Prove: Let S and T be sets. If there exists and injection from |S| --> |T| then there exists an injection from |P(S)|-->|P(T)|
(P(S) and P(T) represent the power sets of each)
Here is a bit of notation. $\displaystyle C \subseteq S \Rightarrow \quad \overrightarrow f (C) = \left\{ {t \in T:\left( {\exists x \in C} \right)\left[ {f(x) = t} \right]} \right\}$
That is called the image set of $\displaystyle C$ under $\displaystyle f$.
Now it is clear that we have $\displaystyle \overrightarrow f :P(S) \to P(T)$, so we must prove that if $\displaystyle f\mbox { is injective then } \overrightarrow f \mbox{ is also injective}$
Of course to do that you must show $\displaystyle \overrightarrow f (C) = \overrightarrow f (D) \Rightarrow \quad C = D$.
No it does not.
Here is the rest of it.
If $\displaystyle \overrightarrow f (C) = \overrightarrow f (D)$ then
$\displaystyle \begin{array}{lcl}
{c \in C} & \Rightarrow & {f(c) \in \overrightarrow f (C)} \\
{} & \Rightarrow & {f(c) \in \overrightarrow f (D)} \\
{} & \Rightarrow & {\left( {\exists d \in D} \right)\left[ {f(d) = f(c)} \right]} \\
{} & \Rightarrow & {d = c} \\
{} & \Rightarrow & {c \in D} \\
\end{array} $
You do the other way by switching D with C.