I have to prove this result:
If $P$ be the power set, and $B$ and $C$ are two sets, then if $B \subseteq C$ prove that $P(B) \subseteq P(C)$.
Now, it seems obvious to me that since all the elements of B are in C, all possible combinations of the elements of C must include those that of B. But other than this straightforward argument, I'm at a loss to see what kind of proof can be given for statements like these.
Can anyone suggest something more "proper"?
Not much more than your "obvious" argument is needed.
To prove that $\mathcal P(B) \subseteq \mathcal P(C)$ you need to take an arbitrary $A\in\mathcal P(B)$ and show that this $A$ is in $\mathcal P(C)$.
But $A\in\mathcal P(B)$ means neither more nor less than $A\subseteq B$, and so we have $A\subseteq B\subseteq C$. Therefore $A\subseteq C$, which is the same as saying $A\in\mathcal P(C)$, which was what we had to prove.
$X\in P(B)\implies X\subset B$ by definition of $P(B)$; by hypothesis, $B\subset C$, so $X\subset C$ and therefore $X\in P(C)$ again by definition of $P(C)$. We have shown that $B\subset C$ implies $P(B)\subset P(C)$.
In other words--you're right, it is obvious, but if you have to write something, just write the obvious. :)
