Basic Topology: Closure, Boundary, Interior, Exterior For the Half-open line topology $(R,H)$.
If $A = R - Q$, compute the closure, boundary, interior, and exterior for $A$.
I have $cl(A) = R$, $int(A) = \emptyset$, $bd(A) = R$, and $ext(A) = \emptyset$.
My logic each is below.  Clearly, if I am wrong please let me kmow but I'm trying to ensure I'm understanding the topology correct and the required computations.
$R - Q$ is neither open nor closed in $(R,H)$, therefore, the smallest closed set containing $A$ is $R$ so $cl(A) = R$.
$int(A)$ is defined as the set of all points $x\in X$ for which there exists an open set $U$ such that $x \in U$ and $U \subseteq A$.
$ext(A)$ is defined as the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U$ and $U \subseteq of X - A$.
$int(A)$ and $ext(A)$ are the empty set.  All irrational numbers are surrounded by rational numbers so there exists no open sets $U$ that will not contain rational numbers. Therefore the open sets would not be a subset of $A$.
$bd(A)$ would be $R$  since $X = int(A) \cup bd(A) \cup ext(A)$ where $int(A)$ and $ext(A)$ are both empty sets.  I know there is a more logical reason for this one but I cannot articulate it to myself.
 A: Your argument regarding the closure needs some work. The fact that $A$ is neither open nor closed in the half-open topology does not imply that the closure of $A$ is $\mathbb R$. After all, $[1,0)$ is neither open nor closed in the stadard topology, but its closure is not $\mathbb R$. A better way is to assume that $A^C$ contains an open set. Without loss of generality, this open set is a basis element $[a,b)$. Show that $[a,b)$ must contain a rational number.
A similar argument formalizes the notion that the exterior and interior of $A$ are empty.
Your reasoning about the boundary is sound. For each point $x\in\mathbb R$, one of the following has to be true:


*

*There exists an open set $B$ so that $x\in B\subset A$ ($x$ is an interior point).

*There exists an open set $B$ so that $x\in B\subset A^C$ ($x$ is an exterior point).

*Every open set containing $x$ intersects nontrivially with $A$ and $A^C$ ($x$ is a boundary point).


Since no points are interior points or exterior points, all points much be boundary points.
