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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.

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1 Answer 1

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The Wikipedia discussion of the relationship between the MacNeille completion and Dedekind cuts makes it appear that you should try $A^{ul}\mapsto\langle A^{ul},A^u\rangle$; all that needs to be shown is that $\langle A^{ul},A^u\rangle$ actually is a cut. Suppose that $a\in A^{ul}\cap A^u$. Then $A\le a\le A^u$, so $a=\sup A$. There are two possibilities.

  1. $a\notin A$: then $\langle A^{ul},A^u\rangle$ is clearly a cut.

  2. $a\in A$: then $\langle A^{ul},A^u\rangle$ fails to be cut (a) because $A^{ul}\cap A^u=\{a\}\ne\varnothing$, and (b) $A^{ul}$ has a greatest element, $a$. Both of these problems are repaired by removing $a$ from $A^{ul}$.

Thus, the correspondence should actually be $A^{ul}\mapsto\langle A^{ul}\setminus A^u,A^u\rangle$.

Added: This gives the expected results for the Dedekind completion of the rationals, for instance. In the setting of arbitrary linear orders, however, one cannot require that the left set of a cut have no largest element. Suppose that $\langle X,\le\rangle$ is a linear order and that $x,y\in X$ are adjacent with $x<y$. The completion should not insert anything between $x$ and $y$, so the cuts corresponding to $x$ and $y$ should be adjacent. The construction given above makes $\big\langle (\leftarrow,x),[x,\to)\big\rangle$ the cut corresponding to $x$ and $\big\langle (\leftarrow,y),[y,\to)\big\rangle=\big\langle (\leftarrow,x],[y,\to)\big\rangle$ the cut corresponding to $y$, which are indeed adjacent. Moreover, the correspond exactly to the scheme for Dedekind cuts in the rationals that makes the cut $\big\langle (\leftarrow,q),[q,\to)\big\rangle$ correspond to the rational $q$.

The only alternative is the scheme that puts an element of $X$ into the left set of its own Dedekind cut; in the rationals this means that no right set has a smallest element, but in the case of my $X$ here it means that the cut corresponding to $x$ will be $\big\langle (\leftarrow,x],(x,\to)\big\rangle=\big\langle (\leftarrow,x],[y,\to)\big\rangle$, and again the halves of the cut will have adjacent endpoints. It simply can’t be avoided.

In particular, it’s possible that $A^{ul}\setminus A^u$ still has a largest element, but if it does, the original linearly ordered set had adjacent elements here.

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Yes, this is the right map. I have just one more problem. How do I know that $A^{u\,l}\!\setminus\!A^u$ does not contain a greatest element? If $A$ contains a greatest element $a$, then $A^{u\,l}\!\setminus\!A^u = a^{u\,l}\!\setminus\!a^u = a^l\!\setminus\!a^u = (-\infty,a)$, and this can have a greatest element. –  Leon Lampret Jun 15 '12 at 14:45
    
@Leon: See if I’ve answered your objection in the addition to my answer. –  Brian M. Scott Jun 16 '12 at 10:37
    
Yes, thank you for your help. Regards! –  Leon Lampret Jun 17 '12 at 7:49

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