A space $(X,\tau)$ is said to be $R\_1$ if for all $x,y\in X$ with $cl(\{x\}) \neq cl(\{y\})$, there are disjoint open sents separating $cl(\{x\})$ and $cl(\{y\})$.


If $X$ is compact and $R\_1$, does this imply that it is regular?



Compact $R_1$-spaces
The answer is Yes.


Suppose $(X,\tau)$ is $R\_1$ and compact. Pick $V$ open and let $x\in V$. For each $y\in X\setminus V$ there are open neighborhoods $U(y)$ of $x$ and $V(y)$ of $y$ such that $U(y)\cap V(y)=\emptyset$. The family $\{V(y): y\in X\setminus V\}$ is an open cover of $X\setminus V$ which is closed, therefore compact. So we can choose finitely many $y\_1,\ldots, y\_n$ such that $X\setminus V = V\_{y\_1}\cup \ldots \cup V\_{y\_n}$. Set $U:=\bigcap\_{i=1}^n U(y\_i)$. Then $x\in U$, $U$ is open and $cl(U)\subseteq V$, so $(X,\tau)$ is regular.


