Q:

Ring of sets and topology

We define a class of set $A$ to be a ring if it is closed under unions and differences. The question:

What is it necessary for a topology $\tau$ to be a ring?

I can see that if the topology $\tau$ is a ring then for a closed set $F$ we have that $X-F$ is open and consequently $F$ have to be in $\tau$. That is the class of the closed set is a subset of the class of the open sets. 
But I don't get nothing more. Any advice?
Thanks!

A:

Every open set is closed, and every closed set is open. This means that if $\mathscr{U}$ is any family of open sets, then it's also a family of closed sets, so $\bigcap\mathscr{U}$ is closed and therefore open. In other words, arbitrary intersections of open sets are open as well as arbitrary unions. In particular, if $x\in X$, and $\mathscr{U}$ is the family of all open nbhds of $x$, then $\bigcap\mathscr{U}$ is an open nbhd of $x$. Clearly it is the smallest open nbhd of $x$, in the sense that it is a subset of every open nbhd of $x$. 
For each $x\in X$ let $B(x)$ be this smallest open nbhd of $x$.

Show that $\mathscr{B}=\{B(x):x\in X\}$ is a partition of $X$.  
Show that $\mathscr{B}$ is a base for the topology.

In particular, it follows that if the topology is $T_0$, then in fact it's the discrete topology.

