Be $X$ an infinite set and $x_0$ a point of $X$. Be $B = \{\{x\} : x \neq x_0\} \cup \{A \subseteq X : x_0 \in A$ and $X -A$ is finite$\}$.
How are the open and closed of this topology? Knowing the basis I can find out who are the open of this topology? I just know they're unions of the base elements. But it's still a little confusing for me.
I assume $B$ is a base (which was answered in your previous question Base of topologic space ). The answer to this question provided by Kavi Rama Murthy in the comments is correct (but the $A$ there should be $U$). Here are the details.
If $x_0\not\in U$ then $U$ is open since $U=\cup\{\{x\}:x\in U\}$ and each $\{x\}$ is open, since $\{x\}\in B$.
If $x_0\in U$ and $X\setminus U$ is finite then $U\in B$, hence $U$ is open.
Conversely, if $x_0\in U$ and $U$ is open then there is some $A\in B$ with $x_0\in A\subseteq U$. Then $X\setminus A$ finite (as it is not possible that $A=\{x\}$ for some $x\not=x_0$). Hence $X\setminus U$ is also finite since $X\setminus U\subseteq X\setminus A$.
Thus, the open sets are exactly those sets that either do not contain $x_0$, or if they contain $x_0$ then their complement must be finite.
Taking complements (of the open sets as described above), the closed sets are all sets that contain $x_0$, as well as all finite sets (regardless whether they contain $x_0$ or not).
