Passing from basis of topology to regular definition of open sets The following is the definition of basis for a topology.
If X is a set, a basis for a topology on X is a collection B of subsets of X such that 
1)For each x $\in X$ there is atleast one basis element B containing X.
2)If x belongs to the intersection of two basis elements, then there is $B_3$ 
containing X such that $B_3$ $\subset B_1 \cap B_2$.
A subset U of x is said to be open in X if for each x $\in$ U, there is a basis element B $\in \beta$ such that x $\in B$ and B $\subset U$.
I was wondering how does this definition satisfies like regular open intervals 
and doesn't satisfy [a,b] ? I mean how come does it satisfy regular definition of open sets that we know of from analysis..?
 A: It is possible for $[a,b]$ to be open according to the definition open that you've given. There are many topolgies on $\Bbb R$. One in which intervals such as $[a,b]$ are open is the discrete topology.  In this topology, all subsets of $\Bbb R$ are open.
A: There seems to be a little confusion on terminology.
You are right that the definition you state is the definition of a basis for a topology. The word "a" is important here, since we did not make a choice for a specific topology here.
What you call the regular definition of open sets in analysis is in fact "a" topology that you can put on the real line. There are many more choices of topologies, as Tim pointed out.
This particular topology has as a basis the open intervals of the form $(a,b)$ for $a,b \in \mathbb{R}$. It could be a useful for you to check that this is indeed a basis, although this is more or less trivial once you see it.
You see the confusion? The analysists made a choice for a topology when they told what they meant with an open set. It is the topologists that generalize this and consider other possible topologies.
A: Just to comment on Tim Rackowski's answer, for the basis definition of open sets, you still need to know what the basis is. For example, the basis on $\mathbb{R}$ which generates the topology you're familiar with is the set $\{(a,b), a,b\in \mathbb{R}, a<b\}$. You can take arbitrary unions of these sets to generate the topology on $\mathbb{R}$. Why doesn't this hold for $[a,b]$? Because $[a,b]$ cannot be written as a union of sets of the form $(c,d)$, since we would have to have an open interval $(c,d)$ containing $b$ or $a$, but this would imply that there are elements in $[a,b]$ that are bigger than $b$ or smaller than $a$.
