Show that $A$ is countable iff $\mathcal{P}_{fin}(A)$ is countable I need to show that $A$ is countable iff $\mathcal{P}_{fin}(A)$ is countable.
where $\mathcal{P}_{fin}(A) = \left \{ x\subseteq A\mid x\;is\;finite \right \} $
I know that if $A$ is countable, I can build an infinite amount of such $x$'s, but its obviously not enough. I am missing the main direction of the proof.
I would appreciate help on this proof
 A: As mentioned in other answers, one direction is completely trivial. Assume that $A$ is countable. There exists a bijection $\phi : A\to\mathbb{N}$. Now, rewrite $\mathcal P_{\text{finite}}(A) = \bigcup_{k=0}^\infty \mathcal P_k$ where $\mathcal P_k$ is the set of $k$ element subsets of $A$. For each $k$, there exists an injection $\phi_k : \mathcal P_k\to \prod_{i=1}^k A$ sending a $k$ element subset to some ordered $k$ tuple of the same elements (there are indeed many choices for the $k$ tuple, but you may choose one of them by, say, ordering each finite subset in $\mathcal P_k$). So, by definition, $|\mathcal P_k|\leq |\prod_{i=1}^k A|$. Further, there exists an obvious bijection between $\prod_{i=1}^k A$ and $\mathbb{N}^k := \prod_{i=1}^k \mathbb{N}$ where $(a_1, a_2, \ldots, a_k)$ is mapped to $(\phi(a_1), \phi(a_2), \ldots, \phi(a_n))$. So, $|\mathcal P_k|\leq |\mathbb{N}^k|$. Putting this together, we obtain $|\mathcal P_{\text{finite}}(A)|\leq |\bigcup_{k=1}^\infty \mathbb{N}^k|$. You could easily show that $|\mathbb{N}^2| = |\mathbb{N}|$ (this is essentially the same argument as for $|\mathbb{Q}| = |\mathbb N|)$.  By a straightforward and simple induction argument (which you should go through if you have not seen it before), we get that $|\mathbb N^k| = |\mathbb N|$. By Cantor's theorem, a countable union of countable sets is countable, hence $\mathcal P_{\text{finite}}$ is (at most) countable. 
A: If $A$ has a finite number of elements $n$, $\mathcal{P}_{fin}(A)=\mathcal{P}(A)$ has $2^n$ elements. If $A$ has $\aleph_0$ elements, $\mathcal{P}_{fin}(A)$ has $\sum_{k\ge0}\aleph_0^k=1+\aleph_0^2=\aleph_0$ elements. If $A$ is uncountable, $\mathcal{P}_{fin}(A)$ has as many $1$-element elements as $A$ has elements, so is uncountable.
A: You could use a naive but intuitive zig-zag argument for the non-trivial direction. (For obviously if the finite subsets are countable the singletons are countable, and the set of singleton subsets of $A$ is equinumerous with $A$.)
Note that, for fixed finite $n$, the $n$-element subsets of $A$ are countable (why?)
Imagine now putting the one element subsets in a (countable) row, starting on the left; beneath them put the two element subsets in a (countable) row; beneath them put the three element subsets; keep on going.
You've got all the finite subsets in a two-dimension array, countably many columns, countably many rows.
Take the empty subset, then zig-zag through from the top left in the usual way to count them all!
A: Hint:  One direction is trivial.
For the other, $\aleph_0^k=\aleph_0$.  This follows from the fact that a countable union of countable sets is countable.
