Existence of large chains provable in ZFC?

Solution 1:

Yes, this is true. Let $\lambda$ be the least cardinal such that $\kappa^\lambda>\kappa$. Let $X={}^{<\lambda}\kappa$, the set of all sequences in $\kappa$ indexed by an ordinal less than $\lambda$. By minimality of $\lambda$, $|X|=\kappa$, so it suffices to find a large chain in $\mathcal{P}(X)$.

To find such a chain, note that we can totally order $X$ lexicographically, and in fact this ordering extends to $X\cup {}^{\lambda}\kappa$. Now let $A\subset\mathcal{P}(X)$ be the collection of all lower sets of $X$: that is, $A$ is the set of all sets $Y\subseteq X$ such that if $y\in Y$ and $x\in X$, $x<y$ implies $x\in Y$ (where $<$ is the lexicographic order). Note that $A$ is totally ordered under inclusion. Every $s\in{}^\lambda\kappa$ determines an element of $A$, namely $\{x\in X:x<s\}$. It is easy to see these sets are different for different values of $s$ (between any two elements of ${}^\lambda\kappa$ there is an element of $X$), so $A$ has cardinality at least $\kappa^\lambda>\kappa$.

In general, the supremum of all lengths of chains in $\mathcal{P}(\kappa)$ is called $\operatorname{ded}(\kappa)$. It can also be described as the supremum of all cardinalities of the set of Dedekind cuts in a totally ordered set of size $\kappa$ (hence the name "ded"), or as the supremum of all cardinalities of totally ordered sets with a dense subset of size $\kappa$. The argument above shows that $\operatorname{ded}(\kappa)\geq\kappa^\lambda$ for the least $\lambda$ such that $\kappa^\lambda>\kappa$. For a bit more discussion and some references, see this question on MO.