Yesterday I came up with this problem and I just cannot get it out of my head:
Problem: Consider a finite $I$ set of integer-valued intervals $I_i=(a_i,b_i)$ $i = 1,\ldots,n; a_i \in \mathbb{N_0}, k \in \mathbb{N}, I=( I_1,\ldots,I_n\ )$
Assume there is a map $\phi: I \longrightarrow \{1,\ldots,k\}$ such that
$(*): \{l, m\} \subset \{1,\ldots,n\}, l \neq m, \phi(I_l)=\phi(I_m) \Rightarrow I_l \cap I_m=\emptyset$
Now if $M=(x,y)$ is an inteval such that $(**): M\cap\left(\bigcap_{i=1}^k\left(\bigcup\{I_j|\phi(I_j) = i\}\right)\right)=\emptyset$. then $M$ can be always appended to $I$ with a new map $\phi'$ such that property $(*)$ holds.
Example: $k=3$, $I=( (0,2), (3,5), (2,4), (5,7), (0,3), (4,6) )$, $\phi(I)=( 1,1,2,2,3,3)$
This corresponds to the following image (edit: the $k$ on the left side is a bit misleading, better ignore it):
As you can see the property $(*)$ is fullfilled (no intervals are overlapping in their time line). We can also see that for the interval $(2,4)$ he have the property $(**)$ i.e. we find spaces to put it into the image:
Now the theorem tells us that there is a new image where the green interval does not have to be broken. And indeed we find $\phi(I \cup (2,4))=(1,2,3,2,2,2,1,1)$. Or in the image:
I hope I could explain the problem well. I am not sure if the theorem is correct but I tried quite a few examples and it always worked out.