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 ?

  • 5
    $\begingroup$ 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. $\endgroup$
    – JMoravitz
    Oct 24 '19 at 17:00
  • $\begingroup$ @JMoravitz do you think i should edit the question to implement your idea $\endgroup$
    – Milan
    Oct 24 '19 at 17:03
  • $\begingroup$ No, but you can post it as an answer. $\endgroup$
    – Berci
    Oct 24 '19 at 17:36

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