Let's write $\DeclareMathOperator{\ch}{ch}\ch({<}\kappa) = \sup\{|\mathcal{C}|\mid \mathcal{C}\text{ is a chain of sets, each of cardinality}<\kappa\}$.
Let $(L,\leq)$ be a linear order. By a cut in $L$, I just mean a downwards-closed set, i.e., $C\subseteq L$ such that for all $x\leq y$, if $y\in C$ then $x\in C$. Let $\DeclareMathOperator{\Cut}{Cut}\Cut(L) = \{C\subseteq L\mid C\text{ is a cut}\}$, and note that $(\Cut(L),\subseteq)$ is a chain of sets.
Now suppose $(\mathcal{C},\subseteq)$ is a chain of sets. For $x,y\in \bigcup\mathcal{C}$, define $x\trianglelefteq y$ iff for all $C\in \mathcal{C}$, if $y\in C$ then $x\in C$. This relation $\trianglelefteq$ is a linear preorder. If we quotient by the induced equivalence relation (i.e., identifying elements of $\bigcup \mathcal{C}$ if they are elements of exactly the same sets in $\mathcal{C}$), we obtain a linear order $(\DeclareMathOperator{\Lin}{Lin}\Lin(\mathcal{C}),\leq)$. The image of each $C\in \mathcal{C}$ under the quotient map is a cut in $\Lin(\mathcal{C})$, so the chain $(\mathcal{C},\leq)$ embeds into $\Cut(\Lin(\mathcal{C}))$.
These observations should make it clear that $\ch({<}\kappa)$ is closely related to the number of cuts in linear orders.
For a cardinal $\kappa$, $\DeclareMathOperator{\ded}{ded}\ded(\kappa)$ is the supremum of $|\Cut(L)|$ over all linear orders $(L,\leq)$ of cardinality $\kappa$.
Here are some facts about $\ded(\kappa)$:
- For all $\kappa$, $\kappa<\ded(\kappa)\leq 2^\kappa$. (So if GCH holds, then $\ded(\kappa) = 2^\kappa = \kappa^+$ for all $\kappa$.)
- $\ded(\aleph_0) = 2^{\aleph_0}$.
- For any cardinal $\kappa$ of uncountable cofinality, it is consistent that $\ded(\kappa)<2^\kappa$.
For more on $\ded(\kappa)$, see the introduction of On the number of Dedekind cuts and two-cardinal models of dependent theories by Chernikov and Shelah.
Similarly, we can write $\ded({<}\kappa) = \sup_{\lambda<\kappa}\ded(\lambda)$, the supremum of $|\Cut(L)|$ over all linear orders $(L,\leq)$ of cardinality $<\kappa$.
Since $\lambda<\ded(\lambda)\leq 2^\lambda$ for all $\lambda<\kappa$, we have $\kappa\leq \ded({<}\kappa)\leq 2^{<\kappa}$.
I claim that for all infinite $\kappa$, $\ch({<}\kappa) = \ded({<}\kappa)$.
In one direction, if $L$ is a linear order with $|L|<\kappa$, then $(\Cut(L),\subseteq)$ is a chain of sets, each of cardinality $\leq |L|<\kappa$, so $\ded({<}\kappa)\leq \ch({<}\kappa)$.
The other direction is not so immediate, because even if each set in $\mathcal{C}$ has cardinality $<\kappa$, we might have $|\Lin(\mathcal{C})| = |\bigcup\mathcal{C}| = \kappa$.
Nevertheless, let $(\mathcal{C},\subseteq)$ is a chain of sets, each of cardinality $<\kappa$. Let $f\colon \alpha\to \mathcal{C}$ be an order-preserving injective map, where $\alpha$ is an ordinal and $\mathrm{im}(f)$ is cofinal in $\mathcal{C}$ (this can be constructed by transfinite recursion).
Note that $\alpha\leq \kappa$. Indeed, if $\alpha>\kappa$, then for each $\beta<\kappa$, we can pick some $x_\beta\in f(\beta+1)\setminus f(\beta)$. The $(x_\beta)_{\beta\in \kappa}$ are distinct, and they are all elements of $f(\kappa)$, so $|f(\kappa)|\geq \kappa$, contradiction.
Now for all $\beta<\alpha$, let $\mathcal{C}_\beta = \{C\in \mathcal{C}\mid C\subseteq f(\beta)\}$, an initial segment of the chain $\mathcal{C}$. Note that $\bigcup\mathcal{C}_\beta = f(\beta)$, so $|\bigcup \mathcal{C}_\beta| <\kappa$. Thus, since $\mathcal{C}_\beta$ embeds in $\Cut(\Lin(C_\beta))$ and $|\Lin(C_\beta)|\leq |\bigcup C_\beta|<\kappa$, we have $|\mathcal{C}_\beta|\leq \ded({<}\kappa)$. Now piecing the whole chain back together, we have
\begin{align*}
|\mathcal{C}| &= |\bigcup_{\beta<\alpha} C_\beta|\\
&\leq |\alpha|\cdot \sup_{\beta<\alpha}|C_\beta|\\
&\leq \kappa\cdot \ded({<}\kappa)\\
&\leq \ded({<}\kappa),
\end{align*}
since, as observed above, $\kappa\leq \ded({<}\kappa)$. This establishes $\ch({<}\kappa)\leq \ded({<}\kappa)$.
Comments:
- If GCH holds, $2^{{<}\kappa} = \kappa$ for all infinite $\kappa$, so $\ch({<}\kappa) = \kappa$.
- $\ch({<}\aleph_0) = \aleph_0$ and $\ch({<}\aleph_1) = 2^{\aleph_0}$.
- The upper bound of $2^{{<}\kappa}$ is a slight improvement to the upper bound $\kappa^{{<}\kappa}$ claimed by David Gao in the comments. These can differ when $\kappa$ is singular.
- It's not clear to me whether the supremum in the definition of $\ch({<}\kappa)$ is always attained, i.e., whether there is always a chain of sets, each of cardinality $<\kappa$, such that the chain has cardinality $\ch({<}\kappa)$.
