Compact sets in regular topological spaces Consider a topological space $ (X, \mathcal{T})$. Suppose $X$ is regular. If $A \subset X$  is compact and $U \in \mathcal{T}$ is such that $A \subset U$. Show that there exists a $V \in \mathcal{T}$ such that $A \subset V \subset \bar V \subset U$.
 A: First note that if $X$ is regular, and $x \in O$, $O$ open, there is an open set $V$ such that $x \in V \subset \overline{V} \subset O$. This property (for all $x$,$O$) is in fact an equivalent reformulation of regularity. Note that what is asked to show, is that we can replace a point $x$ by a compact set ("compact sets behave like points" we were told in university).  
To show the first fact: apply regularity to $x$ and the disjoint closed set $X \setminus O$, to obtain disjoint open sets $U$, $V$ such that $x \in U$, $X \setminus O \subset V$.
Then $x \in U \subset X \setminus V \subset O$ (the former as $U$ and $V$ are disjoint, and the latter follows from $X \setminus O \subset V$). Now as $X\setminus V$ is closed, $\overline{U} \subset \overline{X\setminus V} = X\setminus V \subset O$, showing that $U$ is as required. (I don't need the reverse, but the idea is very similar).
Now, if $A$ is compact and $A \subset U$ with $U$ open, for every point $a \in A$ we pick $V_a$ such that $a \in V_a \subset \overline{V_a} \subset U $ by the above.
Finitely many $V_a$, say $V_{a_1},\ldots,V_{a_k}$ cover $A$ as well, by compactness of $A$.
Then define $V$ as their finite (!) union. As they cover, $A \subset V$ and 
$$\overline{V} = \overline{\cup_{i=1}^k V_{a_i}} = \cup_{i=1}^k\overline{V_{a_i}} \subset O,$$ using that finite unions and closure commute and all $V_a$ are subsets of $O$, hence so is their union. 
A: *

*As $ U $ is an open set, we see that $ X \setminus U $ is a closed set.

*We have $ a \notin X \setminus U $ for each $ a \in A $. Then as $ (X,\mathcal{T}) $ is a regular topological space, we can find


*

*an open neighborhood $ U_{a} $ of $ X \setminus U $ and

*an open neighborhood $ V_{a} $ of $ a $
such that $ U_{a} \cap V_{a} = \varnothing $.

*Observe that $ \{ V_{a} \in \mathcal{T} ~|~ a \in A \} $ is an open cover of $ A $. Then as $ A $ is a compact set, we can find $ a_{1},\ldots,a_{n} \in A $ such that $ \{ V_{a_{k}} \}_{k=1}^{n} $ is a finite sub-cover.

*Clearly, $ \displaystyle \bigcap_{k=1}^{n} U_{a_{k}} $ is an open set. As $ X \setminus U \subseteq U_{a} $ for each $ a \in A $, we see that $ \displaystyle X \setminus U \subseteq \bigcap_{k=1}^{n} U_{a_{k}} $.

*Consequently, $ \displaystyle X \setminus \bigcap_{k=1}^{n} U_{a_{k}} \subseteq X \setminus (X \setminus U) = U $, which means that $ \displaystyle X \setminus \bigcap_{k=1}^{n} U_{a_{k}} $ is a closed set contained in $ U $.

*Observe that $ \displaystyle V := \bigcup_{k=1}^{n} V_{a_{k}} $ is disjoint from $ \displaystyle \bigcap_{k=1}^{n} U_{a_{k}} $. If this were not the case, then there would exist an $ x \in X $ such that $ \displaystyle x \in \bigcup_{k=1}^{n} V_{a_{k}} $ and $ \displaystyle x \in \bigcap_{k=1}^{n} U_{a_{k}} $. Hence, $ x \in V_{a_{k}} $ for some $ k \in \{ 1,\ldots,n \} $, and as $ x \in U_{a_{k}} $ also, we would obtain $ U_{a_{k}} \cap V_{a_{k}} \neq \varnothing $. This contradicts Step (2).

*Consequently, $ \displaystyle V \subseteq X \setminus \bigcap_{k=1}^{n} U_{a_{k}} $.

*As $ \displaystyle X \setminus \bigcap_{k=1}^{n} U_{a_{k}} $ is a closed set contained in $ U $ (by Step (5)), we get $ \displaystyle \overline{V} \subseteq X \setminus \bigcap_{k=1}^{n} U_{a_{k}} \subseteq U $.

*We therefore conclude that $ A \subseteq V \subseteq \overline{V} \subseteq U $, where the first inclusion comes from Step (3).
