# $|A|\le |B|$implies$|P(A)|\le|P(B)|$

If cardinality of set $$A$$ is less or equal to the cardinality of set $$B$$ then cardinality of the power set of $$A$$ is less or equal to the cardinality of power set of $$B$$

This holds for finite sets, i want to know if it does for infinte ones too. I tried to do a proof:

Since $$|A|\le |B|$$ then there exist a one to one function $$f: A \to B$$

To show $$|P(A)|\le|P(B)|$$ we have to find a one to one function $$g: P(A) \to P(B)$$

Define $$g$$ such that it maps $$\{a_k\}~\to \{f(a_k)\}$$

Likewise $$\{a_j,~...~,~a_n\}~\to \{f(a_j),~...~,~f(a_n)\}$$.

$$g$$ is injective because $$f$$ is. I am not sure if $$g$$ is well defined for infinte subsets ?

• Remove the subscripts and you should be better off. Let $f$ be such an injective function from $A$ to $B$. Let $g$ be the function such that a subset $X\in\mathcal{P}(A)$ maps to $\{f(x)~:~x\in X\}\in\mathcal{P}(B)$. This way, we can talk about uncountable sets as well. It remains to show that $g$ is in fact well defined here (that the outputs are in fact actually elements of $\mathcal{P}(B)$) and that it is in fact injective. Oct 24 '19 at 17:00
• @JMoravitz do you think i should edit the question to implement your idea Oct 24 '19 at 17:03
• No, but you can post it as an answer. Oct 24 '19 at 17:36