Given a well ordering of a set $A$ we can define a total order $A^A$ in an obvious way (for $f \neq g$ find the least $i$ such that $f(i) \neq g(i)$ and define $f < g$ if $f(i) < g(i)$)


Does the inverse direction work? Does a total order on the powerset of $A$ give rise to a well ordering of $A$? (without choice, of course, for otherwise the result is obvious)



Total order on the powerset
No. If this were true, then ZF would prove that "every set can be totally ordered" implies "every set can be well-ordered", which (assuming ZF is consistent) it doesn't. I can't find the original citation for this nonimplication, but it's in Howard and Rubin's "Consequences of the Axiom of Choice" for example.


