Non-standard models of arithmetic for Dummies Why is (1) a copy of $\mathbb{N}$ "followed by" a copy of $\mathbb{Z}$ not a (non-standard) model of arithmetic, neither (2) a copy of $\mathbb{N}$ followed by an infinite sequence of copies of $\mathbb{Z}$, but (3) a copy of $\mathbb{N}$ followed by infinitely many densely ordered copies of $\mathbb{Z}$ is? (see the Wikipedia entry on non-standard models)
Can this intuitively be seen, or be explained in laymans terms?
The axioms concerning the successor function hold in all of these (pseudo-)models, don't they?
But the induction axiom really puzzles me! Naively, it can be interpreted as describing essentially an infinite row of dominoes: knocking over the first will let fall all of them. How can this be understood in the non-standard model (3), where there is no immediate "contact" between the building blocks? What's the "true" interpretation of induction then? And why does it work in (3) but not in (2) or (1)?

${\mathbb N}$ is more than an ordered set with successor. You have addition (and multiplication) as well. Let $M$ be a nonstandard model; we can identify its beginning with ${\mathbb N}$. Given $n\in M$ let $[n]$ be the collection of all $m$ that are at finite distance from $n$. Note $[n]={\mathbb N}$ if $n$ is finite, and otherwise $[n]$ has the same order type as ${\mathbb Z}$, because every number has a successor, and every number other than zero has a predecessor. 
If $n$ is an infinite nonstandard number, then $n+n$ is also infinite but, moreover, it is infinitely away from $n$, so just from $M$ being nonstandard we deduce that there is no largest copy of ${\mathbb Z}$ in the order of $M$.
Also, either $n$ or $n+1$ is even, so there is an $m$ such that $m+m=n$ or $m+m=n+1$. This $m$ is infinite, so the copy $[m]$ of ${\mathbb Z}$ is before the copy $[n]$. This shows that there is no first copy of ${\mathbb Z}$ in $M$.
Finally, given $n<k$ and infinitely apart, $n+k$ or $n+k+1$ is even, and if $l$ is its half, then $n<l<k$, and $l$ is infinitely apart from both. This shows that between any two copies of ${\mathbb Z}$ we have another.
We have shown that, removing the original ${\mathbb N}$, and taking the quotient that identifies all elements in a class $[n]$ as a single point, we are left with a linear order that is dense in itself and has no end points. If $M$ is countable, this order is ${\mathbb Q}$. Otherwise, it is even more complicated.
