Find a set $X$ such that $X\cap P(X)\neq\emptyset$ 
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*Find a set $X$ such that $X\cap P(X)\neq\emptyset$


My answer is: Take $X=\left\{ \emptyset \right\}$. Then, $\left\{ \emptyset \right\}\cap P(\left\{ \emptyset \right\})=\left\{ \emptyset \right\}\neq \emptyset$. So we are done.
Can you check my answer and proof-writing?
 A: Any non-empty transitive set would work (if $X$ is transitive then $X \subseteq P(X)$), and in particular any non-zero ordinal would work, and in particular any non-zero natural number would work, and in particular $1 = \{\varnothing\}$ would work, which is your example.
Another collection of solutions is the van Neumann universes with non-zero finite index.
Also, more generally, the transitive closure of any non-empty set would work.

$X \cap P(X) \ne \varnothing$ is equivalent to an element of $X$ being a subset of $X$.
If we start with any set $X$, we could pick any subset $Y \subseteq X$, and then $Y$ would also be a subset of $Z := X \cup \{Y\}$, but now $Y \in Z$ also, so $Z$ would work.
In particular, taking $Y = X$ gives us $Z := X \cup \{X\}$, i.e. the successor of any set works, and in particular the successor of the empty set works, which is also again your example.
A: Only one thing you can be sure of when constructing a power set for some set $X$ that $\emptyset \in P(X)$ and $X\in P(X)$.
So, for our solution we want something to be both in $P(X)$ and $X$ itself. But $X\in X$ is restricted by axiom of foundation. So anything in the form $\{\emptyset,a,b,c,1,2,3,...\}$ would be a solution.
