Proving $\text{PMI}\implies\text{WOP}$ without PCI I used the following proof to show that  $\text{PMI}\implies\text{WOP}$. What I  intend to show, is that from $P(1)$ we can show that $P(n)$ holds for all natural numbers. However, my professor said it is wrong insofar as I'm assuming $n$ satisfies my property, and that I can't generalize that way without pinpointing what $n$ is and why $P(n)$ holds. He proceeded by giving me his proof, but I just realized it makes the same mistake as mine.
Is my proof truly wrong? What about his? Thank you in advance.

Beginning with mine, here are both proofs:

We want to use the Principle of Induction, i.e $$P(1)\land \forall n [P(n)\implies P(n+1)]\implies S=\mathbb N,$$ to show that if a set $A$ is a non-empty subset of the natural numbers, then it has a least element. We will proceed to prove this by contraposition, i.e,$$A\subset\mathbb N \land A\text{ does not have a least element}\implies A=\emptyset$$

Suppose $A$ is a subset of the natural numbers such that it does not contain a least element, and let S be the set complement of A such that $S=\mathbb N-A$. Now let us define a property $P(n)$
  $$P(n):=\{1,2,...,n\}\subset S.$$ Clearly $P(1)$ is true (since $(1\in A\implies 1=min\ A)$ is a contradiction of our premise). Now suppose $P(n)$ is true. Then it similarly follows that $P(n+1)$ is true, because if $n+1$ were in $A$ then it would be minimal. Then, by induction on $S$, it follows $S=\mathbb N$ and thus $A$ is empty.



Let $A$ be a subset of $\mathbb  N$ without a minimum ($\forall x\in A\exists y\in A: y\lt x$), and let $S$ be the set containing all the lower bounds of $A$ ($\forall x\in A\forall y\in S:y\le x$). 
$1\not\in A$ otherwise it would be minimal, so $1\in S$. Now suppose $n\in  S$ and consider $n+1$. 
If $n+1$ were not in $S$, then there would be an $x\in A$, say $x_{0}$, such that $$x_{0}\lt n+1\implies x_{0}=n \lor x_{0}\lt n.$$ However, $x_{0}$cannot be smaller than a lower bound of $A$, and thus it must follow that $x_{0}=n$, so $A\cap S = \{x_{0}\}$. But this would also be a contradiction since, by means of membership of S, $x_{0}$ would be the least element of $A$. 
Then it must follow that $n+1$ is in S, and, by induction on S, $S=\mathbb N$ and $A=\emptyset$. It is proved, then, by contrapositive, that $$A\subset \mathbb N \land A\not=\emptyset \implies A\text{ has a min.}$$

 A: Your proof is perfectly correct.  There is absolutely nothing wrong with assuming $P(n)$ is true, since you are proving that for all $n$, $P(n)$ implies $P(n+1)$.  That means you get to assume $P(n)$ is true, and then need to prove $P(n+1)$.
His proof is essentially correct, but the phrasing of the first sentence is weird.  It is redundant to say 

Let $A$ be a subset of $\mathbb  N$ without a minimum  such that $\forall x\in A\exists y\in A: y\lt x$

since the statement at the end is just a restatement of what it means for $A$ to not have a minimum.  The statement

let $S$ be the set containing all the lower bounds of $A$ such that $\forall x\in A\exists y\in S:y\le x$

appears to have a similar redundancy (the formula at the end is attempting to repeat the first part of the statement), but is actually stated incorrectly.  If you want to say every element of $S$ is a lower bound of $A$, you should say $\forall x\in A{\color{red}\forall} y\in S:y\le x$ instead.  (Note though that this formula still does not mean that $S$ is the set of all lower bounds of $A$, though it is unnecessary to make such a statement at all since you have already defined $S$ to be that set.)
