How to prove there is no bijection between a set and its second power set? Let $X$ be a set. How can I show that there is no bijection between $X$ and $P(P(X))$, the powerset of the powerset of $X$. I know that there is no bijection between $X$ an $P(X)$, due to Cantor's theorem. Since $P(P(X))$ has a 'larger' cardinality than $P(X)$, the result should follow immediately.
I just have trouble writing down the proof.
 A: One way: You can adapt the proof of Cantor's theorem directly: Suppose $f:X\to\mathcal P(\mathcal P(X))$ and consider
$$ \{\{x\}\mid x \in X, \{x\} \notin f(x) \} \subseteq \mathcal P(X)$$
This can't be in the image of $f$. Thus there is no surjection $X\to\mathcal P(\mathcal P(X))$.
Another: It is trivial that there are injections $X\to\mathcal P(X)$ and $\mathcal P(X)\to\mathcal P(\mathcal P(X))$. If $\mathcal P(\mathcal P(X))\cong X$, the latter injection gives us an injection $\mathcal P(X)\to X$, and Bernstein's theorem now concludes $X\cong \mathcal P(X)$ which we know is not the case.
A: Here's another option; we shall use the Cantor-Schroeder-Bernstein Theorem.
Suppose for the sake of a contradiction that you have a bijection $f:\mathcal{P}(\mathcal{P}(X))\rightarrow X$, which in particular is an injection.  Next, $X$ injects into $\mathcal{P}(X)$, so let $g$ be such an injection.  Then $g\circ f:\mathcal{P}(\mathcal{P}(X)) \to \mathcal{P}(X)$ is an injection.  Next, there is an injection $h:\mathcal{P}(X)\rightarrow \mathcal{P}(\mathcal{P}(X))$.  But then by Cantor-Schroeder-Berstein there exists a bijection $j:\mathcal{P}(X)\rightarrow \mathcal{P}(\mathcal{P}(X))$.  
This leads to a contradiction by Cantor's Theorem, and thus no such bijection can exist.
A: Cantor's theorem says that for all sets $X$, $|X|<|\mathcal{P}(X)|$.
Hence for all sets $X$, $|\mathcal{P}(X)|<|\mathcal{P}(\mathcal{P}(X))|$.
So by transitivity, we have that for all sets $X$, $|X| < |\mathcal{P}(\mathcal{P}(X))|.$
So really, the problem is just to show transitivity of $<$. (You can probably find a proof of this online.)
A: Suppose $f\colon X\to \mathcal{P}(\mathcal{P}(X))$ is a bijection and let $f^{-1}\colon \mathcal{P}(\mathcal{P}(X))\to X$ be its inverse. There is an injection $i\colon X\to\mathcal{P}(X)$ given by $i(x)=\{x\}$ for all $x\in X$ and so $i\circ f^{-1}\colon \mathcal{P}(\mathcal{P}(X))\to \mathcal{P}(X)$ is the composition of two injections so is itself an injection.
As there is also an injection from $\mathcal{P}(X)\to\mathcal{P}(\mathcal{P}(X))$ given by mapping a set $A$ to the singleton $\{A\}$ it follows that there is a bijection $\mathcal{P}(X)\to\mathcal{P}(\mathcal{P}(X))$ by Cantor-Shroeder-Bernstein. But this violates Cantor's theorem.
A: It seems like your issue is understanding the implications/definition of bijection.  If a function is a bijection, then it is both injective (1 to 1) and surjective (onto).  Injection means that every element of the domain maps to exactly one element of the range, and surjection means that every element of the range is mapped to by at least one element of the domain.  Assuming both of these properties hold, then it is guaranteed that the domain and range have the same size (cardinality, just think about it for a little bit).
Since you already know that the second power set of any set has a larger cardinality than than the original, the existence of a bijection between them would create a contradiction, because it would imply that their cardinalities are equal. Therefore, a bijection cannot exist.
