Good idea.
Suppose that k is well ordered.
k+1= k U {k}
First of all I'll define an ordering on k+1.
If s and t are both in k then I use the ordering from k.
If one of s and t is k then k>s.
Suppose that S is a nonempty subset of k+1.
Then if it doesn't contain k it has a lowest member.
If...
Hi,
I want to show that there exists a well ordering for every finite set.
(I know if you add axiom of choice you can prove this theorem for infinite sets too but I think the finite sets do not need axiom of choice to become well ordered)