chain A s.t. $|X|<|A|\leq |P(X)|$

Question

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.

Do you also want the inequality given in the title? Is $X$ finite? What is $k$? — Tobias Kildetoft, Jan 06 '14 at 10:20
If $X$ is finite, we may assume $X={1,…,n}$. Let $A:={\emptyset, {1}, {1,2}, {1,2,3},…,{1,2,…,n-1}, {1,…,n}}$. Then $A$ is a chain of length $n+1$. Is this what you are looking for? — , Jan 06 '14 at 10:22
The "in other words" part is stating a different problem. Please make a precise question. — Emanuele Paolini, Jan 06 '14 at 10:32
@AsafKaragila Don't you mean $|X| + 1$ (as per example further up)? — Tobias Kildetoft, Jan 06 '14 at 10:46
I removed my (wrong) answer, but the OP gave there in the comment the following reference which might be useful: math.stackexchange.com/questions/90946/uncountable-chains — Emanuele Paolini, Jan 06 '14 at 10:58
If we assume GCH, it follows that there is a chain A with $|A|=2^{|X|}$ http://math.stackexchange.com/questions/90946/uncountable-chains. — TKM, Jan 06 '14 at 10:58
If you are curious,the original question is at http://math.stackexchange.com/questions/628537/fpx-to-x-property — TKM, Jan 06 '14 at 11:06
Here is the answer:http://math.stackexchange.com/questions/628537/fpx-to-x-property — TKM, Jan 11 '14 at 22:22

score 0 · Answer 1 · edited Apr 13 '17 at 12:19

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 $|A|>\aleph_{1}$.

chain A s.t. $|X|<|A|\leq |P(X)|$

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