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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.

Eric Wofsey
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Paul Stelian
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  • I'm confused. You can of course create an ordering where you arbitrarily declare $a<b$ and go home. But I assume you want an ordering that respects the partial ordering you already have? – probably_someone Jun 17 '17 at 08:57
  • Indeed, I want to respect the preexisting partial ordering and just add one additional rule. I'm asking if it can be done arbitrarily. – Paul Stelian Jun 17 '17 at 08:58
  • I suspect the answer is no, as that would mean there would be no posets, just ordered sets. But I don't have the required set-theoretic expertise to say for sure. – probably_someone Jun 17 '17 at 08:59
  • Well I asked because I used this assumption to prove something in a Data Structures exam (I did get the maximum grade, I'm just curious if it was indeed correct) – Paul Stelian Jun 17 '17 at 09:01
  • Anyway as I said when imposing the comparison between a and b others would be implied and I would obey those. – Paul Stelian Jun 17 '17 at 09:01
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    May be related: https://math.stackexchange.com/questions/271003/every-partial-order-can-be-extended-to-a-linear-ordering – Crostul Jun 17 '17 at 09:02
  • @Crostul Not exactly my question but it does imply that the answer to my question is yes. Especially since I was thinking about finite posets (forgot to specify that) – Paul Stelian Jun 17 '17 at 09:04
  • Well you can try and prove the following claim : given $p,q$ incomparable elements, then $\leq^* = \leq\cup{(p,q')\mid q\leq q' }$ is still a partial ordering. This is what Asaf uses in his answer to the post given by Crostul – Maxime Ramzi Jun 17 '17 at 09:09
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    Order theory is another tag for your question. – William Elliot Jun 17 '17 at 20:56

1 Answers1

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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 $b\preceq a$.

Eric Wofsey
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