Proof that for any set $A$,$\ \ $ $\mathcal{P}\left(A\right)\nsubseteq A$

Question

I was asked, as an exercise in a course on elementary set theory, to prove that for any set $A$,$$\mathcal{P}\left(A\right)\nsubseteq A$$

This is the very beginning of the course, so cardinality arguments are out of the question.

The only apparent contradiction I can draw from the above statement (Using only the axioms and not "Common sense") is that it would imply $A\in A$, but reading some about this it seems this alone is not enough to deduce a contradiction without the axiom of regularity, which was not presented yet. Hints would be greatly appreciated.

Answer 1

Do you know the proof of Cantor's Theorem? It can be adjusted to address this specific problem: let

$$B = \{ X \subseteq A : X \notin X \} \subseteq \mathcal{P}(A).$$

Then either $B \in B$ or $B \notin B$, in either case we get $B \not \subseteq A$, hence $\mathcal{P}(A) \not \subseteq A$.

Answer 2

Try with $\;A=\{1\}\;$ . Then $\;\mathcal P(A)=\{\emptyset,\,A=\{1\}\}\;$ . Can you see now why $\;\mathcal P(A)\not\subset A\;$ ?