Welcome to our community

Be a part of something great, join today!

Problem of the Week #199 - March 22, 2016

Not open for further replies.
  • Thread starter
  • Admin
  • #1


Staff member
Jan 26, 2012
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.

Remember to read the POTW submission guidelines to find out how to submit your answers!


MHB Global Moderator
Staff member
Jun 20, 2014
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.
Not open for further replies.