I've given this a shot, but I think it's a bit jumbled. I decided to try induction:Show that for any $\displaystyle E \subset A$, the following are equivalent:

a). $\displaystyle E \in K$ for every Kuratowski-Inductive $\displaystyle K \subset \mathcal{P}(A)$.

b). $\displaystyle E$ is finite.

The first Kuratowski-Inductive subset of $\displaystyle \mathcal{P}(A)$ is when $\displaystyle K=\{\phi \}$. This is finite since $\displaystyle |K|=1$.

Assume each of the next $\displaystyle n-1$ sets formed by unions with elements $\displaystyle x_1, \ldots , x_{n-1} \in A$ are finite.

ie. $\displaystyle \{ \phi \} \bigcup \{x_1\} \bigcup \ldots \bigcup \{x_{n-1} \}=\{\phi, \ x_1, \ldots, \ x_{n-1} \}$.

If we add another $\displaystyle x_n \in A$ we get:

$\displaystyle \{\phi, x_1, \ldots , \ x_{n-1} \} \bigcup \{x_n \}=\{\phi, x_1, \ldots , \ x_{n} \}$.

This is finite since there are n+1 elements in this set.

Is this right?

Also, do I need to prove the converse? My notes tell me that the converse is by definition.