9

3

More specificaly, is there a haussdorf non-discrete topology on $\mathbb{Z}$ that makes it a topological group with the usual addition operation?

flag
Of course in any of these cases, once you have your topology on Z, then it is useful to contemplate the completion. – Gerald Edgar Mar 17 2010 at 0:04

2 Answers

12

Yes. Take, for example, the subgroups $p^k\mathbb{Z}$, for $k>0$ and a fixed prime $p$, as a basis of neighborhoods of the identity.

link|flag
Sorry, I had those examples and related ones in mind and wanted to prove they were the only ones. I'll think of a well-posed question and post it again. – Cristos A. Ruiz Mar 16 2010 at 22:48
8

There is a topology on $\mathbb Z$ which has the set of all arithmetic sequences as a basis. It shows up in the topological proof of the infinitude of primes, cf. [H. Fürstenberg, On the Infinitude of Primes, Amer. Math. Monthly 62 (1955), 353]

link|flag
This is a nice topology. Another description says that an integer is close to zero iff it is divisible by a large factorial. – Gerald Edgar Mar 17 2010 at 0:03
3 
This is a very natural topology, if I've got it right: I think it's the topology on $\mathbf{Z}$ induced from its inclusion into its completion (give the completion the profinite topology). Thought about this way, you see that an element is close to 0 iff it's in a lot of finite index subgroups. – Kevin Buzzard Mar 17 2010 at 7:10

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.