My set theory notes define a similarity (or order isomorphism) between two $\textbf{well-ordered}$ sets $(X,\leq_{1})$ and $(Y,\leq_{2})$ as a bijection $f:X\longrightarrow Y$ such that $x_1 \leq_1 x_2 \iff f(x_1) \leq_2 f(x_2)$. Now I do not have any explicit examples of these and I am struggling to think of any, as the definition specifies well-ordered sets as opposed to posets.
I tried showing that the function $f:\mathbb{Z}\longrightarrow \mathbb{N}, f(z)=2z (z>0), f(z)=1-2z (z\leq 0)$ was one, and it turns out its not an order isomorphism, but I forgot that under that usual ordering of $\mathbb{Z}$, it is not a well-ordering.
So, can anyone give a simple example of what I'm looking for?
Thanks
You may be having a hard time coming up with suitable examples of well-ordered sets. Here’s a simple concrete example to get you started.
Let $X=\Bbb N\times\{0,1\}$, and let $\preceq$ be the lexicographic order on $X$: $\langle m,i\rangle\preceq\langle n,j\rangle$ iff $m<n$, or $m=n$ and $i\le j$. Thus, $\langle 0,0\rangle$ is the $\preceq$-least element of $X$, and the next few in increasing order are $\langle 0,1\rangle,\langle 1,0\rangle,\langle 1,1\rangle$, and $\langle 2,0\rangle$. It’s a straightforward exercise to verify that this is a well-order and that
$$f:X\to\Bbb N:\langle n,i\rangle\mapsto 2n+i$$
is an order-isomorphism whose inverse is
$$f^{-1}:\Bbb N\to X:n\mapsto\left\langle\left\lfloor\frac{n}2\right\rfloor,n-2\left\lfloor\frac{n}2\right\rfloor\right\rangle\;.$$