I was reading the book "Quantum Computing Since Democritus".
"The set of ordinal numbers has the important property of being well ordered,which means that every subset has a minimum element. This is unlike the integers or the positive real numbers, where any element has another that comes before it."
Unlike integers? Let's consider a set $\{1,2,3\}$ This has a minimum element.
Do you get what does the author wants to say here?
On
When considering an ordinal $\alpha$ (von neumann ordinals), not all its subsets will have the $0$ as the minimum element. I might be mistaken, but I got the impression that perhaps OP might be assuming that to be the case.
So if one considers $\alpha \subseteq \alpha$, its minimum element will be $0$. In other words $\min(\alpha)=0$.
Now say $\alpha=\omega^3$. We might consider a subset $S_1$ defined as: $S_1=\{\, \omega \cdot (1+x) \,|\, x \in \mathrm{Ord} \, \, \wedge x<\omega^2 \, \}$. Similarly we might define $S_2=\{\,\omega^2 \cdot (1+x) \,|\, x \in \mathrm{Ord} \, \, \wedge x<\omega \, \}$. We can also write $S_2=\{\, \omega^2,\, \omega^2 \cdot 2,\, \omega^2 \cdot 3,.... \,\}$. Both $S_1$ and $S_2$ are well-defined subsets of $\alpha=\omega^3$. However, we have $\min(S_1)=\omega$ and $\min(S_2)=\omega^2$.
Edit: An easier example similar to two above. Define $S_0=\{\, 1+x \,|\, x \in \mathrm{Ord} \, \, \wedge x<\omega^3 \, \}$. We get $S_0 \subset \omega^3$ and $\min(S_0)=1$.
The integers mean $\Bbb Z$ here, so there we have sets like $\{-1,-2,-3,-4,\ldots\}$ (or $\Bbb Z$ itself) which do not have a minumum element and every $n$ has an element $n-1$ before it. So there are not well-ordered. The positive integers, i.e. $\Bbb N$, are.