Let A be well ordered by an order relation G. Let A be well ordered by an order relation G'. G ≠ G'.
Then does there exist an isomorphism f between A ordered by G and A ordered by G'?
Plus, is there a generalization of this statement?
Let A be well ordered by an order relation G. Let A be well ordered by an order relation G'. G ≠ G'.
Then does there exist an isomorphism f between A ordered by G and A ordered by G'?
Plus, is there a generalization of this statement?
Not in general. Let $\le$ be the usual well-ordering of $\Bbb N$, and define another well-ordering $\preceq$ on $\Bbb N$ as follows: for $m,n\in\Bbb N$,
$m\preceq n\text{ iff }\begin{cases}1\le m\le n,\text{ or}\\ m=n,\text{ or}\\ m>0\text{ and }n=0\;. \end{cases}$
Schematically, $\langle\Bbb N,\preceq\rangle$ looks like: $1,2,3,4,\dots,0$.
Then $\langle\Bbb N,\preceq\rangle$ has a last element, and $\langle\Bbb N,\le\rangle$ has none. However, $\langle\Bbb N,\le\rangle$ is isomorphic to an initial segment of $\langle\Bbb N,\preceq\rangle$ by the map $n\mapsto n+1$. This is a special case of the fact that given two well-orders, one is always isomorphic to an initial segment of the other.
As stated this is not true in general. Just notice that $\mathbb N$ with the usual order has order type $\omega$.
But you can reorder $\mathbb N$ to get order type $\omega+1$, simply by choosing the order such that $a\preceq 0$ for each $a$ and $a\preceq b$ is the usual ordering for $a,b\ne 0$.
However, there is a closely related result saying that if $A$, $B$ are any well-ordered set then one of them is isomorphic to initial segment of the other one. Some authors call this fundamental theorem for well-ordered sets, e.g. Suppes: Axiomatic Set Theory, p.223.
On the other hand, it is possible to have two different well-orders of $\mathbb N$ which is isomorphic. E.g. take the usual ordering as one of them and an ordering where $0$ and $1$ are interchanged as the other one.