# Basic Topology: Closure, Exterior, Interior, and Boundary of Open Half-Line Topology

I haven't dealt with the Open Half-Line Topology much and I'm really unsure of how to go about calculating the required sections below.

Let $A$ = (-$\infty$,4) $\cup$ [5,$\infty)$ be a subset of $(R,C)$ and $C$ is the Open Half-Line Topology.

(a.) $Cl(A)$.

Closed sets in this topology are of the form $(-\infty,a]$. Thus the smallest closed set that would contain $A$ is $R$.

(b.) $Ext(A)$.

The definition I am using for $Ext(A)$ is "the set of all points $x$ $\in$ $X$ for which there exists an open set $U$ such that $x$ $\in$ $U$ $\subseteq$ $X - A$.

We have also been told that $Ext(A) = Int(X - A)$.

Based on that I have $Ext(A) = (4,5)$. I am not confident that this is correct.

(c.) $Bd(A)$.

The definition I am using for $Bd(A)$ is "the set of all points $x$ $\in$ $X$ for which every open set containing x intersect $A$ and $X - A$.

I believe this is $Bd(A)$ = {$4,5$}.

(d.) $Int(A)$.

The definition I am using for $Int(A)$ is "the set of all points $x$ $\in$ $X$ for which there exists an open set $U$ such that all $x$ $\in$ $U$ $\subseteq$ $A$.

I am confused on this one since I am not confident that the above answers are correct.

For the most part it seems correct. The last part then follows by taking the set that contains every open interval (a,b) such that (a,b) is not a proper subset of (4,5).Hope this helps.

Since exterior, interior, and boundary are all pairwise disjoint, then $\mathbb{R}=Ext(A)\cup Bd(A)\cup Int(A)$. So, we know that any one point cannot be in more than one of these sets. Also, the $Cl(A)=Int(A)\cup Bd(A)$ and $Ext(A)=\mathbb{R}-Cl(A)$.

Let $A=(-\infty,4)\cup[5,\infty)$. $A$ is not open in $C$ since it is not in the form $(a,\infty)$. Likewise, $A$ is not closed in $C$ since it is not of the form $(-\infty,a]$.

(a) $Cl(A)=\mathbb{R}$ because $\mathbb{R}$ is the smallest closed set that contains $A$.

(b) $Ext(A)=\mathbb{R}-Cl(A)=\emptyset$.

(d) $Int(A)$ is the largest open set contained within $A$. If $A$ is open, then $Int(A)=A$. Since we established earlier that $A$ is not open, then the $Int(A)\subset A$. The largest open set of the form $(a,\infty)$ that is contained within $A$ is $(5,\infty)$. So, I believe $Int(A)=(5,\infty)$.

(c) For $Bd(A)$, the open sets you are using such that $x\in U$ and $U\cap A$ and $U\cap(X-A)$ should be considered open with respect to the topology $C$, so they should be of the form $(a,\infty)$. But, $Cl(A)=Int(A)\cup Bd(A)$, so $\mathbb{R}=(5,\infty)\cup Bd(A)$ and since $Bd(A)$ and $Int(A)$ are pairwise disjoint, so $Bd(A)=(-\infty,5]$.