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The partially ordered set $(Y,\le)$ is called a superspace of the partially ordered set $(X,\le')$ iff

  1. $X\subseteq Y$.
  2. $\le'=(\le\cap X^2)$.

and a completion of $(X,\le')$ if in addition

$~~ 3$. $(Y,\le)$ is a complete lattice.

Are two minimal completions of a partially ordered set isomorphic?

(a minimal completion $(Y,\le)$, is a completion that no other completions have it as a superspace).

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I claim that according to your definition, every minimal completion is isomorphic to the Dedekind-MacNeille completion. Furthermore, every Dedekind-MacNeille completion is a minimal completion.

Suppose that $X$ is a poset. If $A\subseteq X$, then define $\uparrow A$ and $\downarrow A$ to be the sets where $x\in\uparrow A$ if $x\geq a$ for each $a\in A$ (i.e. if $x$ is an upper bound of $A$) and where $x\in\downarrow A$ if $x\leq a$ for each $a\in A$ (i.e. if $x$ is a lower bound of $A$). Then let $DM(X)=\{A\subseteq X|A=\downarrow\uparrow A\}$, and let $e:X\rightarrow DM(X)$ be the mapping where $e(x)=\downarrow\{x\}$. Then $DM(X)$ is a complete lattice ordered under inclusion and the mapping $e$ embeds $X$ into $DM(X)$, and $DM(X)$ is known as the Dedekind-MacNeille completion of $X$. The Dedekind-MacNeille completion is the unique up-to isomorphism complete lattice $L$ such that each $l\in L$ is the least upper bound of some subset of $X$ and the greatest lower bound of some subset of $X$. The reader is referred to [1] for more information on the Dedekind-MacNeille completion.

Suppose that $X$ is a poset and $L$ is a minimal completion of $X$. Then let $M=\{\bigvee^{L}R|R\subseteq X\}$. Then $M$ is a complete lattice with partial ordering induced from $L$ with $X\subseteq M$. Therefore, by minimality, $M=L$. Similarly, let $N=\{\bigwedge^{L}R|R\subseteq L\}$. Then $N$ is a complete lattice with $X\subseteq N\subseteq L$. Again, by minimality, we conclude that $N=L$ as well. Therefore, since each $l\in L$ is the least upper bound and greatest lower bound of a subset of $L$, we conclude that $L$ is the Dedekind-MacNeille completion of $X$.

$\textbf{Added 2/18/2015}$

It turns out that the Dedekind-MacNeille completion of a poset is always a minimal completion of a poset, so the minimal completion of a poset always exists. To prove this fact, we will need the following lemma and definition. We say that a subset $A$ of a complete lattice $L$ is join-dense if $L=\{\bigvee^{L}R|R\subseteq A\}$.

$\mathbf{Lemma}$ Suppose that $P$ is a poset and $L$ is the Dedekind-MacNeille completion of $P$. Let $M$ be a complete lattice such that $P$ is join dense in $M$. Then there is a surjective mapping $j:M\rightarrow L$ that preserves all least upper bounds and where $j(p)=p$ for each $p\in P$.

Suppose that $P$ is a poset and $L$ is the Dedekind-MacNeille completion of $P$. Suppose now that $P\subseteq Y\subseteq L$ and $(Y,\leq)$ is a complete lattice. Let $X=\{\bigvee^{Y}R|R\subseteq P\}$. Then $P\subseteq X$, $X$ is a complete lattice, and $P$ is join-dense in $X$. Therefore there is a surjective mapping $j:X\rightarrow L$ that preserves all joins and fixes $P$. Therefore the mapping $j$ has a unique right adjoint $k:L\rightarrow X$ defined by $k(l)=\bigvee\{x|j(x)\leq l\}$ and the mapping $k$ is injective. I claim that $k$ fixes all elements in $P$.

Suppose that $p\in P$. Then $k(p)=\bigvee\{x|j(x)\leq p\}$. We already know that $j(p)\leq p$. Now suppose that $j(x)\leq p$ and $x\not\leq p$. Then since $P$ is join dense in the complete lattice $X$, we have $x=\bigvee^{X}R$ for some $R\subseteq P$. Therefore there is some $r\in R$ with $r\not\leq p$ but clearly $r=j(r)\leq p$, a contradiction. Therefore, if $j(x)\leq p$, then $x\leq p$. We conclude that $k(p)=\bigvee\{x|j(x)\leq p\}$. Let $k^{\sharp}:L\rightarrow L$ be the mapping where $k^{\sharp}(x)=k(x)$ for each $x\in L$. Now suppose that $x\in L$. Then $x=\bigvee^{L}R=\bigwedge^{L}S$ for some subsets $R,S\subseteq L$. However, we have $k^{\sharp}(x)\geq k^{\sharp}(r)=r$ for each $r\in R$, so $k^{\sharp}(x)\geq\bigvee^{L}R=x$. Similarly, $k^{\sharp}(x)\leq k^{\sharp}(s)=s$ for each $s\in S$, so $k^{\sharp}(x)\leq \bigwedge^{L}S=x$. Therefore, we have $k^{\sharp}(x)=x$ for each $x\in L$. However, since $L=k^{\sharp}[L]\subseteq X$, we have $X=L$, so $Y=L$ as well. We therefore conclude that $L$ is a minimal completion of the poset $P$.

  1. Harzheim, Egbert. Ordered Sets. New York: Springer, 2005. Google Books link, Springer
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