Show that the set $A$ is inductive Prove that for $m,n\in \mathbb{N}$, $m>n$ it holds that $m-n\in \mathbb{N}$.
Hint: Consider $A=\left \{n\in \mathbb{N}\mid \forall m\in \mathbb{N}, m>n:m-n \in \mathbb{N}\right \}$ and show that $A$ is inductive.
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I have done the following:
$1\in A$ : Let $m > 1$ and $m-1=r \Rightarrow m = 1 + r$ for some $r \in \mathbb{N}$ . We have that since $\mathbb{N}$ is an inductive set and $r\in \mathbb{N}$ then $r+1\in \mathbb{N}$, i.e. $m\in \mathbb{N}$.
Let $n\in A$. Then let $m > n$ and $m = n + r$ for some $r \in \mathbb{N}$ then $m \in \mathbb{N}$. We want to prove that $n+1\in A$, i.e. that if $m' > n+1$ and $m' = (n+1) + r$ for some $r \in \mathbb{N}$ then $m'\in \mathbb{N}$.
From $m'>n+1$ we have that $m'-1>n$. So we get that $m'-1\in A$. Then $m'-1=n+r$ for some $r\in \mathbb{N}$, so $m'=(n+1)+r$, for the same $r\in \mathbb{N}$ and so $n+1\in A$.
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Is that correct? Or am I confusing the meaning of the set $A$ ?

You seem to be confusing the point is to prove $k\in A$ you must prove for every natural $m > k$ than $m-k = r$ or $m = r+k$ then you must prove that it is !!$r$!!! that is a natural number; not !!$m$!! that is a natural number.
So to prove $1 \in A$ you must prove that if $m > 1$ and $m\in \mathbb N$ that $m-1\in \mathbb N$.  By definition if $m\in \mathbb N$ then either $m = 1$ or $m$ is the successor of a natural number $r$.  That is $m = r+1$, for some $r\in \mathbb N$.
And as $m$ isn't equal to $1$ then $m$ is a successor of some $r$.  And $m-1 = r\in \mathbb N$.  So that proves that $m-1 \in \mathbb N$ for all $m \in \mathbb N; m > 1$ and therefore than $1\in A$.
And to prove $n\in A\implies n+1 \in A$, you know that $n\in A$ so for all $m> n$ then $m-n \in \mathbb N$.  So for all $m > n+1$ (no need to use a second variable-- we never took a specific $m$; we said all $m > n$ and now we are refering to all $m > n+1$. we have by trasitivity that $m > n$ therefore $m-n \in \mathbb N$.  So either $m-n = 1$ or $m-n$ is a successor to some $r \in \mathbb N$. That is $m-n = r+1$.
If $m-n= 1$ then $m=n+1$ and we do not have that $m > n+1$.  So $m-n \ne 1$.  So $m-n=1 + r$ for some $r\in \mathbb N$.  So $m -n - 1 = m-(n+1) = r\in \mathbb R$. So for any $m > n+1$ we have shown that $m-(n+1) \in \mathbb N$.  Therefore $n+1 \in A$.
So $A$ is inductive.
So as $1 \in A$ and $A$ is inductiove then $\mathbb N\subset A$.
So for any $n \in \mathbb M$ and any $m > n$ then $n \in A$ so $m-n \in \mathbb N$.
And that proves our statement.
