Prove the elementary equivalence of the two models

Question

There are two models $\mathfrak A$ and $\mathfrak B$ in class $K$.

$\mathfrak A = <P(\omega), \subseteq>$

$\mathfrak B = <P(\omega), \supseteq>$

Is the $Th(K)$ of a full theory of signature $\sigma = <P>$? P is two place predicate.

So what I made:

$Th(K)$ is full theory $\Leftrightarrow$ $\mathfrak A \equiv \mathfrak B$. I proved it.

I need to prove that $\mathfrak A \equiv \mathfrak B$. I think that $\mathfrak A$ and $\mathfrak B$ are isomorphic. But I did not get the right to construct an isomorphism.

I also have to prove that $\mathfrak M \equiv \mathfrak N$ where $\mathfrak M = <P(\omega), \subseteq>$ and $\mathfrak N = <P(\mathbb{Z}),\subseteq>$. I think that it looks very similar to previous question.

Could you help me please

Answer 1

Hint: The mapping that takes any subset $X$ of $\omega$ to $\omega\setminus X$ is a (partial order) isomorphism from $<P(\omega), \subseteq>$ to $<P(\omega), \supseteq>$.

For the problem about $P(\omega)$ and $P(\mathbb{Z})$, first note that there is a bijection $\varphi$ from $\omega$ to $\mathbb{Z}$. This induces a bijection $\varphi^\ast$ from $P(\omega)$ to $P(\mathbb{Z})$ which preserves $\subseteq$. Much more informally, $\mathbb{Z}$, as a set, can be viewed as $\omega$ with elements renamed. Precisely the same result will hold if $\mathbb{Z}$ is replaced by any countably infinite set.