A basic doubt on open sets and open interval How to prove that in $\Bbb R$ any open set can be written as an union of open intervals ? By an open set set we mean that every point is an interior point of that set. How to construct such open intervals from this information ? If we can construct then the proof is easy. But, I am confused over how to construct such intervals. 
Actually, I want to prove that it can be expressed as a "countable" union. For that I need to find those largest open intervals, then I can use denseness of $\Bbb Q$. Hope I have made my point clear.
 A: Let $O\subseteq\mathbb{R}$ be open. If $O$ is void we are done. Otherwise, select
$x\in O$. Since $O$ is open, we can choose $\epsilon_x >0$ so that 
$(x-\epsilon_x, x + \epsilon_x)\subset O.$  It is not hard to see that
$$O = \bigcup_{x\in O} (x-\epsilon_x, x + \epsilon_x).$$
A: There are two issues here. If all you want is to check that an open set is union of intervals, without any worry about overlapping, repetitions, or redundant information, then this is obvious: Any $x$ in your open set is in some open interval contained in your open set (by definition of open). Say $x\in I_x\subseteq O$, $I_x$ an open interval, for each $x\in O$, $O$ the given open set. Then clearly $O\subseteq \bigcup_{x\in O} I_x$ (since $x\in I_x$ for all $x$). But also $\bigcup_{x\in O} I_x\subseteq O$, since $I_x\subseteq O$ for all $x$.
[Irrelevant remark: This does not use the axiom of choice, as we can pick $I_x$ centered at $x$ and of rational length, and we have an explicit enumeration of the rationals.]

If what you want is to show that $O$ is union of disjoint open intervals, the usual idea is simply an extension of the trivial observation that if $(a,b)$ and $(c,d)$ intersect at all, then $(a,b)\cup(c,d)$ is also an interval.
One way to proceed is to define a relation $x\sim y$ on the elements of the given open set. The relation means that there is an open interval contained in the open set, and such that $x$ and $y$ both belong to the interval. Start by checking that $\sim$ is an equivalence relation, and then verify that the equivalence classes of this relation are open intervals. And since different equivalence classes are always disjoint, we are done.
That $\sim$ is an equivalence relation is straightforward. To see that the equivalence classes are intervals, note that if $C$ is an equivalence class, $\alpha=\inf C$ and $\beta=\sup C$ ($\alpha$ or $\beta$ may be infinite), then for any $t,s$ with $\alpha<t<s<\beta$, we have that $t\sim s$, and therefore $C$ is just $(\alpha,\beta)$. 

Let me close with another side remark: The representation of $O$ as a disjoint union of open intervals only requires countably many intervals (maybe even only a finite number). This is because we can associate each interval with one of its rational numbers, and disjoint intervals are then associated with different rationals. (Contrast this with the fact that there are uncountably many open intervals, so the fact that they are disjoint in this case is crucial.) In general, any pairwise disjoint collection of open subsets of $\mathbb R$ is countable, by the same argument.
In 1920, M. Y. Suslin asked whether this could be used to characterize $\mathbb R$, that is, whether a (non-empty) linearly ordered set $R$ without end-points is order isomorphic to $\mathbb R$ provided that the order is dense in itself, has the least upper bound property, and any pairwise disjoint collection of open intervals is countable. This question is known as Suslin's problem. 
It turns out that this statement is independent of the usual axioms of set theory: It is consistent that it is false (as shown by Jensen's work on Gödel's constructible universe), and it is also consistent that it is true. The technique of forcing axioms in set theory originated with the work of Martin and Solovay showing the latter.
A: Let $A$ be your set. Then for each $x\in A$ there is an open interval $I_x$ with $x\in I_x\subseteq A$. What if you take the union of these? A more interesting statement is that every open set in $\Bbb R$ is the disjoint union of countably many open intervals. To prove this, you'll have to dwell into component intervals, that is, maximal open intervals contained in a set w.r.t. to inclusion. The fact that $\Bbb Q$ is dense and countable will be crucial, too.
A: Suppose $G\subseteq\mathbb R$ is a non-empty open set.  Let $x\in G$.  Let $a_x=\sup\{y\in\mathbb R : [x,y)\subseteq G\}$ and $b_x =\inf\{y\in\mathbb R: (y,x]\subseteq G\}$.
Then try to show that $(b_x,a_x)\subseteq G$ and $G=\bigcup_{x\in G} (b_x,a_x)$.
