Prove that the well ordering principle is equivalent with PMI. So I am supposed to prove that the well ordering principle is equivalent with the maximum principle.
Well ordering principle:
Every nonempty subset of the set of positive integers has a least element.
The maximum principle:
let $T \subset Z_{\geq 0}$ be a nonempty subset which is bounded above. Then $T$ has a greatest element.
Actually, I dont see how I am going to use WOP to prove TMP, I know it might be wrong but since we consider integers, isn't TMP rather obvious? I mean, if it did not contain a greatest element then it would not be bounded above (this is of course not true if we consider real numbers). Am I thinking about this in a wrong way?
 A: Suppose the WOP, and let $\;\emptyset\neq T\subset\Bbb N\;$ be bounded above. Let $\;X\;$ be the set of upper bounds of $\;T\;$ , i.e.:
$$X:=\{x\in\Bbb N\;;\;\forall y\in T\;,\;\;y\le x\}$$ 
Since $\;X\neq\emptyset\;$ (why?), there's an element $\;x\in X\;$ which is minimal, from which it follows that $\;x-1\notin X\;$ ...complete the proof now.
A: Suppose that there is no greatest element of $T$. Let an upperbound of $T$ be denoted as $\zeta$. We choose $\zeta$ to be a positive integer (such an integer always exists by the Archimedean Property of the Number System, and that follows from the WOP) . Now choose any element $t_0$ $\in$ $T$. By assumption since there is no greatest element of $T$ hence we conclude that there must be an infinite number of integers between two integers which in view of The Principle of Mathematical Induction (and its corollary that if for a set of integers $S$, 1. $-1$ $\in$ $S$, 2. $k$ $\in$ $S$ $\implies$ $k-1$ $\in$ $S$ then $S$ is the set of all negative integers) is impossible.
Now note that PMI only states that any two positive (by the corollary that I have stated above, negative also) integers can be reached by a finite number of addition of $1$. But my proof is still incomplete because I have not yet proved that the element $0$ can be reached in finitely many steps from any integer. Now this is trivial because start from any integer $a$, if it is positive then reach $1$ in finitely many steps and then just take one more step to go to $0$. Similar is the argument for negative $a$.
Hence proved.
