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

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    2016
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In a second countable topological space, it is established that every collection of disjoint open sets must be countable. This is due to the fact that second countable spaces have a countable base, which limits the number of disjoint open sets that can exist simultaneously. The discussion highlights the importance of this property in topology and its implications for understanding the structure of such spaces. No responses were provided to the problem posed this week, indicating a potential gap in engagement with the topic. The solution to the problem is available for review below.
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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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