I have a question related to the one found here. I am struggling to understand this part of the proof:
So $n$ must be a product of two integers $a$ and $b$ where $1 < a, b < n$. Since $a$ and $b$ are smaller than the smallest element in $C$
How come $1 < a$ and $b < n$? Where does it come from? And why does it imply that $a$ and $b$ are smaller than the smallest element in $C$ if $C$ is by definition contains all the nonnegative integers greater than $1$?
Upd.: this is not a duplicate of Proof by well ordering: Every positive integer greater than one can be factored as a product of primes. because in this question there is a different statement that needs to be explained, hence Part II in the heading.
Sorry I deleted my comment. The proof attempts to show $C$ has no members at all. It is defined as only the integers greater than $1$ which are not prime and do not factor into primes. Let $D$ be the numbers starting at $2$ which do factor into primes or are prime themself. If two numbers are in $D$ so is their product. If, in fact, $C$ is not empty it has a smallest member $n$ so if $1 \lt m\lt n$ then $m$ is in $D$ but not in $C$ and does factor into primes. But for this smallest $n$ in $C$ , $n$ is not prime so $n$ does factor into $n=ab$ with $2 \leq a \leq b \leq \frac{n}2$ so neither of $a$ or $b$ is in $C$, they are both in $D$ as is their product $n=ab.$ that is a contradiction.