Power set of a subset Proof that if $A \subseteq B$, then ${\mathscr P}(A) \subseteq {\mathscr P}(B)$.
I tried using the definition of a subset: $A \subseteq B = \forall x(x \in A \to x \in B)$, but get stuck as to how to use it to write the power sets of A and B.
 A: Given $X\in {\mathscr P}(A)$ hence $x\in X\implies x\in A$ but we know that $x\in A\implies \cdots$ hence $x\in X\implies \cdots $  as a conclusion $X\in {\mathscr P}(B)$
(Fill in dots)
A: The proposition of the exercise tells you to consider that $A \subseteq B$. Power set of A is defined as follows: if X is in $\mathscr{P}(A)$, then $X \subseteq A$. But as stated before, $A \subseteq B$, so $X \subseteq B$, too. Thus, as $X \subseteq B$, it follows that X is in $\mathscr{P}(B)$.If X is in $\mathscr{P}(A)$ implies that X in $\mathscr{P}(B)$, then $\mathscr{P}(A) \subseteq \mathscr{P}(B)$, q.e.d.
A: If you are happy with a "naïve" set theoretical proof:
$\scr P(A)$ is nothing but $2^A$, the collection of all maps from $A$ to the two-element set $2$(an element of $A$ is either inside a subset, or outside, represented as two "states").
If $B \subset A$, then an inclusion map $\mathscr P(B) \hookrightarrow \scr P(A)$ can be seen by extension of every map $B \mapsto 2$, by a fixed map $A \setminus B \mapsto 2$, into a map $A \mapsto 2$.
