Sign up ×  I need to show specifically using transfinite
induction that given two well-ordered sets
$\left(A,<_{1}\right)$ and $\left(B,<_{2}\right)$ there is only
one isomorphism between them. To do this I want to show by
induction that there is only one way to define the mapping and
that is: $$\forall\, x\in A:
\varphi\left(x\right)=\min\left(B\backslash\varphi\left(\left\{
a\in A\,:\, a<_{1}x\right\} \right)\right)$$ Showing the base of
the induction is no problem and the induction hypothesis would
be that $$\forall\, y<_{1}x\,:\,\varphi\left(y\right)=\min\left(
B\backslash\varphi\left(\left\{ a\in A\,:\, a<_{1}y\right\}
\right)\right)$$ My problem is that I can't seem to manage to
use the hypothesis in order to prove the step. I've found a
couple of ways to show this definition of the mapping is indeed
necessary but none of those used the induction hypothesis. My
plan was to mark
$$c_{0}=\min\left(B\backslash\varphi\left(\left\{ a\in A\,:\,
a<_{1}x\right\} \right)\right)$$ Using the fact $\varphi$ is
surje