# Problem of the Week #199 - March 22, 2016

Status
Not open for further replies.

#### Jameson

Staff member
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.

-----
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.