Is it true that if I add the Hausdorffness condition to any topology on $\mathbb{N}$, then it is a $\sigma$- algebra on $\mathbb{N}$? Once I have tried to prove this, I think that compactness is also a necessary condition.
Asked
Active
Viewed 447 times
0
-
You need the complement of every open set to be open, and the only Hausdorff topology with that property is the discrete one. – Zhen Lin Apr 05 '14 at 09:27
-
That means it is true iff my topology is discrete? – user110706 Apr 05 '14 at 09:29
-
1Not every Hausdorff topology on a countable set is discrete. See http://math.stackexchange.com/questions/16670/are-there-any-countable-hausdorff-connected-spaces – William Apr 05 '14 at 09:42
1 Answers
0
Note that if $(\Bbb N,\tau)$ is a Hausdorff topology, then $\{n\}$ is closed for every $n\in\Bbb N$ and $\Bbb N\setminus\{n\}$ is open.
If $\tau$ is a $\sigma$-algebra it means that the complement of an open set is open. Therefore $\{n\}$ is open for every $n\in\Bbb N$. So $\tau$ is the discrete topology.
Asaf Karagila
- 405,794