Any order isomorphism $F$ between $\mathcal{P}(A)$ and $\mathcal{P}(B)$ W.R.T. $\subseteq$ must be $F:X\mapsto\{f(x)\mid x\in X\}$ for some $f:A\to B$

Question

I'd like to prove that if there exists an order isomorphism $F:\left\langle \mathcal{P}\left(A\right),\subseteq\right\rangle \to\left\langle \mathcal{P}\left(B\right),\subseteq\right\rangle $ for some nonempty $A$ and $B$ then there must be some $f:A\to B\ $ S.T. $F$ is defined by $F:X\mapsto\left\{ f\left(x\right)\mid x\in X\right\} $

My thought was to explicitly build $f$ out of $F$ by defining $f\left(x\right)=F\left(\left\{ x\right\} \right)$, which would work exactly as requested. However, I'm having trouble proving that $F$ must take singletons to singletons. I suspect that in general, $F$ must preserve cardinality, but I was not successful in proving this.

Answer 1

Hint: Singleton sets are the minimal nonzero elements of $\mathcal{P}(X)$ (and $\mathcal{P}(Y)$). Then define $f$ in such a way that $\left\{f(x)\right\}=F(\left\{x\right\})$, as you wanted.

Answer 2

Is clear that $F(\{x\})=\{y\}$ for some $y \in B$ otherwise if $f(\{x\})=Z=\{y_1, y_2, \cdots \}$ for surjective there exist $A \in X$ such that $f(A)=\{y_1\}$, and like $F$ is isomorphism then $A \subset \{x\}$ but $F(\emptyset)=\emptyset$. Thereforce that $f(x)=F(\{x\})$ works.

Answer 3

Proposed solution:

Lemma: $F$ Takes singletons to singletons $(\forall\left\{ a\right\} \in\mathcal{P}\left(A\right)\,\,\,\exists b\in B\,\,\,F\left(\left\{ a\right\} \right)=\left\{ b\right\} )$

Proof: We are assuming $A$ and $B$ are non empty. Note that for any singleton $\left\{ a\right\} \subseteq A$ we cannot have $F\left(\left\{ a\right\} \right)=\emptyset$ since we'd get $F\left(a\right)\subseteq F\left(\emptyset\right)$ but $a\not\subseteq\emptyset$. This gives us that for $\left|A\right|=\left|B\right|=1$ the claim is immediate, so assume $\left|A\right|=\left|B\right|\geq2$. Given some singleton $\left\{ a_{1}\right\} \subset A$ we cannot have $F\left(\left\{ a_{1}\right\} \right)\geq2$ since then $F\left(\left\{ a_{1}\right\} \right)=\left\{ b_{1},b_{2}\ldots\right\} \implies\left\{ b_{1}\right\} \subseteq F\left(\left\{ a_{1}\right\} \right)$ but $\left\{ b_{1}\right\} =F^{-1}\left(a'\right)$ for some $a'\neq a_{1}$ ($F^{-1}$ is injective) but $\left\{ a'\right\} \not\subseteq\left\{ a_{1}\right\}$

Now we are ready to define $f$. Given $a\in A\left\{ f\left(a\right)\right\} =F\left(\left\{ a\right\} \right)$

We claim $F$ must take $X\mapsto\left\{ f\left(x\right)\mid x\in X\right\}$ . This is true since otherwise either for some $\left\{ x\right\} \subseteq X$ we have $\left\{ f\left(x\right)\right\} \not\subseteq F\left(X\right)$ or for some $\left\{ f\left(x\right)\right\} \subseteq F\left(X\right)$ we have $\left\{ x\right\} \not\subseteq X$. Both of these contradicting $F$ being an isomorphism.