About extending an order, comparing two uncomparable elements

Question

Suppose you have a partially ordered set $(X, <)$, and let $a,b \in X$ two uncomparable elements, i.e. nor $a <b$ or $b<a$ hold.

My question is:

Deos it exist an extension $<'$ of the order $<$, for which $a <'b$?

In other words: I want to extend the order by imposing that $a<b$. This is just an arbitrary choice I made, however I am not sure whether this is always possible.

MY TRY: I extend my relation $<$ inductively, by defining successive extensions $<_0, <_1, <_2, \dots$. The union of all these will be $<'$.

Start by $<_0:=<$.

Then $<_1$ is defined by adding the relations $a<_1b$ and all the relations of the form $$s<_1b \ \ \ \ \ \mbox{ whenever } s<_0a$$ and $$a<_1t \ \ \ \ \ \mbox{ whenever } b<_0t$$

And then I continue extending $<_1$ with a relation $<_2$ and so on... However this looks quite complicated and inelegant.

I am looking for an elegant argument, possibly working in ZF.

Answer 1

Define $<'$ as the transitive closure of $< \cup \{(a,b)\},$ which is the intersection of all transitive relations $R \in X\times X$ for which $< \cup \{(a,b)\} \subseteq R.$ This will always be transitive, so you just need to check it's a partial order (i.e. you just need to check it is still irreflexive, which is just the property that $(b,a)\not\in <$)

The above is perfectly fine in ZF. If, however, you want a constructive definition, then it basically amounts to what you've written.

Answer 2

Simply define $x\le'y$ if and only if either $x\le y$ or else $x\le a$ and $y\le b$. It is completely straightforward to verify that $\le'$ is a partial order relation extending $\le$ and that $a\le'b$. For example, to verify the transitive law, you only have to consider four easy cases:

If $x\le y$ and $y\le z$, then $x\le z$, so $x\le' z$.

If $x\le y$ and $y\le a$, $b\le z$, then $x\le a$ and $b\le z$, so $x\le' z$.

If $x\le a$, $b\le y$, and $y\le z$, then $x\le a$ and $b\le z$, so $x\le'z$.

If $x\le a$, $b\le y$, and $y\le a$, $b\le z$

If $x\le a$, $b\le y$, and $y\le a$, $b\le z$, then $b\le a$, contradicting the assumption that $a$ and $b$ were incomparable.