Proof of Well Ordering Principle Please help me in Validating the below proof for Well Ordering Principle.
Theorem:
Every non-empty set S of non-negative integers contains a least element. 
Proof:
Let $W=\{0,1,2,3,4,5, \ldots\}$ and $S$ be a non-empty subset of $W$.
Let us assume that $S$ has no least member.
Hence if some $a$ belongs to $S$, $a\leq b$, for some $b$, belongs to $S$, then there exist an $M$ in $S$, such that $M\leq a$.
(I.e I can always find an $M$ in $S$ for every $a$ in $S$ such that $a\leq b$ and $M \leq a$ for some $b \in S$).
So we have $M_n$ for every $a_n \in S$ such that $a_n \leq b$ some $b \in S$.($M_1$ for $a_1 \leq b$,
$M_2$ for $a_2 \leq b \ldots$ So on till some $M_n$)
With the sequence of $M_n$'s such that satisfying below statement :
$M_1 < M_2 < \ldots < M_n$.
Above sequence will have  lower bound of $0$, that is the least value of $\{M_n\}$ sequence as the $S$ is subset of $W$.
Thus there will an $M$ in $S$ such that it will be minimum of all $\{M_n\}$.
Hence our assumption was wrong that $S$ have no least element, as we have a bound on $M$ in $S$ such that it belongs to $W$ and $S$ both.
Please help me to pinpoint where I am wrong and what step will rectify the error.
 A: It looks that your proof is a bit unclear, but actually if you clear it up I think there's no actual error as such in the proof or idea. If I understood you right.
Basically you note that you can since $S$ is non empty it has an element $b$ and the set $S_b = \{x\in S| x\le b\}$ is both non-empty and finite and therefore has a minimal element. That element is also the minimum of $S$.
But what you're missing is pointing out that $S_b$ has these properties, especially finiteness. You just point out "and so on" until $M_n$...
Also it's a bit overelaborate as you actually doesn't need $M_n$ to work it. You just need the finite sequence $a_n \in S_b$. Also calling the set of natural numbers $W$ instead of $\mathbb N$ makes the proof just a bit less straight forward to read.
Also the use of reductio ad absurdum is formally correct, but redundant as the assumption of no least argument arrives at the contradiction that it in fact has, but you never actually used the assumption and you would arrive at that fact nevertheless and having a direct proof.
