usual construction of everything from $\emptyset$ in set theory, a natural number has finite von Neumann rank (namely, "itself") whereas each real number has von Neumann rank $\omega+1$, so strictly speaking $\mathbb N\not\subseteq\mathbb R$. See related discussion here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.