This question is probably answered some where but I couldn't find it: I have a question with an argument Dudley (Real analysis and Probability) used in his proof of AC implies WO. He did not explain very clearly from my sight.
Given set X, there is a choice function $c: 2^X\setminus\{\emptyset\}\to X$, such that $c(S)\in S$. Let $(A,<^A)$ be well ordered subset of X such that $x=c(X\setminus L^A_x)$ for all $x\in A$, where $L^A_x=\{y\in A,y<^Ax\}$ (Such sets exist, e.g, $\{c(X)\}$.), call such set A c-WO.
For two such sets, A, B, assuming wlog that A is not an initial segment (IS) of B (S is an initial segment of $(B,<^B)$ if $S\subset B$ and for all $x\in S$, $\{y\in B,y<^Bx\}\subset S$ ), let $m\in A$ be the $<^A$-least element such that $\bar{L}^A_m=\{y\in A, y\le^Am\}$ is not an IS of $(B,<^B)$, this exists since if no such m exists, then $\cup_{x\in A}\bar{L}^A_x=A$ is an IS of $(B,<^B)$ (contradiction), and since set of such x's is a subset of the WO set $(A,<^A)$, it has a least element m.
Let $D=L^A_m=\bigcup_{x\in A,x<^Am}\bar{L}^A_x$, then $D$ is an IS of A and B since each set in the union is an IS of A and B by construction. Now Dudley claims that $<^A$ and $<^B$ agree on $D$, which I can't see why. I guess I need to use the c-WO of A and B and thus D, but how to show the c-order is unique on D by construction?
Suppose $x<^A y$ and $x\in D$. Then $S=\bar{L}_x^A$ is an initial segment of $B$, with $x\in S$ and $y\not\in S$. This implies $x<^B y$, since otherwise $y$ would need to be in $S$ in order for $S$ to be an initial segment. Thus $<^A$ and $<^B$ agree on $D$.