Does the Base of a topology form itself a topology? I am studying topology,and in the beginning I came to know about bases of a topology and I feel problem there. By topology  in a special class of subsets of a set X such that they satisfy some specific properties. A base is a 
sub-collection of the topology such that any open set i.e any set from the topology can be written as a union of elements of the base.
Now my question is can the base itself be a topology?
The question arises to me in solving the following problem.
Let $\mathbf{R}$ be the set of reals and $B=\{[a,b):b>a\}$.


*

*Is $B$ a topology?

*Is $B$ a base of a topology?
Now we know that $B$ is a base for the lower limit topology on $\mathbf R$.
While solving this I thought that $\emptyset $ doesn't belong to $B$, so it is not a topology, but then I thought that the $\emptyset $ is always a subset of any set so $B$ must be a topology.
Again while proving B to be a base  for the lower limit topology,taking any open set $G$ and and taking any point $x \in G$ we see that $x$ belongs to $[x,x+1)$ which belongs​ to $B$, hence it is a base.
Am I wrong​ in thinking?. If so hope to get corrected.
Thank You.
 A: First of all, recall the definition of base.
Definition. Let $(X,\tau)$ be a topological space. A family $\mathscr{B}\subseteq \tau$ of open sets is called a base for $\tau$ if (1) $\mathscr{B}$ is an open cover of $X$ and (2) for any $U,V\in \mathscr{B}$ and for any $x\in U\cap V$ there exists $W\in \mathscr{B}$ such that $x\in W\subseteq U\cap V$.
I think you can easily see that your example is a base with respect to this definition. But it can not be a topology on $\mathbb{R}$, the real line. 
The reason is the following. If $\mathscr{B}=\{[a,b):a<b\}$ were a topology, it would be closed under infinite unions. But what about that?
$$ \bigcup_{n>0} [1/n,2)$$
A: If you start with some topology $\tau$ then you can discern a base for $\tau$ as done in the answer of Caligula. 
But if you start without a topology then also some reasonable things can be said about bases.

Let $X$ be a set and let $\mathcal B\subseteq\wp(X)$. Then $\mathcal B$ is a base for a topology iff:


*

*The intersection of two elements of $\mathcal B$ can be written as a union of elements of $\mathcal B$.

*For every $x\in X$ there is a $B\in\mathcal B$ with $x\in B$.


Now observe that a topology will satisfy both conditions, so it is formally correct to say that a topology is also a base for a topology.
You can wonder: if $\mathcal B$ is indeed a base for a topology, then is this topology unique? And how does it look like? Yes, it is unique and can be described as the collection of all subsets of $X$ that can be written as union of elements of $\mathcal B$. 
Denoting this collection by $\mathcal B^{\bigcup}$  it can be shown that: $$\mathcal B\text{ is a base for a topology }\iff\mathcal B^{\bigcup}\text{ is a topology}$$
Note that for a base $\mathcal B$ we have: $$\mathcal B^{\bigcup}=\mathcal B\iff\mathcal B\text{ is a topology}$$ This because a topology is closed under unions. 
