# Total order on the powerset

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)

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What about considering elements of $A$ as one-element subsets? – darij grinberg Jan 22 '11 at 14:37
This does necessarily work. You can fix a bijection from 2^Q to R such that {q} is mapped to q, and the usual ordering of R is total, but Q is not well ordered... – mathahada Jan 22 '11 at 15:03
Ah, you want a well ordering. Should learn to read. – darij grinberg Jan 22 '11 at 17:09