Mapping between a set and its power set

Question

For a set $X$, let $P(X)$ be the set of all subsets of $X$ and $\Omega(X)$ be the set of all functions from $f:X \to \{0,1\}$. Then

$(1)$ If $X$ is finite , then $P(X)$ is finite

$(2)$ if $X$ and $Y$ are finite and if there is a one-one correspondence between $P(X)$ and $P(Y)$, then there is a one-one correspondence between $X$ and $Y$

$(3)$ There is no one-one correspondence between $X$ and $P(X)$

$(4)$ There is a one_one correspondence between $\Omega(X)$ and $P(X)$

My attempt:-

$(1)$ is true.

$(3)$ True by Cantor's Theorem

I know there are answers to $(4)$ on this site but I am specially interested in $(2)$ which I know to be true.

Let $f:P(X) \to P(Y)$ be a one-one correspondence. Since $X$ and $Y$ are finite we have $|P(X)|=|P(Y)|\Rightarrow |X|=|Y|$ otherwise $|P(X)|\neq P(Y)|$. Since X and Y are finite sets with same no. of elements , there is a bijection . Is this a good enough proof? Can this be generalized to two infinite sets having a one-one correspondence between their power sets.

Thanks for your thoughts and time.

Answer 1

The proof is not good enough, you have said that $|P(X)|=|P(Y)|\Rightarrow |X|=|Y|$, otherwise $|P(X)|≠|P(Y)|$, why is that "otherwise" part true?

you can prove that for finite $X$, we have $|P(X)|=2^{|X|}$, then you really get that $|X|≠|Y|$ implies $|P(X)|≠|P(Y)|$

This result cannot work on infinite sets, as it is consistent(by Easton's theorem) that it is even possible that $P(\aleph_n)=P(\aleph_k)$ for all $n,k∈\Bbb N$

Answer 2

I think this answer will help you a lot.

Given that $~~P(X):$ set of all subsets of $X$
$~~~~~~~~~~~~~~~~~~~Ω(X):$ set of all functions $~f:X→\{0,1\}$

For option $\bf(1)$, if $X$ is finite, let $|X|=n$, then $|P(X)|=2^n$ which is again finite. Hence option $(1)$ is true.

For option $\bf(2)$, let $X$ and $Y$ are finite and there is one-one correspondence between $P(X)$ and $P(Y)$.$$\implies |P(X)|=|P(Y)|\implies 2^{|X|}=2^{|Y|}\implies |X|=|Y|$$Hence there exists one-one correspondence between $X$ and $Y$. Thus option $(2)$ is true.

For option $\bf(3)$, since $~|P(X)|=2^{|X|}\gt|X|~$
Thus there is no one-one correspondence between $X$ and $P(X)$. Therefore option $(3)$ is true.

For option $\bf(4)$, since $~|P(X)|=2^{|X|}=|\Omega(X)|~$
Therefore there exists a one-one correspondence between $\Omega(X)$ and $P(X)$. Therefore option $(4)$ is true.