For a linear order $(X,\le)$, define a cut to be any downwards closed subset of $X$.
In the case $X = \mathbb Q$, there are $2^{\aleph_0}$ cuts, which is more than there are elements in $\mathbb Q$.
My question is: for which cardinals $\kappa$ besides $\aleph_0$ is there a linear order on $\kappa$ with more than $\kappa$ cuts?
Partial answer: Assume GCH and let $\kappa$ be an uncountable cardinal. Let $\le_{\text {Lex}}$ be the lexicographical order on ${\mathbb Q}^\kappa$, the set of functions from $\kappa$ to $\mathbb Q$, which has cardinality $2^\kappa$. Let $A \subseteq {\mathbb Q}^\kappa$ be the set of sequences that are eventually $0$. Notice that $A$ is dense in ${\mathbb Q}^\kappa$: if $a\le_{\text {Lex}}b$ and $\alpha$ is minimal such that $a_\alpha \ne b_\alpha$, define $c_\beta = a_\beta$ for $\beta < \alpha$, $c_\alpha = \frac{a_\alpha + b_\alpha}{2}$ and $c_\gamma = 0$ for $\gamma > \alpha$. Then $a \le_{\text {Lex}} c \le_{\text {Lex}} b$
Because $A$ is dense in ${\mathbb Q}^\kappa$, $(A,\le_{\text {Lex}}\restriction_A)$ has $2^\kappa$ cuts. If $\kappa$ is a successor cardinal, then each element of $A$ corresponds to a function $f:\kappa^- \to \mathbb Q$ and an ordinal $\alpha < \kappa$ (because $|\kappa \cap \alpha | \le \kappa^-$ for all $\alpha < \kappa$), so $|A| = \aleph_0^{\kappa^-} \cdot \kappa\stackrel{GCH}{=} \kappa$.
This shows that if $\kappa$ is a successor cardinal, it is consistent that the statement holds for $\kappa$. In particular there can be an order on $\mathbb R$ with more than $\mathfrak c$ cuts. I'm not sure what can happen to the cardinality of $A$ when $\kappa$ is a limit cardinal, or if it is consistent that the statement fails. Also, while my main question is in ZFC, I would also be interested in results in the absence of choice.
Best Answer
This exists for any $\kappa$ in ZFC.
Let $\lambda \leq \kappa$ be least so that $2^{\lambda} > \kappa$. Then the lexicographical order on the set $S$ of binary sequences on $\lambda$ which are eventually constant has size $\leq \kappa$, but it has $2^\lambda > \kappa$ many cuts. If $\vert S \vert < \kappa$ just add some dummy elements on the side (for example a copy of the well-order $\kappa$). Then we are guaranteed an order of size exactly $\kappa$ with more than $\kappa$ many cuts.