Cofinite topology: find interior, closure and boundary The problem statement, all variables and given/known data.
Let $X$ be a set and let $\tau=\{U \in \mathcal P(X) : X \setminus U \text{is finite}\} \cup \{\emptyset\}$. This is called the cofinite topology on $X$. Describe the interior, the closure and the boundary of the subsets of $X$ with respect to $\tau$.
The attempt at a solution.
If $A \subset X$, then $A^{\circ}=\{x \in A: \exists U \in \tau : x \in U \subset A\}$. Now, as $U \in \tau$, then $X \setminus U$ is finite. I don't know what this is telling me about $A^{\circ}$.
The definition of closure of a set $A$ is $\overline{A}=\cap_{F \text{closed}, A \subset F} F$. But $F^c=U$, an open set which belongs to $\tau$. Again, I don't know how to use this to describe the closure of $A$.
Now, the boundary of $A$ is $∂A=\overline{A}-A^{\circ}$.
I am pretty lost, I would appreciate if someone could help me to characterize these three in terms of the given $\tau$.
 A: The interior of $A$ is defined as the largest open set containing $A$ (this is equivalent to your definition since union of open sets are open). If $A$ is open (i.e. if $X \backslash A$ is finite), it is a general fact that open sets are equal to their interior, so you should have $A^{\circ} = A$. Otherwise $X \backslash A$ is infinite, and if $B \subseteq A$, then $X \backslash B$ has greater cardinality than $X \backslash A$, so it is also infinite. This shows that $A^{\circ} = $? 
Closed sets are finite sets (and $X$). The closure of $A$ is the smallest closed set containing $A$ (this is equivalent to your definition, since intersections of closed sets are closed). If $A$ is finite, then it is closed, and again closed sets are always equal to their closure in a topology. If $A$ is not finite, then what are the closed sets containing $A$? Take their intersection (it is easily computed). 
If you computed the two previous steps using my hints, the boundary should be no problem ; split in cases whether $A$ and/or its complement are finite or not.
Hope that helps,
