Is Every Collection of Disjoint Open Sets Countable in a Second Countable Space?

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    2016
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SUMMARY

In a second countable topological space, every collection of disjoint open sets is indeed countable. This conclusion is established through the properties of second countable spaces, which possess a countable base for their topology. The proof relies on the definition of second countability and the nature of open sets within these spaces. The discussion emphasizes the importance of understanding topological concepts to grasp this theorem fully.

PREREQUISITES
  • Understanding of second countable topological spaces
  • Familiarity with open sets in topology
  • Knowledge of countable and uncountable sets
  • Basic principles of set theory
NEXT STEPS
  • Study the properties of second countable spaces in detail
  • Explore the concept of bases in topology
  • Investigate the implications of disjoint open sets in various topological contexts
  • Learn about different types of topological spaces, such as first countable and compact spaces
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Mathematicians, students of topology, and anyone interested in advanced set theory and its applications in mathematical analysis.

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Euge is feeling under the weather so I'm filling in for this week. Huge thanks to him for his hard work keeping on top of these every week!

Problem: Prove that in a second countable topological space, every collection of disjoint open sets is countable.

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No one answered this week's problem. You can read my solution below.

Let $X$ be a second countable space, and let $\{U_\alpha\}_{\alpha\in J}$ be a collection of disjoint, open subsets of $X$. Let $\{B_n\}_{n\in \Bbb N}$ be a countable basis for $X$. For each $\alpha\in J$, there exists $n\in \Bbb N$ such that $B_n\subset U_{\alpha}$. So there is a function $f : J\to \Bbb N$ such that $f(\alpha) = \min\{n\in \Bbb N: B_n\subset U_{\alpha}\}$, for all $\alpha\in J$. If $f(\beta) = f(\gamma)$, then $B_{f(\beta)} \subset B_\beta$ and $B_{f(\beta)}\subset U_\gamma$; since the $U's$ are disjoint, this can only occur if $\beta = \gamma$. Thus, $f$ is an injection from $J$ into $\Bbb N$. Hence, $J$ is countable and consequently the collection is countable.
 

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