6
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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 $a=\varnothing$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.