Discrete Math - Set Theory - Power Set I am stuck on a problem in my discrete mathematics textbook at the moment. The problem, as written in the textbook, is:
For a certain set $A$, the power set of $A$ is $\mathcal{P}(A) = \{\aleph_0, \{0\}, B\}$, where $B$ is a set. What is $A$?
My confusion here is that I was under the impression that for any set, let's say $D$, that $|\mathcal{P}(D)| = 2^n$. If this is the case, I don't really understand how the power set of $A$ from the problem above can contain only three elements. If set $A$ has two elements, then its power set will have fourelements. If set $A$ has one element, then its power set will have two elements. I know I missing something here, and I appreciate any hints, help, or guidance. 
 A: You are perfectly correct: there is no such set, for precisely the reason that you give. If the power set of $A$ is finite, then so is $A$, and in that case $|\wp(A)|=2^{|A|}$; $3$ is finite and not a finite power of $2$, so $\{\aleph_0,\{0\},B\}$ is not the power set of any set.
A: Even worse, if $\aleph_0$ is taken as a representative of the first infinite cardinality (e.g., the usual $\aleph_0=\omega$), then since $\aleph_0\in P(A)$, $\aleph_0\subset A$, so $A$ is infinite. An infinite set cannot have a finite power set.
A: I typed, "for a certain set A, the power set of A is" into Google Books, and it seems that where you have written $\aleph_0$, what's actually in the book is $\emptyset$ (thanks, Asaf). Can you see how to do the problem now?
EDIT: OP having confirmed that what was posted as $\aleph_0$ was really meant to be the empty set, we still have to work out how a power set can have 3 elements. Well, it can't have 3 distinct elements, so two of the listed elements have to be the same. Since all we are told about $B$ is that it's a set, we are both able and required to take $B$ to be either $\{\,0\,\}$ or the empty set --- it doesn't matter which. Now the power set has only 2 distinct elements, $A$ has to be $\{\,0\,\}$, and all is right in the world. 
[Of course, in the light of the clarification by OP, all the other answers are wrong....]
A: $B$ must be $\emptyset$ since $\emptyset \subset A$. But then $\{ 1  \} \subset \aleph_0 \subset A$ and $\{ 1\} \neq \mathcal{P}(A)$, a contradiction.
