I'm testing my understanding through the exercises in my topology textbook and I am unsure of the answer for this one:
Consider the excluded point topology $EPX_p$ on a set $X$. Determine the $Int(A)$ and $Cl(A)$ for sets $A$ containing $p$ and for sets $A$ not containing $p$.
My thoughts:
I know that the interior is the union of all open sets in $A$ and that the closure is the union of all closed sets in $A$. I get a little confused when I start applying this onto topologies other than the usual topology.
I'm thinking that when $p\in A$, then $Int(A)=A$ and $Cl(A)=A$. Then when $A$ does not contain $p$ $Int(A)=A-\{p\}$ and $Cl(A)=A-\{p\}$. However, I know that $A$ is not in the topology, since the topology only contains subsets that exclude p. How would that fact play a role?
Any explanations would be helpful.
Best Answer
Assume $p \in A$. Then as you note, $\operatorname{int} A$ is the union of all open sets contained in $A$. $A \setminus \{p \} \subseteq A$ is open and is clearly the largest open set contained within $A$, so $\operatorname{int} A = A \setminus \{ p \}$.
The closed sets of this topology are $\varnothing$ and the sets containing $p$. Thus, $A$ is closed and is therefore its own closure.
Now assume $p \notin A$. Then $A$ is open, so it is its own interior. $A \cup \{ p \}$ is closed. Any closed non-empty set must contain $p$, so the intersection of all closed sets containing $A$ must contain $A \cup \{p \}$. Also, $A \cup \{ p \}$ is closed, so the intersection of all such closed sets must fall within $A \cup \{ p \}$. Thus, $\overline{A}= A \cup \{ p \}$.