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Problem of the Week #199 - March 22, 2016

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Jameson

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Jan 26, 2012
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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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 

Euge

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Jun 20, 2014
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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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