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??
$\begingroup$
$\endgroup$
$\begingroup$
$\endgroup$
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 $C$.
answered Mar 25 '13 at 13:18
$\begingroup$
$\endgroup$
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 $\overline K$.
answered Mar 25 '13 at 14:30
-
$\begingroup$ Thanks for the answer, the last statement however, I couldn't comprehend. $\endgroup$ – Vishesh Mar 25 '13 at 15:02
-
$\begingroup$ @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 $\overline K$. $\endgroup$ – Stefan Hamcke Mar 25 '13 at 15:07
-
$\begingroup$ 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 $\endgroup$ – James Well Oct 10 '16 at 15:20
-
$\begingroup$ Actually isn't this a bit far fetched ? Since every singleton is compact, being regular implies being Hausdorff, thus compact sets are all closed. $\endgroup$ – James Well Oct 10 '16 at 15:40
-
$\begingroup$ @JamesWell. I don't understand your reasoning. Could you add more details and explain which of my statements you are using where? $\endgroup$ – Stefan Hamcke Oct 10 '16 at 17:48
