Confusion between finite set and countable set? I wanted to find set $A$ such that the cardinality of power set is 
(1)finite, 
(2)countable and 
(3) uncountable. 
My efforts
For finite set, it is very easy as we can take $A$ to be any finite set and cardinality of $P(A)$, power set of $A$ would be $2^{|A|}$
For uncountable set it is again very easy as I can take $A=\mathbb{N}$ and result follows.
Point of confusion
Since I have already shown the result for finite set, does the result follows from there as every finite set is countable. 

A set S is countable if there exists an injective function $f$ from $S$ to
  the natural numbers $N = \{0, 1, 2, 3, ...\}$

Now if I take $S$ to be finite set, say $\{a_1, a_2,\cdots,a_n\}$ and take $f$ to be a map that sends $$a_1\mapsto 1, a_2\mapsto 2, \dots, a_n\mapsto n $$
This map is clearly injective.
But I am having second thought as my gut says "countable" term is used when we talk about infinity. 
May be I am confused in the terminology. 
 A: Please correct me if I'm wrong; this is only what I've heard, I've never studied cardinalities.
I believe there is no set whose power set is countable. Roughly,The $P(A)$ has stricly greater cardinality than $A$. If $A$ is finite, $P(A)$ is finite. If $A$ is countable, $P(A)$ is not countable. And there is no cardinality between finite and countable, so there cannot be a set whose power set is countable.
A: Trust in the definitions! 
You are right with the injection you gave; any finite set is trivially countable, hence you can use the same set for part 1 and 2. When we are talking about infinite countable sets (eg $\mathbb{N}$) we use the terminology countably infinite (in which case, as shown by cantor, there does not exist an injection from the power set to $\mathbb{N}$, and hence it is not countable). Equivalently, there does not exist a power set which is countably infinite.
For example (as I said in the comments) consider the set $S=\{1,2\}$. This is countable via the injection $f:n\to n+1$. Its power set $\mathcal{P}(S)= \bigl\{\emptyset, \{1\},\{2\},\{1,2\}\bigr\}$. We can now construct an explicit injection $g:\mathcal{P}(S) \to \mathbb{N}$ by saying $f(\{1\})=1$, $f(\{2\})=2$, $f(\{1,2\})=3$, and $f(\emptyset)= 4$. (Of course this is one of (countably!) infinite possible injections.) Since we have constructed the injection, we have by definition that $\mathcal{P}(S)$ is countable.
