Proving that a compact space is regular

Question

I am studying a book and I am stagnating on a what should be a straightfoward proof:

Show that if $X$ is compact, $V\subset X$ is open and $x\in V$, then there exists an open set $U$ in $X$ with $x\in U\subset \bar U\subset V$

I don't know how to find the appropriate set $U$. I am guessing you need to do something like taking the intersection with $V$ of a finite subcover of X and then show that its closure is contained in $V$ ...

Could someone nudge me in the right direction?

Answer 1

Hint: For any $y\in X\setminus V$ we have $x\neq y$, there exists $W_y$ a neighborhood of $y$ and $U_y$ a neighborhood of $x$ s.t $U_y\cap W_y=\emptyset$.

Now $W_y (y\in X\setminus V)$ is a cover of $X\setminus V$ which is compact (closed in Hausdorff compact space ), Let $W_{y_1},\ldots,W_{y_n}$ be a finite cover of $X\setminus V$.

Try with $U=\cap_{i=1}^nU_{y_i}$ this is a neighborhood of $x$.

Answer 2

So we need to show that $X$ is regular (and as other users importantly pointed out, we need the Hausdorff property). Let $A \subset X$ be a closed set and fix $x \in X\backslash A$. Now let $\{V_a \}_{a \in A}$ and $\{U_a \}_{a \in A}$ be collections of open sets generated by the Hausdorff property such that $x \in V_a$, $a \in U_a$ and $V_a \cap U_a = \emptyset$. Note that $\{U_a \}_{a \in A}$ is an open cover of $A$. Since closed subsets of compact spaces are compact, we know there exists a finite subcover, say $\{U_{a_i} \}_{i=1}^n$. Similarly, we can extract finitely many of the $V_a$'s from our arbitrary collection to create the collection $\{V_{a_i} \}_{i=1}^n$ where $x \in V_{a_{i}}$, $a \in U_{a_i}$ and $V_{a_{i}} \cap U_{a_{i}}= \emptyset$.

Now see if you can take those two finite collections to find open sets $U,V$ with $x \in V$, $A\subset U$ and $U \cap V = \emptyset$. Once done, we will have shown that $X$ is regular. Then, you can site the fact that this is an equivalent statement to what you are trying to show.