I believe the following statement is true but I can't find or figure out a proof:
For any partition of the set $\mathbb{Q}$ of rationals into a finite number of parts, there is a part containing an isomorphic copy of $\mathbb{Q}$ (preserving the < relation)
This is hinted at in passing in a paper by Hasson, Kojman and Onshuus (www.math.bgu.ac.il/~kojman/symindiv.pdf) who call this property 'indivisibility'.
I think IF I can prove (and it is true) that any finite partition of $\mathbb{Q}$ must include a part containing an interval $\mathbb{Q}\cap(a,b)$ then I can use the function $f(x)=a + \frac{1+x/(1+|x|)}{2(b-a)}$.
- Is this function an isomorphism $f:\mathbb{Q}→\mathbb{Q}\cap(0,1)$ preserving $<$ ?
- Is it true that in a partition of finitely many parts, one must contain an interval $\mathbb{Q} \cap(a,b)$?
- Is there any other way of proving the statement?
This feels like it should be easy and I'm just missing something, so any help would be appreciated. Thanks!
It suffices to prove the result for a partition into two parts, $A$ and $B$, say. Also, it suffices to show that one of $A,B$ contains a countable dense subset without endpoints (cf. ''Brian M. Scott's'' comment). Of course, the endpoint condition is harmless as we can simply remove them (we want to show only that one of the parts contains an isomorphic copy of $\mathbb Q$).
If one parts avoids an interval, say $A\cap (a,b)=\emptyset$, then $(a,b)\cap\mathbb Q\subseteq B$ and we are done. This includes the case that one of the sets is bounded from above or below or even finite. Hence $A$ (as well as $B$) can be assumed as infinite. It is also dense, for if $a,b\in A$ with $a<b$ then as just seen $(a,b)\cap A$ is not empty.