chain A s.t. $|X|<|A|\leq |P(X)|$ Can we prove that there exists at least one chain $A$ in P(X), where X is a non-empty set (finite or infinite), s.t. $ |X|<|A|\leq |P(X)|$? 
If you can't solve it, ideas/possible directions are welcome :)
The two links mentioned above do not contain answers.
 A: All we need to prove is that there exists a chain A in P(X) with $|A|>|X|$ so that at least two of it's elements have to map to the same one.
If X is finite, we may assume $X=\{1,2,...,n\}$ and so $A=\{\varnothing,\{1\},\{1,2\},...,X\}$,which has $|A|=n+1$.
If X is countable then we may assume $X=\{1,2,...\}$. 
One example is the chain of the sets we get from repeated Dedekind cuts adopted from here. From $|P(\omega)|=2^{\aleph_{0}}=|\mathbb{R}|$ and $|\omega|=\aleph_{0}=|\mathbb{Q}|\Rightarrow$ we set a isomorphism between them. We associate to $x\in \mathbb{R}$ the pair $(A_{x},B_{x})$,where $A_{x}=\{r\in\mathbb{Q}:r\leq x\}$ and $B_{x}=\{r\in\mathbb{Q}:x<r\}$. Thus, $x\leq y\Rightarrow A_{x}\subseteq A_{y}$. This chain has $|\{A_{k}\}_{k\in\mathbb{R}}|=2^{\aleph_{0}}$ and so from the isomorphism we get a chain of sets in $P(\omega)$
I am not sure for the uncountable.
For $|X|= \aleph_{1}$, since $2^{\aleph_{0}}\geq \aleph_{1}$,we can use the same chain from the countable case extended to a maximal chain so that $|A|>\aleph_{1}$.
