Show that there is no set satisfying the following properties. Show that there is no set satisfying the following properties.
(a) If $X\in A$, then $\mathcal P(X)\in A$
(b) If $T\subseteq A$, then $\bigcup T\in A$
I wonder how to prove this problem.
According to the note that I saw this problem, this problem is called Hilbert's Paradox. But when I searched for Hilbert's paradox, I could only find something about the infinite-hotel. Is it related to this problem?
 A: I'm glad to see my notes are in use. I should make the caveat that they were written for my students, who previously took introduction to set theory with Azriel Levy and myself (where a lot of material was covered).

Hilbert's paradox predates the axiom of regularity by quite a few years. So appealing to it in the solution is admittedly overshooting. The same can be said about replacement (which is used for transfinite recursion).
The solution, instead, is due to Zermelo, and it is quite clever: If $S$ satisfies the two properties, then $\bigcup S\in S$, and therefore $\mathcal P(\bigcup S)\in S$. This now implies that $\mathcal P(\bigcup S)\subseteq\bigcup S$. This is impossible, as the previous exercise in the notes points out, since $\{x\in X\mid x\notin x\}\in\mathcal P(X)\setminus X$ for any set $X$.
Of the "strong axioms", we only need separation and union, since power set is implicit in the properties of the set. Of the separation axioms, we only need one; and we only need union to apply to $S$ itself.
A: What you've stumbled upon is so called von Neumann universe, at least partially.
Let $A_0:=\emptyset$. Note that since $\emptyset\subseteq A$ and $\bigcup\emptyset=\emptyset$ then $\emptyset\in A$ by  (b). Now for any ordinal $\alpha$ define
$$A_{\alpha+1}=P(A_\alpha)$$
and for a limit ordinal $\lambda$ define
$$A_{\lambda}=\bigcup_{\alpha<\lambda}A_\alpha$$
By the transfinite induction and (a)+(b) we get that $A_\alpha\in A$ for any ordinal $\alpha$. But these are pairwise distinct sets. Meaning $A$ is at least as big as all ordinals, and so it cannot be a set by the Burali-Forti paradox.
A: I proved this with a hint of TheLast Cipher's comment in the original post. 
Note that $A \subseteq A \Rightarrow \bigcup A \in A \Rightarrow \mathcal{P}(\cup A) \in A \Rightarrow \mathcal{P} (\mathcal{P}(\cup A)) \in A$. Since $A\subseteq \mathcal{P}(\cup A)$ we get $A\in P(P(∪A))\in A$. And this is contradictory to axiom of regularity. 
