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.
Best Answer
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.