Finding a largest chain I am trying to find the cardinality of the largest chain in $(P(\omega),\subseteq)$ and $(P(\omega_{1}),\subseteq)$. So for the first one I found one with cardinality $2^{\aleph_{0}}$ by bijecting it to the reals and using dedekind cuts. But I think there must be bigger ones (like in finite sets). How do I go about finding them? Just hints :).
Thanks
 A: The maximum cardinality of a chain in $P(\omega)$ is at least $2^{\aleph_0}$ as you showed using Dedekind cuts, and it's no bigger than that because that's the cardinality of $P(\omega)$ itself, so it's exactly $2^{\aleph_0}$.
The same problem for $P(\omega_1)$ is much harder. The maximum cardinality of a chain in $P(\omega_1)$ is at least $2^{\aleph_0}$ because $P(\omega)\subset P(\omega_1)$, and it's at most $2^{\aleph_1}$ because $|P(\omega_1)|=2^{\aleph_1}$, but there could be lots of cardinals between $2^{\aleph_0}$ and $2^{\aleph_1}$. Actually, what I loosely referred to as "the maximum" may not exist: there is certainly a least cardinal $\lambda$ such that $P(\omega_1)$ does not contain a chain of cardinality $\lambda$, but I see no obvious reason why $\lambda$ can't be a limit cardinal. At least, the cofinality of $\lambda$ must be greater than $\omega_1$: it's easy to see that, if $P(\omega_1)$ contains a chain of cardinality $\kappa_\alpha$ for each $\alpha\lt\omega_1$, then it also contains a chain of cardinality $\kappa=\sum_{\alpha\lt\omega_1}\kappa_\alpha$.
If $2^{\aleph_0}=\aleph_1$, then there is a chain of cardinality $2^{\aleph_1}$ in $P(\omega_1)$. Hint: it's like the construction you used for $P(\omega)$, but with $\{0,1\}^{\omega_1}$ (ordered lexicographically) playing the role of $\mathbb R$, and the elements with countably many nonzero coordinates playing the role of rational numbers.
Therefore, we can prove in ZFC that there is a chain of cardinality $\aleph_2$ in $P(\omega_1)$. It's an odd proof by cases, where we use one construction if $2^{\aleph_0}=\aleph_1$, and another construction if  $2^{\aleph_0}\ge\aleph_2$. I wonder if there's a more elegant proof.
So, if either $2^{\aleph_0}=\aleph_1$ or $2^{\aleph_0}=2^{\aleph_1}$, then there is a chain of cardinality $2^{\aleph_1}$ in $P(\omega_1)$. If $\aleph_1\lt 2^{\aleph_0}\lt2^{\aleph_1}$. we seem to be in a gray area. According to comments by Ashutosh on this question at Math Overflow, William Mitchell constructed a model of set theory in which $P(\omega_1)$ does not contain a chain of cardinality $2^{\aleph_1}$, in his paper "Aronszajn trees and the independence of the transfer property", Ann. Math. Logic 5 (1972), 21-46.
There may be some relevant information in James E. Baumgartner's paper "Almost-disjoint sets, the dense set problem and the partition calculus", Ann. Math. Logic 9 (1976), 401-439.
