Minimal linear extension of a poset Let $(X,R)$ be a poset, and let $\bar{R}$ be a total order on $X$, s.t. $R \subseteq \bar{R}$. I would call such a total order a minimal extension of $R$ if all the relations in between are not total. If such a minimal extension exists, is it unique (up to something)? 
I'm asking because I would like to have a justification for taking as monoidal product in the category of finite totally ordered sets the lexicographical ordering, and I would feel much better if this would be up to ?isomorphism? the minimal extension of the very natural product order of posets.
 A: No. Take $X=\{0,1\}$ with the discrete ordering $R$. Then there are exactly two linear orders extending it. 
You might argue that they are still isomorphic. They are, but not "with respect to $X$". And in fact, any two linear orders on a finite set of a given size are isomorphic, so you can see that in this context it's not the correct question. 
For a perhaps better example, you might want to look at $\omega\times 2$ with the product order ($(x,y) \leq (x',y') \iff x\leq x' \land y\leq y'$). 
The two different lexicographic orderings extend it, but they're not even isomorphic (one of them yields the ordinal $\omega 2 = \omega + \omega$ and the other yields $2\omega = \omega$)
Note that your condition "all relations inbetween are not total" is redundant : if $R$ is a linear order, then any strict subrelation is not total : if $R'\subset R$ and $(x,y)\in R\setminus R'$, then $R'$ doesn't decide whether $x\leq y$ or $y\leq x$; so in a sense all linear orders are minimal as linear orders (and maximal, for obvious reasons)
