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 $a,b \in I_x, a<b$. 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?


1 Answer 1


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.

Also your proof here is not quite convincing.

It is clear that $I_x$ is an interval (pf: take any $a,b\in I_x$,$a<b$. For any $c\in(a,b)$, $c$ lies in $I_x$.)

Hint. Try to find subcollection picking intervals wich corresponds to some rational points.

  • $\begingroup$ 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. $\endgroup$
    – Koro
    May 21 at 6:46
  • $\begingroup$ @Koro, this looks right. In other words, if a in some interval and b in the other, then these two intervals have common point x that's why union will be an interval again. $\endgroup$
    – Danil
    May 21 at 6:50

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