TITLE: Confused about a well-ordering lemma
QUESTION [5 upvotes]: I happened to stumble across the following lemma in Kenneth Kunen's set theory book: 
$\textbf{Lemma:}$ Let $\langle$ $A,R$ $\rangle$ be a well ordering. Then for all $x \in A$, $\langle$ $A,R$ $\rangle$ is not isomorphic to 
$\hspace {15mm}$ $\langle$ pred$(A,x,R),R$ $\rangle$.
Where $A$ is a set, $R$ is a strict total order on $A$, and pred$(A,x,R)$ = { $y \in A: yRx$ } where $x \in A$.  
But then what if we let $A=\mathbb{N}$, $R$ be the standard greater than relation >, and $x=0$?
$A$ would be the set $\{0,1,2......\}$ 
pred$(A,x,R)$ would be the set $\{1,2,3......\}$ 
I am pretty that for both sets, > is still a strict total order and every subset has a least element under this order. So couldn't we just set an isomorphism $f: A \rightarrow$ pred$(A,0,R)$ where $f(x)=x+1$?
Sorry if I happen to be incredibly stupid right now but I can't spot my error.

REPLY [4 votes]: No, the set of predecessors are those "smaller" than $x$. So in the case of $\Bbb N$ and $x=1$ this is just $\{0\}$. For $x=0$ it's $\varnothing$. Neither is isomorphic to $\Bbb N$ itself.
If you reverse the order, then it's not a well-ordering anymore, since there is no minimal element.