Is the Axiom of Choice Necessary to Well-Order Finite Sets?

[Csharp]

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)

 

[Deveno]

Have you tried using induction on the cardinality of the set?

 

[Csharp]

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 it contains k then S-{k} has a lowest member which is also lower than k itself.

 

[Evgeny.Makarov]

I want to show that there exists a well ordering for every finite set.

Every total order on a finite set is a well-ordering.