Turning a poset into a fully ordered set Given a poset P and two incomparable elements a and b, can one have two posets P1 where a < b and P2 where b < a, with the other elements comparing the same way if they were comparable before? That is, can an arbitrary order be imposed between incomparable elements of a poset?
(I am aware that imposing one comparison might imply/force a few other previously incomparable elements to compare one way or another by transitivity)
Also can one apply the correct tags for me? I can't seem to find them myself.
EDIT: The new poset must also preserve the preexisting ordering, in addition to the new imposed comparison.
EDIT 2: The posets are finite.
 A: Yes.  More precisely, the following theorem is true:
Theorem: Let $(P,\leq)$ be a poset and let $a,b\in P$ be elements such that $b\not\leq a$.  Then there is a partial order $\preceq$ on $P$ which contains $\leq$ and such that $a\preceq b$.
Proof: The idea is to let $\preceq$ contain $\leq$ and $(a,b)$, and also all other elements it is forced to contain by transitivity.  So define $\preceq$ to be the union of $\leq$ and the set $$\{(c,d):c\leq a\text{ and }b\leq d\}.$$
Clearly $\preceq$ contains $\leq$ and $a\preceq b$ (we can take $c=a$ and $d=b$).  So we just have to check that $\preceq$ is a partial order.
For transitivity, suppose $x\preceq y$ and $y\preceq z$.  If $x\leq y$ and $y\leq z$, then $x\leq z$ and so $x\preceq z$.  So let us assume that $x\not\leq y$ (the case $y\not\leq z$ is similar).  We then have $x\leq a$ and $b\leq y$.  Now just note that $y\preceq z$ implies either $y\leq z$ or $b\leq z$, and in either case $b\leq z$ is true.  Thus we have $x\leq a$ and $b\leq z$, and hence $x\preceq z$.
For antisymmetry, suppose $x\preceq y$ and $y\preceq x$.  If $x\leq y$ and $y\leq x$ then $x=y$, so we may assume without loss of generality that $x\not\leq y$.  This means $x\leq a$ and $b\leq y$.  But now as in the previous paragraph, $y\preceq x$ implies $b\leq x$.  Thus $b\leq x\leq a$, contradicting the assumption that $b\not\leq a$.

In particular, if $a$ and $b$ are incomparable, then the Theorem can be applied to $a$ and $b$ in either order to get extensions of $\leq$ in which either $a\preceq b$ or $b\preceq a$.
