Welcome to our community

Be a part of something great, join today!

Problem Of The Week # 352 - Jul 23, 2019

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,925
Here is this week's POTW:

-----
Suppose $X$ is compact Hausdorff. If $S$ is a subset of $X$ and $O$ is an open set in $X$ with $\overline{S} \subset O$, prove that there is another open set $V$ in $X$ with $\overline{S} \subset V \subset \overline{V} \subset O$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,925
This week's problem was solved by Opalg and Olinguito . You can read Olinguito 's solution below.


We need two results about compact Hausdorff spaces. First, every closed subset of a compact space is compact.

Let $C$ be a closed subset of the compact space $X$ and let ${\frak U}=\{U_\lambda:\lambda\in\Lambda\}$ be an open cover of $C$. Then ${\frak U}\cup\{C^c\}$ (where $C^c=X\setminus C$) is an open cover of $X$. As $X$ is compact, there is a finite subcover $\{U_1,\ldots,U_n,C^c\}$. Then $\{U_1,\ldots,U_n\}$ is an open cover of $C$ and a finite subcover of $\frak U$, and so $C$ is compact.

The second result we need is that any two disjoint compact subsets of a Hausdorff space can be separated by disjoint open sets each containing one of the compact subsets.

Let $A,B$ be disjoint compact subsets of the Hausdorff space $X$. We want to show that there are open sets $U,W$ such that $U\cap W=\emptyset$ and $A\subseteq U$ and $B\subseteq W$. If $A$ and $B$ are both empty, we can take $U=W=\emptyset$; if $A$ is empty and $B$ non-empty, take $U=\emptyset$ and $W=X$; and if $A$ is non-empty and $B$ is empty, take $U=X$ and $W=\emptyset$. So assume that $A$ and $B$ are both non-empty.

Fix $a\in A$. Then, by Hausdorff, for each $b\in B$, there exist disjoint open sets $U_{a,b},W_{a,b}$ such that $a\in U_{a,b},b\in W_{a,b}$. The collection $\{W_{a,b}:b\in B\}$ is an open cover of $B$; by compactness of $B$, there is a finite subcover $\{W_{a,b_1},\ldots,W_{a,b_n}\}$. Set $\displaystyle W_a=\bigcup_{i=1}^nW_{a,b_i}$ and $\displaystyle U_a=\bigcap_{i=1}^nU_{a,b_i}$; thus $a\in U_a$, $B\subseteq W_a$, and $U_a,W_a$ are open and disjoint. Now, as we let $a$ range over $A$, the collection $\{U_a:a\in A\}$ is an open cover of $A$; by compactness of $A$, there is a finite subcover $\{U_{a_1},\ldots,U_{a_m}\}$. Set $\displaystyle U=\bigcup_{j=1}^mU_{a_j}$ and $\displaystyle W=\bigcap_{j=1}^mW_{a_j}$. Then $A\subseteq U,B\subseteq W$, both $U$ and $W$ are open and $U\cap W=\emptyset$ as required.

Now, to the problem itself. The subsets $\overline S$ and $O^c=X\setminus O$ are disjoint and closed, hence compact; thus there are disjoint open sets $V,Y$ such that $\overline S\subseteq V$ and $O^c\subseteq Y$, i.e. $Y^c=X\setminus Y\subseteq O$. As $V$ and $Y$ are disjoint, $V\subseteq Y^c$; also $Y^c$ is closed and so $\overline V\subseteq Y^c$ since $\overline V$ is the smallest closed subset of $X$ containing $V$. Thus we have
$$\overline S\ \subseteq\ V\subseteq\ \overline V\ \subseteq O$$
as required.
 
Status
Not open for further replies.