Power sets : Do the relations between P(A) and P(B) always mirror the relations between the sets A and B? If I am correct it is true that: 
(1) "$P(A)$ is included in $P(B)$" implies "$A$ is included in $B$". 
(2) "$P(A) = P(B)$" implies "$A = B$". 
Might I conclude from this that the power sets of two sets always have the same relations as these two sets have with one another? 
Are there classical counterexamples to this (hasty) generalization? 
I can think of this as a counterexample : 
The fact that $A$ and $B$ are disjoint does NOT imply that $P(A)$ and $P(B)$ are disjoint. 

Attenpt to prove (1) using the theorem : "$\{ x \}$ belongs to $P(S)$" $\Longleftrightarrow$ "$x$ belongs to $S$.
Let's admit that : $P(A)$ is included in $P(B)$. 
Now, suppose (in view of refutation) that $A$ is not included in $B$. 
It means that there exists an x such that x belongs to $A$ but not to $B$. And consequently that there is an $x$ such that $\{ x \}$ belongs to $P(A)$ but not to $P(B)$. If this were true, there would be a set $S$ such that $S$ belongs to $P(A)$ but not to $P(B)$. This contradicts our hypothesis according to which $P(A)$ is included in $P(B)$. 
Conclusion: "$P(A)$ is included in $P(B)$" implies "$A$ is included in $B$". 

Let's take these sets as an example:
$$
\begin{align}
A &= \{1\} \\
B &= \{2\} \\
P(A) &= \{\emptyset, \{1\}\} \\
P(B) &= \{\emptyset, \{2\}\}
\end{align}
$$
A relation is in this context a function that takes two sets and gives true or false.
A simple counterexample relation is:
$$f(X, Y) = X \text{ contains a set}$$
Then we have:
$$
\begin{align}
f(A, B) &= \text{false} \\
f(P(A), P(B)) &= \text{true} \\
\end{align}
$$
