# Isomorphism between well ordered sets

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?

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## 2 Answers

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.

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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.

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