Do all possible subsets of a partially ordered set map to every element of its MacNeille completion?

Question

Every partial order can be completed with respect to arbitrary meets and joins as a complete lattice by embedding it into its MacNeille completion. My question is the following: does this mean that we have effectively constructed a 1-1 correspondence (bijection) between the powerset of the partial order and the elements of its MacNeille completion? I know that the embedding preserves all meets and joins that exist in the partial order, so I would guess that the answer is that the bijection is in fact an implicit part of the construction (since it is modeled as generalized Dedekind cuts). The reason I am asking is because there are other "bigger" completions possible for a partial order, when "size" is properly defined, such that the MacNeille completion can be described in a way as the "smallest". Is the answer obvious? Thank you.

Answer 1

Clearly not: as bof noted in the comments, the MacNeille completion can be (isomorphic to) the original partial order.

Let $\langle X,\le\rangle$ be a partial order, and let $\langle X^\#,\subseteq\rangle$ be its MacNeille completion. Let $$i:X\to X^\#:x\mapsto\downarrow\!\!x$$ be the canonical embedding. It’s true that for any $S\subseteq X$, $i[S]$ has a join $\bigvee i[S]\in X^\#$, but the map $S\mapsto\bigvee i[S]$ is not in general an injection, so it doesn’t give you a bijection from $\wp(X)$ to $X^\#$.

For an example in which $\langle X,\le\rangle$ is not already complete, let $X$ be the set of irrational numbers with the usual linear order. The MacNeille completion in this case is just the Dedekind completion together with an endpoint at each end; this is order-isomorphic to the extended real line $\overline{\Bbb R}=\Bbb R\cup\{-\infty,\infty\}$, and since $|\wp(X)|>|X|=|\Bbb R|$, it’s clear from the cardinalities alone that we can’t have a bijection between $\wp(X)$ and $\Bbb R$.

If we do the same thing with $X=\Bbb Q$, cardinalities no longer rule out the possibility that the construction of the completion produces a bijection between $\wp(\Bbb Q)$ and $\Bbb R$, but an examination of the orders themselves quickly shows that many different subsets of $\Bbb Q$ can have the same meet and join in $\Bbb R$.