Number of Dedekind Cuts in a dense linearly ordered set? I’m looking for a proof or counterexample to the following: for any dense linearly ordered set of cardinality $\kappa$, if there are exactly $\lambda$-many Dedekind Cuts of such a set, then $\lambda >\kappa$. 
My intuition is that this is true since there will be at least $\kappa ^ {cf \kappa}$-many Dedekind Cuts, since Dedekind Cuts correspond to certain increasing/decreasing sequences of the relevant set... 
 A: This is not true. Let $(X; \preceq)$ be a linear order. Recall that a Dedekind cut of $X$ is a partition $X = A \dot{\cup} B$ such that $A \preceq B$ (i.e. every element of $A$ is $\preceq$-less than every element of $B$) and $A$ has no $\preceq$-maximal element.
Now let $(X;\preceq) := (\kappa; \le)$ - the natural order on $\kappa$ for some infinite cardinal $\kappa$. The Dedekind cuts of $\kappa$ are precisely of the form $\eta \dot \cup (\kappa \setminus \eta)$ for limit ordinals $\eta$ (where I regard $0$ to be a limit ordinal). Hence there exactly $\kappa$ many Dedekind cuts of $(\kappa; \le)$.

You changed your question to ask only about dense linear orders. If $\kappa \ge 2^{\aleph_0}$, the answer remains 'no':
Let $\kappa$ be an infinite cardinal. Consider the dense linear order $(\kappa \times \mathbb Q; \preceq)$, where
$$
(\alpha, m) \preceq (\beta, n) \iff \alpha < \beta \vee (\alpha = \beta \wedge m \le n).
$$
(I.e. $\preceq$ is the lexicographical order). Since $(X; \preceq)$ is nothing but a row of $\kappa$ many copies of $\mathbb Q$ laid end to end, it's easy to see that there are exactly $\kappa \cdot 2^{\aleph_0}$ many Dedekind cuts. Hence, if $\kappa \ge 2^{\aleph_0}$, there are exactly $\kappa$ many.
