# totally ordered chain in the powerset with big cardinality

Let $B$ be some set. The problem is to find a set $A\subset\mathcal{P}(B)$ of subsets of $B$ which is totally ordered by inclusion and such that there exists a bijection $A\leftrightarrow \mathcal{P}(B)$.

This is an easy exercise if $B$ is countable where one can explicitly construct such a set (one identifies $B$ with $\mathbb{Q}$ and takes $A$ to be the set of sets of the form $(-\infty, r)$ for all $r\in \mathbb{R}$)

My question now is the following: For which sets $B$ can the existence of such an $A$ be shown using any combination of (reasonable) additional hypotheses (e.g. AC, CH, GCH, ...)? Under which circumstances can one find $B$'s that don't admit such an $A$?

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If I remember correctly, William Mitchell has shown in his thesis that if you add $\omega_{\omega_1}$ Cohen reals to the universe then in the resulting model there is no subset of $\mathcal{P}(\omega_1)$ of size $2^{\omega_1}$ which is linearly ordered under inclusion. –  Ashutosh Feb 17 '13 at 2:09
Here's a reference: William Mitchell - Aronszajn trees and the independence of the transfer property, Annals of Math. Logic., Vol. 5, Issue 1, 1972, pp. 21-46 –  Ashutosh Feb 17 '13 at 2:12

Let's think about the countable case like this: think of the binary tree $2^{\lt\omega}$, which has size $\omega$, but has $2^\omega$ many branches. Each branch describes a cut in the natural lexical order on the nodes, and so we have a countable linear order with $2^\omega$ many cuts.

So consider a cardinal $\kappa$ and the tree $2^{\lt\kappa}$, which admits a similar lexical order. If this tree has size $\kappa$, then we get a linear order of size $\kappa$ with $2^\kappa$ many cuts. And such a family of cuts turns into a chain just as you point out in the question.

Thus, if $B$ has size $2^{\lt\kappa}$, then we can find $A$ of size $2^\kappa$. In particular, whenever $2^{\lt\kappa}=\kappa$, then $P(\kappa)$ has a chain of size $2^\kappa$, as you desire.

In particular, under the GCH, the phenomenon will occur for every infinite cardinal, since GCH implies $2^{\lt\kappa}=\kappa$ for all infinite cardinals $\kappa$.

Lastly, let me point out that this argument method can be turned into a characterization. Namely, the sets $B$ for which there is an $A$ as you desire are exactly the sets $B$ of size $\kappa$ for which there is a linear order on $\kappa$ with $2^\kappa$ many cuts. The one direction we've established, and for the converse, when there is such a chain $A$ in $P(B)$ of that size, then we may place a pre-order on $B$ according to the order in which points are added to sets in the chain (and extend this to a linear order). Each element of $A$ gives a cut in this order, and so we have a linear order on $\kappa$ with $2^\kappa$ many cuts.

So the phenomenon occurs for exactly those sets $B$ of size $\kappa$ for which there is a linear order on $\kappa$ with $2^\kappa$ many cuts.

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Note that the $2^{\lt\kappa}=\kappa$ property, while sufficient, is not necessary. For example, it could be that $2^\beta=2^\kappa$ for some $\beta\lt\kappa$ with $2^{\lt\beta}\leq\kappa$, and in this case, we would still have the tree $2^{\lt\beta}$ of size at most $\kappa$, but still having $2^\kappa$ many cuts. –  Joel David Hamkins Feb 17 '13 at 1:53