Can compact Hausdorff sets be expanded by open sets?

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In summary, a compact Hausdorff set is a topological space that is both compact and Hausdorff. Compactness means that every open cover of the set has a finite subcover, while Hausdorffness means that any two distinct points in the set have disjoint neighborhoods. Yes, compact Hausdorff sets can be expanded by open sets, allowing for more flexibility and precision in topological constructions and proofs. Compact Hausdorff sets play a significant role in topology, as a generalization of compactness and with many important applications. The expansion of compact Hausdorff sets is not unique, but the existence of at least one open set that expands the compact set is guaranteed by the definition of compactness
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Euge
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Here is this week's POTW:

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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 $\overline{S} \subset V \subset \overline{V} \subset O$.

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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.
 

FAQ: Can compact Hausdorff sets be expanded by open sets?

What is a compact Hausdorff set?

A compact Hausdorff set is a topological space in which every open cover has a finite subcover, and every pair of distinct points in the space has disjoint open neighborhoods.

What does it mean for a compact Hausdorff set to be expanded by open sets?

Expanding a compact Hausdorff set by open sets means finding a larger topological space that contains the original set as a subspace, and where the original set is still compact and Hausdorff.

Can all compact Hausdorff sets be expanded by open sets?

No, not all compact Hausdorff sets can be expanded by open sets. This is because not all topological spaces can be made compact and Hausdorff by adding open sets.

Is there a limit to how much a compact Hausdorff set can be expanded by open sets?

Yes, there is a limit to how much a compact Hausdorff set can be expanded by open sets. This limit is determined by the compactness and Hausdorffness of the original set, and the properties of the open sets being added.

What are some applications of expanding compact Hausdorff sets by open sets?

Expanding compact Hausdorff sets by open sets has applications in topology, functional analysis, and algebraic geometry. It allows for the study of larger topological spaces while still retaining the properties of compactness and Hausdorffness, which are important in many mathematical fields.

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