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

Proof:

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?

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    $\begingroup$ Possibly duplicated this link $\endgroup$
    – Zhanxiong
    Commented Aug 10, 2015 at 4:38
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    $\begingroup$ 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
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    $\begingroup$ 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

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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.

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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.

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