Cardinality of the set of chains of $(\mathcal P(\mathbb N), \subseteq)$

Question

In that article, I describe some chains of $(\mathcal P(\mathbb N), \subseteq)$. Some of those are having the cardinality of the continuum.

As for any permutation of $\mathbb Q$, we can define in a one-to-one way a chain (having the cardinality of the continuum) of $(\mathcal P(\mathbb N), \subseteq)$ using Dedekind cuts, we have at least a continuum of chains of $(\mathcal P(\mathbb N), \subseteq)$.

But at the end, what is the cardinality of the set chains of $(\mathcal P(\mathbb N), \subseteq)$?

Answer 1

Given your chain of cardinality $2^{\aleph_0}$; any subset of that is a chain. That gives $2^{2^{\aleph_0}}$ chains.