Basic Topology: Cofinite Topology $A$ = $R$ - {$(n+1)/(n+2):n \in Z+$}
The definition of the cofinite topology in my text is as follows:
Let $X$ be any set.  The collections $T$ of all subsets $U$ of $X$ such that either $U = \varnothing$ or $X - U$ is a finite set is a topology for $X$.
Given A above, determine $Cl(A), Int(A), Ext(A)$ and $Bd(A)$.
Initially, I determined that $R$ - ($R$ - {$(n+1)/(n+2):n \in Z+$}) = {$2/3, 3/4, 4/5, 5/6,...$}.
(a) $Cl(A)$: The smallest closed set that will contain this set is $R$.  Is that correct?
(d) $Bd(A)$:  My thought is that $Bd(A)$ would be {$2/3, 3/4, 4/5, 5/6,...$}.  I feel like if I can understand $Bd(A)$ then I can work backwards to understand $Int(A)$ and $Ext(A)$.
 A: (a) The Cl$(A)$ is indeed $\mathbb{R}$, since a closed set is either finite or all of $\mathbb{R}$; and no finite set contains $A$.
(b) First notice that the boundary of a subset is closed, and hence finite or all of $\mathbb{R}$, so it cannot be $\{2/3,3/4,\cdots\}$.
Since Int$(A) \subset A$, we get $\mathbb{R} -$ Int$(A)\supset \mathbb{R} - A$. But  $\mathbb{R} -$ Int$(A) = $Cl$(A) - $Int$(A) = $Bd$(A)$. Therefore, Bd$(A) \supset \mathbb{R} - A = \{2/3,3/4,\cdots\}$ and hence is infinite. Since Bd$(A)$ is closed, it must therefore be all of $\mathbb{R}$. And so Int$(A) = \emptyset$.
A: Def'n: A subset $S$ of a space $X$ is dense in $X$ iff $\bar S=X.$
In the co-finite topology on an infinite set $X,$ every infinite subset of $X$ is dense in $X,$ because if $Y$ is a non-empty open subset of $X$ then $Y\cap S\ne \phi.$ And if $S$ is an infinite, co-infinite subset of $X$, then $S$ and its complement have empty interior, because no non-empty open set can be a subset of either of them.
So both $A$ and its complement are dense in $R,$ and each of them has empty interior.
