# If τ is the finite-closed topology, then the semi-open sets are precisely the open sets

If τ is the finite-closed topology, then the semi-open sets are precisely the open sets.

We can assume the set is infinite, otherwise the proof is trivial. I can easily show that open sets are semi-open. Any open set A are element of X without a finite number of points. If we remove another point, it is still an open set, O, and is a subset of A. The closure of O is the entire set because that is the smallest closed set that contains O since every closed set is finite. So every open set fits the criterion for being semi open. Now I need to show that if there are any other semi open sets in the topology, is is just part of the open sets. I am not sure how to show this.

Let $$X$$ be an infinite space with the cofinite topology. Let $$A\subseteq X$$ be semi-open, meaning that there is an open $$U$$ in $$X$$ such that $$U\subseteq A\subseteq\operatorname{cl}U$$. If $$U=\varnothing$$, then $$\operatorname{cl}U=\varnothing$$, so $$A=\varnothing$$, which is open. If $$U$$ is non-empty, then $$U=X\setminus F$$ for some finite $$F\subseteq X$$, and $$\operatorname{cl}U=X$$, so $$X\setminus F\subseteq A\subseteq X$$. But then $$X\setminus A\subseteq F$$, so $$X\setminus A$$ is finite, and $$A$$ is open.