Subtraction in $\Bbb N$ as an intersection of all inductive sets

Question

Let $\Bbb R$ be axiomatically defined as a complete ordered field. Consider proved the basic properties of sum, product and order like $1>0$, $-(-a)=a$, etc.

Now, an inductive subset of $\Bbb R$ is a set $I$ such that $1\in I$ and $\forall n(n\in I\to n+1\in I)$.

$\Bbb N$ is defined as the intersection of all inductive subsets. I have proved so far:

• $\Bbb N\subset[1,\infty)$. (Easy)
• For every $x,y\in\Bbb N$ we have that $x+y,xy\in\Bbb N$. (I have done this by defining the sets $A=\{z\in\Bbb R:x+z\in\Bbb N\}$ and $B=\{z\in\Bbb R:xz\in\Bbb N\}$. I have proved that $A$ and $B$ are inductive.)

Now I want to prove that if $x,y\in\Bbb N$ and $x<y$ then $y-x\in\Bbb N$. I have tried doing this in a similar way as with the sum and the product, but no result so far. Any ideas?

Answer 1

Let $$ A=\{\,x\in \Bbb R\mid \forall y\in \Bbb N\colon y>x\to y-x\in\Bbb N\,\}$$ and show that $A$ is inductive.

Answer 2

Lemma: If $y\in\Bbb N$ and $y>1$ then $y-1\in\Bbb N$.

Proof: Assume that there is some $z\in\Bbb N$ such that $z>1$ and $z-1\notin\Bbb N$. Let $A=\Bbb N\setminus\{z\}$. Since $z\neq 1$, $1\in A$. Now let $t\in A$. Then $t+1\in \Bbb N$.

If $t+1\notin A$ then $t+1=z$, or $t=z-1$, which is not natural. This is a contradiction, because $t\in A\subset \Bbb N$, so $t+1\in A$.

Since $1\in A$ and for all $t$, $t\in A$ implies $t+1\in A$ we have that $A$ is inductive, so $z\in\Bbb N\subset A$, another contradiction. This time it implies that there is no $z>1$ in $\Bbb N$ such that $z-1\notin\Bbb N$, q.e.d.

Theorem (?): if $x,y\in\Bbb N$ and $x<y$ then $y-x\in\Bbb N$.

Proof: As it has been suggested, let $$A=\{z\in\Bbb N:\forall t(t\in\Bbb N\wedge t>z)\to t-z\in\Bbb N\}$$ By Lemma, $1\in A$.

Let $z\in A$. Let $t\in\Bbb N$, $t>z+1$. Clearly $z+1>1$, so $t>1$. Again by Lemma, $t-1$ is natural. More over, $t-1>z$. Since $z\in A$ and $t-1>z$ then $t-1-z$ is natural, that is $t-(z+1)$ is natural. Thus, $t-(z+1)$ is natural for every natural $t>z+1$. So $z+1\in A$. Then $A$ is inductive, and this completes the proof.