Let $X$ be a $T1$ topological space. Call $x\in X$ isolated point if $\{x\}$ is open in $X$.
Assume there is an open, finite subset $\emptyset \ne V \subseteq X$. Does $X$ admit an isolated point?
If necessary, I can assume that $X$ is compact, Hausdorff and extremally disconnected (the closure of every open is again open) (this is the case in the scenario where I need it).
Attempt: I feel like I should somehow be able to prove that $V$ contains an isolated point. I also have the impression that I'm missing something obvious.
Let $U$ be a minimal open subset of $V$ that is open in $V$. Since $V$ is finite and $X$ is $T1$, we have that $G:=U^c \cap V$ is open. Hence, $V = U \cup G$. Somehow I want to show that $U$ is a singelton.
Best Answer
If $x \in V$ then $F:=V\setminus \{x\}$ is finite (as a subset of $V$) and so closed (as $X$ is $T_1$). So $\{x\} = V\cap (X\setminus F)$ is open in $X$ and so $x$ is isolated.