Open set of $\Bbb R$ which is not bounded below can be written as atmost countable collection of disjoint segments Suppose I have a open set of $\Bbb R$ which is not bounded below but bounded above. Now, I want to show that that open set can be written as atmost countable collection of open intervals. 
I have done the proof for the case when it is both bounded above and below. Now, , in this case, I want to first identify the open interval which is not bounded below. How to do that ? Then I can repeat the process for the bounded above and below case. 
For the case where the open set $E$ is bounded below and above I do the following : choose any $x \in E$ and consider the sets $\{a|(a,x) \subseteq E\}$ and $\{a|(x,a) \subseteq E\}$ and I know that infimum($y_i$) and supremum ($w_i$) exists for them respectively. So, I can write the open set as union of the open intervals $(y_i, w_i)$. Then I use use the denseness of $\Bbb Q$ in $\Bbb R$ to prove countability. But, the problem is what is the process by which I can write the open set as a union of open intervals when $E$ is not bounded below. If I start with arbitrary $x$ in $E$, then how shall I get these open intervals $y_i, w_i$. How shall I know that $\{a|(a,x) \subseteq E\}$ exists or not ?
 A: There need not be a single (semi-)unbounded interval contained in your open set $U$, just think of $\mathbb R\setminus\mathbb Z$.
However, your proof for the case of a bounded open set $U$ should work just as well for unbounded $U$. After all $\mathbb R$ is homeomorphic to $(0,1)$, which would take you back to the bounded case.
To write $U$ as a disjoint union of countably many intervals, you can proceed as follows:
Enumerate $\mathbb Q$ as $q_1,q_2,\ldots$.Now define open intervals $I_n$ recursively:
If $q_n\in U$ and $q_n\notin I_k$ for all $k<n$, then let $I_n=(\sup ((\mathbb R\setminus U)\cap (-\infty,q_n)), \inf ((\mathbb R\setminus U)\cap (q_n,\infty)))$; otherwise let $I_n=\emptyset$.
Now show that $I_n\cap I_m=\emptyset$ for $n\ne m$ and that $U=\bigcup_{n\in\mathbb N}I_n$. You may drop the empty intervals from this union if you like.
A: It's possible that your open set has infinitely many disjoint open intervals extending toward $-\infty$, so you can't necessarily easily reduce to the case of bounded above and below. What you can do is note that for any $a \in A$ in your open set, you can find a "rational" interval $I_a = (q_1,q_2) \subset A$ where $q_1,q_2$ are rational and $a \in I_a$. Then the set of all $I_a$ is countable (because the rationals are countable) and $\cup_{a \in A} I_a = A$. If you want intervals to be disjoint, then just note if you enumerate the intervals you obtained as $I_1,I_2,\ldots$, then you can iterate for $n=1,2,\ldots$, and if $I_n$ intersects any interval you have already created when considering $I_1,\ldots,I_{n-1}$ then merge $I_n$ with that interval to obtain a unioned interval that covers the same points. This process will make all intervals disjoint.
