# Show that if $\mathcal{P}(A) \subseteq A$ then $\mathcal{P}(A) \in A$.

I was working on a revision worksheet and I came across this question, and I was not sure to answer it. Can anyone help me out with this? Help will be much appreciated.

A falsity implies anything. And the condition is always false.

Due to Cantor's theorem, $|\mathcal{P}(S)| > |S|$ always. But $\mathcal{P}(S) \subseteq S$ implies $|\mathcal{P}(S)| \leq |S|$ (which is false!), and a falsity implies literally everything, including $\mathcal{P}(S) \in S$.

• It may be interesting to note that this is an especially simple instance of Cantor's theorem (which can also be seen as a version of Russell's paradox) because the set to consider is just $\{x \in S : x \notin x\}$. Apr 26, 2015 at 20:39

If $\mathcal{P}(A) \subseteq A$, then $\mathcal{P}(A) \in \mathcal{P}(A) \subseteq A$, because $\mathcal{P}(A)$ is a subset of $A$ and hence an element of $\mathcal{P}(A)$. Therefore, $\mathcal{P}(A) \in A$.

Note: The Axiom of Regularity is necessarily violated here (thanks, Trevor, for pointing out the need to rephrase).

• From a purely formal point of view I understand your answer; nevertheless, the problem is highly counter-intuitive. Could you please produce an explicit example of such set? Apr 26, 2015 at 20:26
• @Alex: This doesn’t happen in $\mathsf{ZFC}$ because it violates the Axiom of Regularity. I guess the OP was requesting for a purely formal argument. Apr 26, 2015 at 20:27
• I don't see how the axiom of regularity is relevant; can't we always diagonalize to get a subset of $A$ that is not an element of $A$? (I am not the downvoter, by the way.) Apr 26, 2015 at 20:30
• @Trevor: Hi Trevor. Yes, of course. All I’m saying is that if we assume that $\mathcal{P}(A) \subseteq A$, then it contradicts our assumption that a set cannot be an element of itself. Apr 26, 2015 at 20:33
• @Trevor: By the way, Trevor, don’t worry about that ephemeral downvote. I’m used to being in a sea of anonymous downvoters. Apr 26, 2015 at 20:38