Let ${\cal P}(\omega)$ denote the power-set of $\omega$. We order it by set inclusion $\subseteq$ and say that ${\cal C}\subseteq {\cal P}(\omega)$ is a *chain* if for all $A, B\in {\cal C}$ we have $A\subseteq B$ or $B \subseteq A$.


Interestingly, [chains in ${\cal P}(\omega)$ can have length $2^{\aleph\_0}$](https://mathoverflow.net/questions/395002/chains-of-length-2-kappa-in-cal-p-kappa). Zorn's Lemma implies that every chain is contained in a maximal chain with respect to set inclusion.


**Question.** What is the cardinality of the collection of maximal chains in ${\cal P}(\omega)$ having uncountable cardinality?



Maximal uncountable chains in ${\cal P}(\omega)$
Any maximal chain in $\mathcal{P}(\omega)$ is determined by (that is, is the unique maximal chain containing) a *countable* subchain; this is basically just a restatement of the fact that $\omega\_1$ doesn't embed into $\mathcal{P}(\omega)$. This gives an upper bound of $2^{\aleph\_0}$ for the number of distinct maximal chains.


On the other hand, it's easy to show that there are at least $2^{\aleph\_0}$ distinct size-continuum maximal chains in $\mathcal{P}(\omega)$ - each bijection $\mathbb{N}\rightarrow\mathbb{Q}$ gives rise to such a chain via Dedekind cuts, and different bijections give rise to incompatible chains in this way - so this bound is sharp.


