Working through Schröder Mathematical Analysis right now. Assuming existence & uniqueness of $\Bbb R$ (as complete ordered field), the text is now trying to prove the various properties of $\Bbb N$ understood as the unique subset of $\Bbb R$ satisfying the following:
(i) $\Bbb N$ is a successor set (i.e. $1 \in \Bbb N$ and $n \in \Bbb N \implies n+1 \in \Bbb N$), and
(ii) For $S$ any successor set, $S \subseteq \Bbb N \implies S = \Bbb N$.
Is it possible from here to prove the following fact about the natural numbers (Prop. 1.25 in Schröder)?
$$(\forall m,n \in \Bbb N) \space n < m \implies m - n \in \Bbb N $$
So far the only other property of $\Bbb N$ proven is that it is closed under addition and multiplication. The proof outlined in the text has two parts:
- Step one is to prove that $(\forall m \in \Bbb N) \space m - 1 \in \Bbb N \space \text{or} \space m = 1$. No problems here.
- For step two, the idea is to define $S := \{ n \in \Bbb N \space:\space (\forall m \in \Bbb N)\space n<m \implies m-n \in \Bbb N \} \subseteq \Bbb N$, prove that this is a successor set as defined above, and then use $S \subseteq \Bbb N$ and the Principle of Induction ((ii) above) to force equality.
It gets really gritty but I believe there's a case that isn't considered in the text's proof: it's easy enough to show that $1 \in S$, but then in showing that $n \in S \implies n+1 \in S$, letting $n \in S$, then letting $m \in \Bbb N$ with $(n+1) < m$, you have to consider two cases: either $m - 1 \in \Bbb N$ or $ m = 1$. The first case presents no problems, but I can't see how to proceed for the case where $m = 1$. Taking $m = 1$, it boils down to needed to prove that $n<0 \implies -n \in \Bbb N$. Can't see how to do this given what we know already.
Is anyone else familiar with this text!? My first post. Not a terribly interesting question, would just like to see if I am missing something. Thanks for any help offered.