I am doing some self study before I begin an intro to proofs class. This question is really bugging me, and I think I need someone to look over my proof for it. The P is for power set, undercase is an element, and capital is a set.
Thank you for any and all help.
Suppose $A \subseteq P(A)$. Prove that $P(A)\subseteq P(P(A))$
Proof.Suppose A ⊆ P(A), z ∈ Y, Y ∈ X, and X ∈ P(A). Since X ∈ P(A), itfollows that X⊆A. Thus according to A ⊆ P(A), X ⊆ P(A). Since Y∈X, it follows that Y∈ P(A) which is equivalent to Y⊆A. Finally, since z∈Y, we have z∈A. Therefore,P(A)⊆ P(P(A)) QED
It is perfectly true that if $A\subseteq\wp(A)$ and $z\in Y\in X\in\wp(A)$, then $z\in A$, but this does not show that $\wp(A)\subseteq\wp(\wp(A))$. To show that $\wp(A)\subseteq\wp(\wp(A))$, you need to let $X$ be an arbitrary element of $\wp(A)$ and show that $X\in\wp(\wp(A))$. And all of the necessary work is actually there in your argument, along with a fair bit of irrelevant material: if $X\in\wp(A)$, then $X\subseteq A\subseteq\wp(A)$, so $X\subseteq\wp(A)$, and therefore by definition $X\in\wp(\wp(A))$. And that’s it: $X$ was an arbitrary member of $\wp(A)$, so we’ve shown that $\wp(A)\subseteq\wp(\wp(A))$.