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If one assumes GCH, then it is not hard to prove that for every infinite cardinal $\kappa$, there exists a chain of $2^\kappa$ subsets of $\kappa$ relative to set inclusion (then of course, such a chain of subsets exists for any set of size $\kappa$). My question is the following: can one prove in ZFC that for every infinite cardinal $\kappa$, there is a chain of $\kappa^+$ subsets of $\kappa$? This may be a standard result of combinatorial set theory (if true), but I'm having a hard time finding a reference.

Greg Oman
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    See Problems 18.4 (For any cardinal $\kappa\ge\omega$ there are $\kappa^+$ subsets of $\kappa$ so that selecting any two of them, one includes the other) and 6.90 (For every infinite cardinal $\kappa$ there is an ordered set of cardinality $\kappa$ that has more than $\kappa$ initial segments) in Problems and Theorems of Classical Set Theory by Péter Komjáth and Vilmos Totik. – bof Dec 31 '16 at 06:23

1 Answers1

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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.

Eric Wofsey
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