Equinumerous well ordered sets are isomorphic

Question

Suppose U and V are well ordered sets that are equinumerous with each other, ie there is a bijection between U and V. Is it true that U and V are isomorphic, ie there is a bijection f between U and V so that $\ x \le_U y$ iff $\ f(x) \le_V f(y)$?

If it is true, why?

Answer 1

It's not entirely clear to me what the question is. I think you are asking whether well-ordered sets of the same cardinality are isomorphic. The answer to this is no. For instance there are uncountably many isomorphism types of countably infinite well-ordered sets. As a concrete example, $$A=\{0\}\cup\{-1/n:n\in\Bbb N\}$$ is a well ordered set (in standard ordering), equinumerous, but not order-isomorphic to $\Bbb N$, as $A$ has a largest element.

See the theory of ordinals for many more examples.

Answer 2

By definition, if U and V are equinumerous,
then there is a bijection between them.

$\omega_0$ and $\omega_0$ + 1 are two equinumerous
ordinals that are not order isomorphic.