TITLE: Is every model of PA well-ordered?
QUESTION [5 upvotes]: Assume first-order Peano arithmetic is consistent and $N$ is its model, we know that every subset of $N$ contains a minimal element. It's a second-order property so I am not sure if it hold in nonstandard models, or specifically, in countable nonstandard models?
Here we define $x<y$ as $\exists z\ne0(x+z=y)$. Anyone can provide some ideas will be appreciated. Thanks!

REPLY [5 votes]: Here is a useful fact, which is not hard to prove, that answers the question (and so it gives a self-contained proof that no nonstandard model of PA is well-ordered). 
Fact: In any nonstandard model $M$ the set of standard natural numbers is an initial segment of $M$, but it is never definable in $M$, even with parameters from $M$. 
Proof: assume the set of standard numbers is definable by a formula $\phi(n)$ that may have parameters from $M$. Then $\phi(x)$ holds for each $x \in M$ if and only if $x$ is standard, so $\psi(y) = (\forall x \leq y)\phi(x)$ also holds exactly when $y$ is standard, because the set of standard numbers is downward closed. Thus $M$ satisfies $\psi(0)$ and $M$ satisfies $(\forall y)[\psi(y) \to \psi(y+1)]$, but $M$ does not satisfy $(\forall y)\psi(y)$, because $M$ is nonstandard and $\psi$ only holds for standard numbers. Thus the axiom scheme of induction, which is part of PA, cannot hold in $M$, because the instance of the scheme for $\psi$ fails. This is a contradiction, because $M$ satisfies PA. 
Now to answer the question. If $M$ is nonstandard, the set $$\{n \in M : n \text{ is not standard }\}$$ cannot have a least element $x$. If it did, then $\{y \in M : y < x\}$ would be the set of standard numbers, and this set would be definable using $x$ as a parameter, but the set of standard numbers is not definable in $M$.