Which is the topology generated by the neighborhood system $V(x)=\{\{x\}\}$ My question is the following:
Which is the topology generated by the neighborhood system $V(x)=\{\{x\}\}$ ? 
I say that is the coarse topology but I don't know how is the mechanism to generate a topology from a basis
sorry if this is so trivial but I always have this doubt
thank you
 A: By definition a set $U$ is open if and only if for every $x\in U$, there is a neighborhood $B_x\in V(x)$ such that $B_x\subset U$. Particularly, $V(x)\ni\{x\}\subseteq\{x\}$.. You should recognize that this is a standard topology on a set after some thought.
A: Say your underlying set for which you define a neighborhood system of its elements is called $X$. A neighborhood system is the collection of all neighborhoods of a point $x$. Thus the only open neighborhood of $x$ is $\{x\}$. A set $U$ is open if and only for each $x \in U$ there exists a basis element $B$ contained in $U$ such that $x\in B$. Therefore, it is easy to see that basis elements are of the form $\{x\}$ such that $x\in X$. Therefore a basis for this topology is: $\mathcal{B} = \{\{x\}\mid x \in X\}$. But assuming that there are two distinct points $x,y\in X$, $\{x\}\cup \{y\}$ is open by definition. Therefore $\{x,y\} \in V(y)$, a contradiction. Conclude that there is not two distinct points in $X$ and that this neighborhood system defines the trivial topology on a set $X$ with $1$ or less elements.  
