Open set of reals as union of open intervals Can we write any open set of reals as a countable Union of disjoint open intervals ?! And can any open set of reals can be written as countable  union of disjoint closed interval !? 
 A: We claim that $(0, 1)$ cannot be written as a countable disjoint union of closed intervals.
Proof. Let $(K_n)_{n=1}^{\infty}$ be any sequence of disjoint closed subintervals of $(0, 1)$. Define $(I_n)_{n=1}^{\infty}$ by $I_1 = (0, 1) \setminus (K_1 \cup \cdots \cup K_{n-1})$ and $I_{\infty} := \cap_{n=1}^{\infty} I_n$. We claim that $I_{\infty}$ is non-empty.
To this end, notice that each $I_n$ is a union of $n$ disjoint open intervals, say $J_{n,1}, \cdots, J_{n,n}$. Let $E_n$ be the union of endpoints of $J_{n,1}, \cdots, J_{n,n}$. Then the closure $\overline{I_n}$ of $I_n$ is simply the disjoint union of $I_n$ and $E_n$.
Now let $C = \cap_{n=1}^{\infty} \overline{I_n}$. Since $C$ is the intersection of non-empty compact sets, $C$ is also compact. Moreover, $C$ has no isolated points. Indeed, if $x \in C$, then for each $n$ there is $i_n \in \{1, \cdots, n\}$ such that $x \in \overline{J_{n,i_n}}$. For convenience, let us write $\overline{J_{n,i_n}} = [a_n, b_n]$. Then a crucial observation is that $a_n, b_n \in C$ for all $n$. From this,


*

*If $\inf (b_n - a_n) = 0$, then both $(a_n)$ and $(b_n)$ converge to $x$. Since $(a_n, b_n) = J_{n, i_n}$ is non-empty, we have $b_n > a_n$. This shows that either $a_n < x$ for all $n$ or $b_n > x$ for all $n$. Thus $x$ is not isolated.

*If $\inf (b_n - a_n) > 0$, then $\cap_{n=1}^{\infty} [a_n, b_n] \subseteq C$ is an interval of positive length and contains $x$. Thus $x$ is not isolated.
Thus $C$ is a non-empty perfect set. It is well-known that any non-empty perfect subset of $\Bbb{R}$ is uncountable. Thus in view of the identity
$$ I_{\infty}
= \bigcap_{n=1}^{\infty} (\overline{I_n} \setminus E_n)
= \bigg( \bigcap_{n=1}^{\infty} \overline{I_n} \bigg) \setminus \bigcup_{n=1}^{\infty} E_n
= C \setminus \bigcup_{n=1}^{\infty} E_n $$
together with the fact that $\cup_{n=1}^{\infty} E_n$ is countable, it follows that $I_{\infty}$ is uncountable as well. Therefore the claim follows.
A: The first is true.  Let $U \subset \mathbb{R}$ be an open set.  Basically the idea is, if we choose a point $x_1 \in U$, there is a certain "maximal" interval $I_{x_1} = (x \! - \! \delta, \ x \! + \! \varepsilon)$ such that $\delta$ and $\varepsilon$ cannot be made any larger for us to also have $I_{x_1} \subset U$.  Once we've constructed such an interval that contains $x_1$, choose another $x_2 \in U$, $\ x_2 \notin I_{x_1}$ and do the same thing again.  Do this repeatedly until we can write $\displaystyle U = \bigcup_{k} I_{x_k}$.  
We are guaranteed that this collection of intervals is countable since, within each interval $I_{x_n}$, we can find a rational number $q_n \in I_{x_n}$ with which to "label" it.  Since the rationals are countable and the labeling set is a subset of the rationals, we see that the collection of intervals is countable.

I'm not sure how to argue the latter question, but I'll keep thinking about it.  Maybe someone can lend some help in another answer?  It suffices to show that this is (im)possible for a single open interval since open sets decompose into a countable union of open intervals, and a countable union of countable sets is countable.
