Power set of any set. Question:

Let $A$ be any set. Let $\mathbb{P}(A)$ be the power set of $A$. Then which one is true?

*

*$\mathbb{P}(A)=\emptyset$ for some $A$.


*$\mathbb{P}(A)$ is finite for some $A$.


*$\mathbb{P}(A)$ is countable for some $A$.


*$\mathbb{P}(A)$ is uncountable for some $A$.

Now 1. is not true, since if $A=\emptyset$, $\mathbb{P}(A)=\{\emptyset\}\neq \emptyset$. 3. is not true since let $A=\mathbb{N}$, then $\mathbb{P}(A)=2^\mathbb{N}$ is uncountable. So answer will be 2 and 4.
But when I saw the answer book, it says answer is (2,4) or (2,3,4). That or is for answer is not decided yet.
My question is why is their doubt!!! The answer should be $(2,4)$.... this is obvious, isn't? I am very bad at set theoritic arguments. Can someone clarify it?
 A: A countable set is a set which cardinality is at most $\aleph_0$, the cardinal of $\mathbb{N}$. In particular, any finite set is countable. So, since $2$ is true, so is $3$: taking a finite set $A$, $\mathcal{P}(A)$ is finite (hence $2$ is true) and thus countable (hence $3$ is true).
Your argument for 1 is a bit off. The power set of $\emptyset$ is $\{\emptyset\}$, not $\{\{\emptyset\}\}$.
Edit:
As R. Suwalski pointed out, countable may mean of cardinality $\aleph_0$. So if that's the case, then no there is no $A$ such that $\text{card}{\mathcal{P}(A)}=\aleph_0$. Indeed, by contradiction, we have according to Cantor's theorem, $\text{card}A<\text{card}\mathcal{P}(A)=\aleph_0$. Thus $A$ is finite and hence so is $\mathcal{P}(A)$, contradiction. 
A: Assuming countable means "countably infinite" (so finite sets are not considered countable), 3 is false. If $A$ is finite, then $\mathbb{P}(A)$ is finite.
If $A$ is infinite it contains a countably infinite subset $B$ (mild form of choice used here). And then $2^{\mathbb{N}} = |\mathbb{P}(B)| \ge |\mathbb{P}(A)|$ so that $\mathbb{P}(A)$ is uncountable. So a power set cannot be countable.
A: There are two conventions on the meaning of the word "countable" that lead to different answers in the third case.
Some use "countable" as an adjective that refers specifically to sets with the same cardinality as the natural numbers. With this usage, statement 3 would be false, since no set has a power set with the same cardinality as the natural number.
Some use "countable" as an adjective that refers to sets whose cardinality is less than or equal to the natural numbers. With this usage, statement 3 would be true, since every finite set is countable, and any finite set has finite power set.

I think originally, the word "countable" was used for the second convention; the intuition behind the name is that they are sets you can count with natural numbers. If one wanted to specifically refer to the sets where you had to exhaust the natural numbers, one would use the phrase "countably infinite".
And in some contexts, you really do want an adjective that covers both the case of finite sets and sets with the same cardinality as the naturals.
But some contexts speak only about infinite sets, so by both conventions "countable" uniquely refers to sets with the same cardinality as the naturals. 
In other contexts, the gulf between finite and infinite cardinals is that having a word that lumps together the cardinality of the natural numbers with finite cardinalities is not useful. So the term "countable" is only ever used in the phrase "countably infinite", and the latter term eventually got shortened to simply "countable". 
A: Your answer to 1 is a bit off. It shows that the empty set cannot be such an A, but it does not prove there is no A. For that, suppose there is an A such that 2^A is empty. The empty set is a subset of all sets, so it is a subset of A, so it is a member of 2^A, so 2^A is not empty, a contradiction. Or more briefly: the empty set is a member of all powersets, so no powerset can be the empty set.
Your argument against 3 is a bit off, in a similar way. The proposition is that the statement is true for SOME A. You then say: look it is not true for this particular A. But, so what, there might be some other A for which it IS true! If the proposition had been for ALL A, then yes, one counterexample is enough. But here it only claims for SOME (and just one will do).
