Dedekind MacNeille completion. Definition of supremum as $\left(\bigcup M\right)^{ul}$. Counterexample for definition as $\bigcup M$?

Question

I didn't found a mathematical text for Dedekind MacNeille completion , so I "defined" supremum in the completion the following way: $$\sup M := \left(\bigcup M\right)^{ul},$$ where M is a family of some subsets of $S_{DM}$, the completion of $(S,<)$. (okay, supremum is always defined, yes, here as term $(\iota x \in S_{DM}: M\leqslant x)$, but then sometimes it is interpreted as proper partial function on the support of the first order model.)

And it works perfectly, indeed a supremum.

But when I tried $\sup M := (\bigcup M)$ as definition, I got stuck in the proof. So I suspect, that there is a counterexample for such definition.

Could you please either provide such counterexample, or actually prove that the second definition is also good?

Answer 1

I think that the smallest counterexample is the following:
let $S=\{\bot,a,b,c\}$, where $\bot$ is the least element and $\{a,b,c\}$ forms an anti-chain.

Then, $S_{DM}$ just adds a top element. Following the definition, the elements of $S_{DM}$ are $$\{\bot\}, \{\bot,a\}, \{\bot,b\},\{\bot,c\},\{\bot,a,b,c\},$$ where $\{\bot,a,b,c\}$ is the top element.

Now let $M=\{ \{\bot,a\}, \{\bot,b\} \}$; it follows that $\bigcup M =\{\bot,a,b\}$ (which is not a member of $S_{DM}$), but $$\left(\bigcup M\right)^{ul} = \{\bot,a,b,c\}.$$