# A topology over $\Bbb N$ based on convergence of series.

Define $$\tau=\{U\subseteq \Bbb N:U\in\{\Bbb N,\emptyset\}\vee\sum_{n\notin U}n^{-1}<\infty\}$$. In other words, a subset of $$\Bbb N$$ is closed iff it is $$\Bbb N$$ or the sum of the inverses of its elements converges. I have proved so far (without much effort):

• $$\tau$$ is a topology over $$\Bbb N$$.
• Singletons are closed.
• A set is compact iff it is finite.
• A sequence converges iff it is eventually constant (this is the "hardest" fact I proved about this topology; not too hard, though).
• The space is not Hausdorff, but it is connected.

I'd like to know if there is more than at first sight in this topology, that is, if it has deeper properties and if it has some use in number theory or something at all.

• More generally, this can be done with any measure on $\mathbb N$, where a set $U$ is open if $\mu(U)<+\infty.$ – Thomas Andrews May 13 at 13:09
• Doh! You added what I was about to inquire about (connectedness). You might want to check local connectedness, too. – Cameron Williams May 13 at 13:17
• I have the same now... But I already added a proof. – RMWGNE96 May 13 at 13:18

The topological space $$(\mathbb{N},\tau)$$ is connected. Proof. suppose that $$\mathbb{N}=A\sqcup B$$ with $$A,B\neq\varnothing$$ and $$A,B\in\tau$$. By definition of $$\tau$$, we simultaneously have $$\sum_{n\in A}n^{-1}<\infty,\,\sum_{n\in B}n^{-1}<\infty$$ which is clearly impossible, because it is well known that the harmonic series $$\sum_{n\in\mathbb{N}}n^{-1}$$ diverges.