Open Sets in $\mathbb{R}$ I was wondering what the general form of an open set is in the real numbers. Is it just an interval of the form $(a,b)$; $a,b \in \mathbb{R}$. 
 A: The set of open intervals of the type $(a,b)$ is a basis for the usual topology on $\mathbb{R}$. This means that all open sets of this topology can be written as countable unions or finite intersections of open intervals.
As an exercise, you can take $a<b<c$ and search all the open sets that you can define from these three points; e.g.
$(a,b) \cup (b,c)$ is a finite union and it's open. 
Also the set that you can obtain as countable union of the countable family of intervals $\{(a,kb): k \in \mathbb{N}^+\}$ is open, and note that it is  $(a,+\infty)$.
And, if you want to see that infinite intersections can give not open sets, you can use the family $\{(a-\frac{b}{k},a+\frac{c}{k}): k \in \mathbb{N}^+\}$ whose union is $(a-b,a+c)$  but  the intersection is the set $\{a\}$. 
But, note that any finite intersection of open intervals gives an open set. A special case is $(a,b) \cap (b,c)=\emptyset$ that is open.
A: This has been stated in the comments. 
Consider $A=(1,2)\cup(3,4)$.
$x\in A$ implies $x\in (1,2)$ or $(3,4)$. Suppose $x\in (1,2)$. As $(1,2)$ is open, $ \exists \epsilon>0 \text{ such that } B(x,\epsilon)\subset (1,2)\subset A$. Similarly, consider the case when $x\in (3,4)$. So, $A$ is open.
In other words, open sets in $\mathbb{R}$ need not be in the form of an interval.
A: We define a primitive open set on the reals ($\mathbb{R}$) as:
$$a,b,x\in \mathbb{R} : a<x<b$$
and we write $(a,b)$.
A union of two open sets is also open. Let $A$ and $B$ be two open sets, their union is open whether $A\cap B=\emptyset$ or not. Any finite union of primitive open sets is open.
Any countable union of open sets is also open, and it is possible for an uncountable union of open sets to be open, for example:
$$A_n=(x,1-x);\; x\in\mathbb{R},0\lt x\lt\frac{1}{2}$$
The union is $(0,1)$.
