While working through Velleman, I proved that if $A \subseteq P(A)$, then $P(A) \subseteq P(P(A))$.

One example where this may be the case is when $A = \emptyset$. Another may be when $\emptyset \in A$.

I cannot think of any other example though. Supposing that $x$ and $y$ are two arbitrary elements of $A$, then $P(A)$ will always enclose those elements in a new set, thus $x,y \notin P(A) $

Thus my question is:

Is there an example of a set, $A$, where $A \subseteq P(A)$ wand $A$ is non-empty and does not contain the empty set.


There is not.

Suppose that $A\subseteq\wp(A)$, where $A\ne\varnothing$. By the axiom of regularity there is an $a\in A$ such that $a\cap A=\varnothing$. But $a\in A\subseteq\wp(A)$, so $a\in\wp(A)$, and therefore $a\subseteq A$. It follows that if $x\in a$, then $x\in a\cap A$ and hence $x\cap A\ne\varnothing$, so it must be the case that $a=\varnothing$.

  • $\begingroup$ This proof is incorrect. What's wrong with $a$ being the empty set? If we let $A=\{emptyset\}$ then $A$ is nonempty and $A\subseteq P(A)$. EDIT: I see that the OP added the proviso "and does not contain the empty set", which does not appear in the title of the question. So the proof is ok, sorry. $\endgroup$ – Alon Amit Dec 21 '16 at 5:57

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