The elements of topological space can be not open? I am really confused. Today I saw something called "final topology" and saw the following:
$\sigma(X,{f}=\{{\emptyset,]-\infty,0],]0,+\infty[,]-\infty,+\infty[}\}$ 
It is said that we can equip this topology. 
I do not understand, how can it be topology, if in the definition of topology we say that the elements of topological space are open sets ($]-\infty,0]$) is not open).
 A: A topology on $X$ is given by defining a collection of subsets of $X$ to be open. Providing this collection satisfies the appropriate axioms, it doesn't matter what the sets in it are. In particular, they don't have to be open in a metric space sense, or in some other standard topology.
A: Let $X$ be a set. For any $\mathcal{V}\subset\wp\left(X\right)$
define: $$\widehat{\mathcal{V}}:=\left\{ \cap\mathcal{V}'\mid\mathcal{V}'\subset\mathcal{V}\wedge\mathcal{V}'\text{ is finite}\right\} $$
and: $$\mathcal{V}^{\cup}:=\left\{ \cup\mathcal{V}'\mid\mathcal{V}'\subset\mathcal{V}\right\} $$
In this context $\cap\emptyset:=X$ so that $X\in\widehat{\mathcal{V}}$
for any $\mathcal{V}$. Also note that $\left(\mathcal{V}^{\cup}\right)^{\cup}=\mathcal{V}^{\cup}$.
The topology generated by $\mathcal{V}$ is: $$\mathcal{O}:=\widehat{\mathcal{V}}^{\cup}$$
It is the smallest topology on $X$ that contains $\mathcal{V}$ and
$\mathcal{V}$ is a subbasis of $\mathcal{O}$. 
If $\widehat{\mathcal{V}}\subset\mathcal{V}^{\cup}$
then $\mathcal{O}=\widehat{\mathcal{V}}^{\cup}\subset\left(\mathcal{V}^{\cup}\right)^{\cup}=\mathcal{V}^{\cup}\subset \mathcal{O}$ hence $\mathcal{O}=\mathcal{V}^{\cup}$, and $\mathcal{V}$ is a basis of $\mathcal{O}$. 
Remark: 
I dislike the notation $\widehat{\mathcal{V}}$ and prefer something like $\mathcal{V}^{\cap}$ with the 'f' of finite under the cap. 
I am not capable of constructing that, but maybe someone else is. Thanks in advance.
