Could someone please verify what I've done on interior and closure of open or closed subsets in the Fort Topology?
We recall that the topological space $(X, \tau)$ is called Fort Topology if $X$ is infinite, $x_0 \in X$, and $\tau = \{A \subset X; x_0 \notin A\}$ $\cup$ $\{A \subset X; X - A$ is finite$\}$, i.e., it is the union of the excluded point topology with the cofinite topology.
Given $A \subset X$, we are asked to find $int(A)$ and $\bar{A}$. We suppose $A$ is a proper nonempty subset, otherwise it's trivial.
We first prove a
Lemma: if $x \neq x_0$, then the set $\{x\}$ is both open and closed in X ("clopen").
Proof: $x_0 \notin \{x\}$ and therefore it is open. But it's complement $X - \{x\}$ is also open, since $X - (X - \{x\}) = \{x\}$ if finite. The lemma follows.
Now we have:
Closure of A: If $A$ is closed, we have $\bar{A} = A$.
If $A$ is open, we need to verify three cases:
i) $x_{0} \notin A$ and $X - A$ is finite
We must have $A = X - \{x_0, x_1, ..., x_k\}$, for some positive integer $k$. From lemma, for each $i = 1, 2, ..., k$, the set $\{x_i\}$ is open (and closed), with $x_i \in \{x_i\}$, but $\{x_i\} \cap A = \emptyset$. Therefore, by definition of a closure point, $x_i \notin \bar{A}$. Now, for any open set $U$, if $x_0 \in U$, we must have that $X - U$ is finite. But $U \cap A \not\ni x_0$ is open as well, and $X - (U \cap A) = (X - U) \cup (X - A)$ is finite as well, and therefore $U \cap A$ can't be finite, because $X$ is infinite, particularly $U \cap A \neq \emptyset$. It follows that $x_0 \in \bar{A}$ and therefore $\bar{A} = A \cup \{x_0\}$.
ii) $x_{0} \notin A$ and $X - A$ is infinite
Similarly to part i), for each point $x \notin A$ ($x \neq x_0$), $\{x\}$ will be an open set containing $x$ with $A \cap \{x\} = \emptyset$, and therefore $x \notin \bar{A}$. Again, for every open set $U \ni x_0$, $U \cap A \not\ni x_0$ is open as well. De Morgan's law tell us that $X - (U \cap A) = (X - U) \cup (X - A)$ is infinite. If $U \cap A$ was empty, we should have $U = (X - A)$, which is closed, a contradiction because $U$ is an arbitrary open subset and $A$ is a proper non-empty subset. Therefore, $U \cap A \neq \emptyset$, and follows that $x_0 \in \bar{A}$, and $\bar{A} = A \cup \{x_0\}$.
iii) $x_{0} \in A$ and $X - A$ is finite follows easily from case i)
Interior of A If $A$ is open, $int(A) = A$.
If $A$ is closed, we must have $x_0 \in A$ and $X - A$ is infinite.
Now suppose that $x_0 \in int(A)$. Therefore, by definition of interior point, there exists an open set $V \ni x_0$ such that $V \subset A$. But than $X - A \subset X - V$, which is an absurd, since $X - A$ is infinite and $X - V$ must be finite. Hence, $x_0 \not\in int(A)$. We now have $int(A) \subset A - \{x_0\}$. Now take $y \in A - \{x_0\}$. Since $y \neq x_0$, by the previous lemma, $\{y\} \ni y$ is open contained in $A$. It follows that $A - \{x_0\} \subset int(A)$, and therefore, $int(A) = A - \{x_0\}$.
I appreciate any corrections (specially on part ii)).
Thank you very much!
Edit: spelling
The boring (somewhat redundant way) is indeed to check cases bases on sizes of $A$ and its complement and whether it contains the compactifying point $x_0$:
If $A$ is finite, $A$ is closed (as $X$ is a $T_1$ space), and so $\overline{A} = A$ and if $x_0 \notin A$, $A$ is open as well, so $\operatorname{int}(A) = A$, while if $x_0 \in A$, $\operatorname{int}(A) = A\setminus\{x_0\}$ as $A$ cannot be a neighbourhood of $x_0$.
If $A$ is infinite with infinite complement:
If $x_0 \notin A$, $A$ is open and $\operatorname{int}(A) = A$, and $x_0 \in \overline{A}$ as every open neighbourhood of $x_0$ is cofinite and thus intersects $A$. So $\overline{A} = A \cup \{x_0\}$.
If $x_0 \in A$, then $x_0 \notin \operatorname{int}(A)$ and the other points are isolated, so $\operatorname{int}(A) = A \setminus \{x_0\}$. And also, if $x_0 \in A$,its complement is open, so that $\overline{A} = A$.
If $A$ is infinite with finite complement:
If $x_0 \in A$, $A$ is open so $\operatorname{int}(A) = A$. It's complement is open (as it does not contain $x_0$ and hence is open), and so $\overline{A} = A$.
If $x_0 \notin A$, $A$ is also open so $\operatorname{int}(A) = A$ and also $x_0 \in \overline{A}$ and thus $\overline{A} = A \cup \{x_0\}$.