# Closure of a compact set in Regular space $X$

I am trying to show that closure of a compact set in a Regular space is compact, but I am hitting some hurdles as a compact set in a regular space need not be closed. This is what I started off with: If $C$ is any compact subset, let $U$ be any open neighbourhood containing $C$. Then by regularity of $X$, $\forall x \in C, \exists V_x : \overset-V_x \subseteq U$ where $V_x$ is a neighbourhood of $x$. So $\lbrace V_x:x \in C\rbrace$ is an open cover for C, which by compactness of C , simplifies to $\lbrace V_{x_i}:x_i \in C\,\,for\,i=1,...,n\rbrace$. Now I end up looking at $\cup \overset-V_{x_i}$ which should contain $\overset-C$. But I get stuck here. Should I look at the intersection rather??

Hint: Suppose that $\mathcal{U} = \{ U_i : i \in I \}$ is any cover of $\overline{C}$ by open subsets of $X$. For each $i \in I$ and $x \in U_i$ by regularity fix an open set $V_{i,x}$ such that $x \in V_{i,x} \subseteq \overline{V_{i,x}} \subseteq U_i$. Note that $$\mathcal{V} = \{ V_{i,x} : i \in I, x \in U_i \}$$ is also a cover of $\overline{C}$ by open subsets of $X$. More importantly, $\mathcal{V}$ covers the compact set $C$.

This following holds in a regular space $X$.

Lemma: Every open neighborhood $U$ of a compact set $K$ contains the closure $\overline K$ as well as a closed neighborhood of $K$.
Proof: Since $X$ is regular, every $x\in K$ has an open neighborhood $V_x$ such that $\overline {V_x}\subseteq U$. The family $\{V_x\mid x\in K\}$ is an open cover of $K$ which by compactness has a finite subcover $\{V_i,\dots,V_n\}$. Since closure commutes with finite unions, we know that $\bigcup_{i=1}^n V_i$ is a neighborhood of $K$ whose closure $\overline{\bigcup_{i=1}^n V_i}=\bigcup_{i=1}^n\overline {V_i}$ is a subset of $U$ and contains $\overline K$.

Corollary: The closure of a compact set $K$ is also compact.
Proof: An open cover of $\overline K$ is an open cover of $K$, so it has a finite subcover which by the lemma also covers $\overline K$.

• Thanks for the answer, the last statement however, I couldn't comprehend. – Vishesh Mar 25 '13 at 15:02
• @Vishesh: A cover of $\overline K$ is also a cover of $K$, which has a finite subcover. This subcover is an open set around $K$ and by the above lemma it covers $\overline K$. – Stefan Hamcke Mar 25 '13 at 15:07
• Please, all great except I don't understand why every point in a compact set has a neighbourhood whose CLOSURE in inside the open neighbourhood of the compact set. Thx – James Well Oct 10 '16 at 15:20
• Actually isn't this a bit far fetched ? Since every singleton is compact, being regular implies being Hausdorff, thus compact sets are all closed. – James Well Oct 10 '16 at 15:40
• @JamesWell. I don't understand your reasoning. Could you add more details and explain which of my statements you are using where? – Stefan Hamcke Oct 10 '16 at 17:48