Is every order isomorphism of sets induced by a bijection?

Question

Let $f$ is an order isomorphism $\mathscr{P}A \rightarrow \mathscr{P}B$ (where $A$ and $B$ are some sets, the order is set-inclusion $\subseteq$).

Is it true that it always exist a bijection $F: A \rightarrow B$ such that $f(X) = F[X]$ for every $X\in \mathscr{P}A$?

Answer 1

Since $f$ is an order isomorphism, show that $f(\{a\})$ is a singleton. Now take $F(a)$ to be the unique $b$ such that $f(\{a\})=\{b\}$, and show that $F$ is a bijection.

Answer 2

This is basically the same answer as Asaf's, but since I've added a little more details, I guess there is not much harm in posting it. (I did not want to throw it away, when the post was already prepared.)

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If $f$ is order isomorphism, then it must map atoms of one ordered set to atoms of the other one.

In this case it implies that singletons are maps to singletons, i.e. $f(\{x\})$ is a singleton for every $x\in A$. Thus the only possibility for the function $F$ is $$F(x)=y \Leftrightarrow f(\{x\})=\{y\}.$$

Since $\mathcal P(X)$ is an atomic lattice, every element of $\mathcal P(X)$ is uniquely determined by atoms which are below it. This shows that images of singletons uniquely determine an order-isomorphism. (And so every order isomorphism is indeed of this form.)