Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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?
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.
