Set theory: Show these are equivalent.

Quote:

Show that the following are equivalent:

a). $A$ is finite
b). $A$ admits a well order <whose inverse> is also a well-order.
c). The structure $(\mathcal{P}(A), \subset_{\mathcal{P}(A)})$ is well-founded: that is every $S \subset \mathcal{P}(A)$ has a $\subset$ -minimal element.

I decided to go for $a) \Rightarrow b) \Rightarrow c) \Rightarrow a)$ .

For $a) \Rightarrow b)$ :

Assign each element of $A$ an ordinal number. Since $A$ is finite, and there are an infinite number of ordinals, we will always be able to do this. We can then order the ordinals (and the elements of $A$ they represent) in the following way:

ie. $\phi$ , $\{ \phi \}$ , $\{\phi, \{ \phi \} \} \ \ldots$

or: $\phi, \ \mathcal{P}(\phi), \ \mathcal{P}\mathcal{P} (\phi ), \ \ldots$

This is a partial order and since every element contains $\phi$ , every non-empty subset contains a least element.

For $b) \Rightarrow c)$

If we take any $S \subset \mathcal{P}(A)$ , $S$ contains sets of elements of $A$ .
Since $A$ is well-ordered, each of the subsets of $S$ has a least element.
We can take all these least elements and make a new set.
We can repeat this argument until we get a singleton set (the number of steps required to do this is equal to $|S|$ ).
This singleton set is the $\subset$ -minimal element.

For $c) \Rightarrow a)$ I run into some trouble.

I figured the best way was to use a proof by contradiction. So suppose $A$ is infinite.
Let $S$ be any subset of $\mathcal{P}(A)$ and let $B \subset \mathcal{P}(A)$ be the $\subset$ -minimal element of $S$ .

I think I need to show that $\mathcal{P}(A)$ is finite which is equivalent to a finite number of $B \subset \mathcal{P}(A)$ . Are there any tips for this?