Prove that every open set in $\mathbb{R}^1$ is the union of an at most countable collection of disjoint intervals.


Let $\mathbb{R}^1\supset G$ is an open set. Then $\forall$ $x\in G$ $\exists \varepsilon_x>0:$ $I_x=(x-\varepsilon_x, x+\varepsilon_x)\subset G$. Then $$G=\cup_{x\in G}I_x.$$

But $\mathbb{R}^1$ is separable then it has a countable base then any open cover has a countable subcover. Then $$G=\cup_{i\geqslant}I_i.$$

How to turn these intervals to disjoint intervals?

  • 2
    $\begingroup$ Possibly duplicated this link $\endgroup$
    – Zhanxiong
    Commented Aug 10, 2015 at 4:38
  • 1
    $\begingroup$ You know I saw all these links and learned many proofs of this theorem. But I want to know how to continue my proof because it seems to me very interesting. $\endgroup$
    – RFZ
    Commented Aug 10, 2015 at 4:47
  • $\begingroup$ Reading your link I didn't find any useful fact that can help to my proof. $\endgroup$
    – RFZ
    Commented Aug 10, 2015 at 4:53
  • 1
    $\begingroup$ You should be a bit more careful in phrasing things. Not every separable space is second-countable (or even hereditarily Lindelöf, which is what you are using). (For example, the Niemytzki/Moore plane.) Although in metric spaces separability and second countability (and Lindelöfness, etc.) coincide. $\endgroup$
    – user642796
    Commented Aug 10, 2015 at 6:39
  • $\begingroup$ Thank you very much Arthur Fischer! Every separable metric space $(X, d)$ has a countable base $\Rightarrow$ it has a Lindelof property. Your example is very interesting. $\endgroup$
    – RFZ
    Commented Aug 10, 2015 at 6:44

2 Answers 2


For the record, an alternative topological approach that may be useful to someone:

Take your open set $U$. Consider its partition in its connected components. Since they are connected, they must be intervals.

Now, it is easy to see that a disjoint family of intervals must be enumerable: take any rational in its midsts as the enumeration.


Your solution is almost correct, but you need to make an additional step to make the intervals disjoint.

If $G$ is empty, then the claim is trivial (the empty collection of intervals is at most countable).

If $G$ is not empty, let, for each $x\in G$, $$\mathcal I_x\equiv\{I\,|\,I\text{ is an open interval, }x\in I,\text{ and }I\subseteq G\}.$$ Since $G$ is open, $\mathcal I_x\neq\varnothing$. Moreover, $$I^{\star}_x\equiv\bigcup_{I\in\mathcal I_x}I$$ is open and it is also an interval: I leave it to you to check that $$I^{\star}_x=\left]\inf_{I\in\mathcal I_x}\{\inf I\},\sup_{I\in\mathcal I_x}\{\sup I\}\right[\in\mathcal I_x.$$

Now, if $x,y\in G$, then $I_x^{\star}$ and $I_y^{\star}$ are either identical or disjoint. To see this, suppose that there exists some $z\in I_x^{\star}\cap I_y^{\star}$. Then, $I_x^{\star}$ and $I_y^{\star}$ are open intervals containing a common element, so that $\hat{I}\equiv I_x^{\star}\cup I_y^{\star}$ is an open interval, too. Since $x\in I_x^{\star}$, $y\in I_y^{\star}$, $I_x^{\star}\subseteq G$, and $I_y^{\star}\subseteq G$, it follows that $x\in \hat I$ and $\hat I \subseteq G$, so that $\hat I\in\mathcal I_x $. Analogously, $\hat I\in\mathcal I_y$. Therefore, $\hat I\subseteq I_x^{\star}$ and $\hat I\subseteq I_y^{\star}$, which, together with the definition of $\hat{I}\equiv I_x^{\star}\cup I_y^{\star}$, is possible only if $I_x^{\star}=I_y^{\star}$.

Finally note that, $$G=\bigcup_{x\in G} I_x^{\star}.$$ By the foregoing, any two members of the union that are not the same must be disjoint. Now use the Lindelöf property as you did before.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .