Let $(X, \leq)$ be a totally ordered set. A downset of $X$ is a subset $A$ of $X$ with the property $$ \forall x \in X \forall a \in A : ( x \leq a \Rightarrow x \in A ) $$ Denote $\mathcal{D}(X)$ for the set of downsets of $X$. We trivially have $\lvert \mathcal{D}(X) \rvert \leq 2^{\lvert X \rvert}$ and equality is possible, e.g. if $X = \mathbb{Q}$. If $X$ is a finite set, then $\lvert \mathcal{D}(X) \rvert = \lvert X \rvert + 1$.
If $X$ is an infinite, complete totally ordered set, then $\lvert \mathcal{D}(X) \rvert = \lvert X \rvert$, because all downsets are of the form $$ \lbrace x \in X \mid x < a \rbrace \text{ or } \lbrace x \in X \mid x \leq a \rbrace $$ for some $a \in X$. I was wondering if for an infinite, non-complete totally ordered set, there are general results on the cardinality of $\lvert \mathcal{D}(X) \rvert$, or criteria to determine whether $\lvert \mathcal{D}(X) \rvert < 2^{\lvert X \rvert}$ or $\lvert X \rvert < \lvert \mathcal{D}(X) \rvert$.