If $X$ is any partially ordered set with $A\!\subseteq\!X$ and $x\!\in\!X$, define $x\!\leq\!A :\Leftrightarrow \forall a\!\in\!A\!: x\!\leq\!a$ and $A\!\leq\!x :\Leftrightarrow \forall a\!\in\!A\!: a\!\leq\!x$, and denote $A^u\!:=\!\{x\!\in\!X; A\!\leq\!x\}$ and $A^l\!:=\!\{x\!\in\!X; x\!\leq\!A\}$ and $A^\downarrow\!:=\!\{x\!\in\!X; \exists a\!\in\!A\!:x\!\leq\!a\}$ and $A^\uparrow\!:=\!\{x\!\in\!X; \exists a\!\in\!A\!:a\!\leq\!x\}$. The Macneille completion of $X$ is $(\{A^{u\,l}; A\!\subseteq\!X\},\subseteq)$.
If $X$ is any totally ordered set, a cut in $X$ is any pair $(A,B)$ such that $X\!=\!A\!\cup\!B$ is a partition with $A\!=\!A^\downarrow$ and $B\!=\!B^\uparrow$ (hence $A\!\leq\!B$) and $A$ contains no greatest element. A cut $(A,B)$ is proper when $A\!\neq\!\emptyset$ and $B\!\neq\!\emptyset$. The completion by cuts of $X$ is $\{\text{proper cuts in }X\}\!\cup\!\{(\emptyset,X),(X,\emptyset)\}$, where $(A,B)\!\leq\!(A',B'):\Leftrightarrow A\!\subseteq\!A' \Leftrightarrow B\!\supseteq\!B'$.
Wikipedia suggests that the MacNeille completion is the generalization of the completion by cuts. If $X$ is totally ordered, how can I prove that $\{A^{u\,l}; A\!\subseteq\!X\}\:\cong\:\{\text{proper cuts in }X\}$?
I've tried the maps $A^{u\,l}\!\mapsto\!(X\!\setminus\!A^\uparrow,A^\uparrow)$ and $A^{u\,l}\!\mapsto\!(A^{u\,l},X\!\setminus\!A^{u\,l})$, but didn't come far.