# Every collection of non-trivial intervals has a countable subcollection with the same union.

Let $$\scr A$$ be a collection of non trivial intervals (i.e., an interval with at least two points) in $$R$$. Then, it is to be proven that there exists a countable subset $$A\subset \scr A$$ such that $$\bigcup_{I\in \scr A} I= \bigcup_{I\in A} I$$, where $$I\in \scr A$$.

I tried to prove it the following way but got stuck at a step:

Let $$G:= \bigcup_{I\in \scr A} I$$. Take any $$x\in G$$. Let $$\scr T_x\subset \scr A$$ be the subcollection of non trivial intervals that contain $$x$$.

Define $$I_x=\bigcup_{I\in \scr T_x} I$$. It is clear that $$I_x$$ is an interval (pf: take any $$a,b \in I_x, a. For any $$c\in (a,b)$$, $$c$$ lies in $$I_x$$.)

Suppose that $$I_x\in \scr A$$ for every $$x\in G$$. I'm not sure how to prove this bolded part but assuming it proves the statement as the following shows:

Define $$\sim$$ on $$G$$ as $$x\sim y\iff I_x=I_y$$.

$$\sim$$ is an equivalence relation.

Equivalence classes are $$I_x, x\in G$$.

We can write $$G$$ as the union of distinct equivalence classes. So $$G=\sqcup_x I_x$$. The union is countable (pf: represent each of the distinct classes by a rational no. in it and then map it to $$Q$$, the set of rationals to get an injection of the equivalence classes into $$Q$$.). As it was assumed that $$I_x\in \scr A$$ for all $$x\in G$$. This proves the statement.

I am not sure how to prove the bolded statement above.

$$I_x=\bigcup_{I\in \scr T_x} I$$. Suppose on the contrary that $$I_x\not\in \scr A$$.

How to get contradiction from here?

You take an interval $$I_x = \cup \mathcal I_x$$ and want to show that $$I_x \in\mathcal A$$. But it will be true if your collection $$\mathcal A$$ have the following property: if $$I_\alpha \in \mathcal A$$ for any $$\alpha$$ (some indexes) then there exists $$I_0\in\mathcal A$$ s.t. $$I_0 = \bigcup I_\alpha$$. I think that we can find a counterexample for this situation.
It is clear that $$I_x$$ is an interval (pf: take any $$a,b\in I_x$$,$$a. For any $$c\in(a,b)$$, $$c$$ lies in $$I_x$$.)
• Thanks for the inputs. :). I think you are right, the claim $I_x\in \scr A$ may not be true. So the method suggested in the post may not work. About $I_x$ being an interval, I didn't write enough details there. But I think the proof could go along the following lines: case 1) a<x<b so c is in (a,x) or in (x,b). In either case, c is in an interval containing (a,x) or (x,b) case 2) x<a<b etc.