Set theory: Show these are equivalent.

Quote:

Show that the following are equivalent:

a). $\displaystyle A$ is finite

b). $\displaystyle A$ admits a well order <whose inverse> is also a well-order.

c). The structure $\displaystyle (\mathcal{P}(A), \subset_{\mathcal{P}(A)})$ is well-founded: that is every $\displaystyle S \subset \mathcal{P}(A)$ has a $\displaystyle \subset$-minimal element.

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

For $\displaystyle a) \Rightarrow b)$:

Assign each element of $\displaystyle A$ an ordinal number. Since $\displaystyle A$ 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 $\displaystyle A$ they represent) in the following way:

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

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

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

For $\displaystyle b) \Rightarrow c)$

If we take any $\displaystyle S \subset \mathcal{P}(A)$, $\displaystyle S$ contains sets of elements of $\displaystyle A$.

Since $\displaystyle A$ is well-ordered, each of the subsets of $\displaystyle S$ 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 $\displaystyle |S|$).

This singleton set is the $\displaystyle \subset$-minimal element.

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

I figured the best way was to use a proof by contradiction. So suppose $\displaystyle A$ is infinite.

Let $\displaystyle S$ be any subset of $\displaystyle \mathcal{P}(A)$ and let $\displaystyle B \subset \mathcal{P}(A)$ be the $\displaystyle \subset$-minimal element of $\displaystyle S$.

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