Understanding the Co-finite Topology on R I'm looking to gain a better understanding of how the cofinite topology applies to R. 
I know the definition for this topology but I'm specifically looking to find some properties such as the closure, interior, set of limit points, or the boundary set and how these change based on whether a subset A in R is closed, open, or clopen. 
Any help would be appreciated. 
Note: I have only the most basic of definitions for closure, interior, etc. 
Thank you! 
 A: In the cofinite topology, '$A$ is closed' means '$A$ is finite or is $\mathbb{R}$' and '$A$ is open' means '$A$ has finite complement or is empty'. In particular...


*

*If $A \subseteq \mathbb{R}$ is infinite then the only closed set containing $A$ is $\mathbb{R}$, and hence $\operatorname{cl} A = \mathbb{R}$. In particular, if $A$ is infinite, every point of $\mathbb{R} - A$ is a limit point of $A$. Hence if $A$ is infinite then $\partial A = \mathbb{R} - A$.

*If $A \subseteq \mathbb{R}$ has infinite compliment then its only open subset is $\varnothing$, so $\operatorname{int} A = \varnothing$.

*The only clopen sets are $\mathbb{R}$ and $\varnothing$.

A: The co-finite topology has a good property: It's the minimal $T_1$ topology, so every other property (Hausdorff, Regular, etc) you might expect to fail.
For example, every closed set is either finite or $\mathbb{R}$. So, the closure of any set is the set iself (if it's finite) or $\mathbb{R}$ if it is infinite.
Every open set is infinite or the empty set, so the interior of any finite set is $\emptyset$. 
So there are no clopen sets (except $\mathbb{R}$ and $\emptyset$)
The boundary $\partial A = \overline{A} \cap \overline{\mathbb{R}\setminus A}$, so, if $A$ is open, its boundary is precisely $\mathbb{R}\setminus A$. If $A$ is closed, the closure of $\mathbb{R}\setminus A$ is $\mathbb{R}$ and then $\partial A = \overline{A} = A$
The set of limit points of $A$ is $\emptyset$ if $A$ is closed, and is the set itself if it's open.
A: So to answer your first question if $O$ is an open set that isn't empty in the co-finite topology then by definition $\bar{O}$ is the smallest closed set that contains $O$. Now we know that $O$ has infinitly many elements and the only closed set that doesn't only have finitely many elements is $\mathbb{R}$ so it must be $\bar{O}$.
The boundary of an open set $O$ is $\bar{O} \backslash O$ so if $O$ isn't empty then in this topology its boundary is $\mathbb{R} \backslash O$.
