If $|A|\leq|B|$, then $|\mathscr{P}(A)|\leq|\mathscr{P}(B)|$ I'm trying to prove that if $|A|\leq|B|$, then $|\mathscr{P}(A)|\leq|\mathscr{P}(B)|$.
I know that $|A|\leq|B|$ implies that there is an injection $f:A\to B$. I also know that for any set $S$, $|S|&lt;|\mathscr{P}(S)|$.
Clearly I need to show that there is also an injection from $\mathscr{P}(A)$ to $\mathscr{P}(B)$ but I'm really struggling to think of a way to get there.
I'm not asking for a solution but if anybody has a small hint to point me in the right direction I would really appreciate it. Thank you!
It is enough to Know that $|\mathscr{P}(A)|=2^{|A|}$ so $$ |\mathscr{P}(A)|=2^{|A|}\leq 2^{|B|}=|\mathscr{P}(B)|$$ It remains to show  $$ |\mathscr{P}(A)|=2^{|A|}$$
define a bijection function $$f\colon \mathscr{P}(A)\to 2^{A}$$ by $f(C)=\chi_{C}$ where $\chi$ is the characteristic function of $C$, that is , $\chi_{C} (x)=1$ for $x\in C$ and $\chi_{C} (x)=0$   for $x\in A\setminus C.$ I hope it is clear now
It is enough to Know that $|\mathscr{P}(A)|=2^{|A|}$ so $$ |\mathscr{P}(A)|=2^{|A|}\leq 2^{|B|}=|\mathscr{P}(B)|$$ It remains to show  $$ |\mathscr{P}(A)|=2^{|A|}$$
define a bijection function $$f\colon \mathscr{P}(A)\to 2^{A}$$ by $f(C)=\chi_{C}$ where $\chi$ is the characteristic function of $C$, that is , $\chi_{C} (x)=1$ for $x\in C$ and $\chi_{C} (x)=0$   for $x\in A\setminus C.$ I hope it is clear now