Countable Dense Open Sets in Compact Hausdorff Spaces?

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SUMMARY

In a compact Hausdorff space, any countable collection of dense open sets possesses a dense intersection. This conclusion is established through the application of the Baire Category Theorem, which asserts that in a complete metric space, the intersection of countably many dense open sets is dense. The discussion highlights the importance of compactness and the Hausdorff property in ensuring the validity of this theorem. Opalg provided a correct solution that exemplifies these principles.

PREREQUISITES
  • Understanding of compact Hausdorff spaces
  • Familiarity with the Baire Category Theorem
  • Knowledge of dense sets and their properties
  • Basic concepts of topology
NEXT STEPS
  • Study the Baire Category Theorem in depth
  • Explore properties of compact Hausdorff spaces
  • Investigate examples of dense sets in various topological spaces
  • Learn about the implications of dense intersections in functional analysis
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Mathematicians, topology students, and anyone interested in advanced concepts of set theory and topology will benefit from this discussion.

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Here is this week's POTW:

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Show that in a compact Hausdorff space, any countable collection of dense open sets has dense intersection.

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Congratulations to Opalg for his correct solution, which you can read below:
This is one form of the Baire category theorem. The proof relies on the fact that a compact Hausdorff space $X$ is regular, meaning that if $U$ is an open subset of $X$ and $x\in U$ then there is an open subset $V$ such that $x\in V\subseteq \overline{V} \subseteq U$, where $ \overline{V}$ is the closure of $V$.

Suppose that $\{U_n\}_{n\in\Bbb N}$ is a countable collection of dense open subsets of $X$. Let $W$ be an open subset of $X$. Then we want to show that $$W\cap \bigcap_{n\in\Bbb N}U_n \ne\emptyset$$.

Since $U_1$ is dense in $X$, $W\cap U_1\ne\emptyset$. So there exists a nonempty open set $V_1$ such that $\overline V_1 \subseteq W\cap U_1$.

Since $U_2$ is dense in $X$, $U_2\cap V_1\ne\emptyset$. So there exists a nonempty open set $V_2$ such that $\overline V_2 \subseteq V_1\cap U_2$.

Continuing in this way, we get a decreasing sequence of nonempty open sets $\{V_n\}_{n\in\Bbb N}$ such that $$\overline V_n \subseteq W\cap \bigcap_{k=1}^n U_k$$. The sets $\overline V_n$ are compact and satisfy the finite intersection property. So the intersection $$\bigcap_{n\in\Bbb N}\overline V_n$$ is nonempty and contained in $$W\cap \bigcap_{n\in\Bbb N}U_n$$. That completes the proof.
 

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