0
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.