# Linearly ordering the power set of a well ordered set with ZF (without AC)

As the title says, my question is, how one can use only ZF-theory to prove that the power set of A, whereby (A, <) is a well-ordering, can be linearly ordered?

• Lexicograpohically, by first difference? Where's the snag? – bof Jan 21 '19 at 6:20

• Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$. – bof Jan 21 '19 at 6:22
• Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x \in X$. – Ross Millikan Jan 21 '19 at 16:46
• You don't "just need a total order", you need a well-order. How would you order the power set of $\mathbb R$? Which comes first, $\mathbb Q$ or $\mathbb R\setminus\mathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $\mathbb R$. – bof Jan 22 '19 at 1:11