9
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Of course in any of these cases, once you have your topology on Z, then it is useful to contemplate the completion. $\endgroup$ Mar 17, 2010 at 0:04
  • 1
    $\begingroup$ How about p-adic topology? $\endgroup$ May 11, 2014 at 11:36

3 Answers 3

14
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Mar 16, 2010 at 22:48
15
$\begingroup$

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]

$\endgroup$
2
  • $\begingroup$ This is a nice topology. Another description says that an integer is close to zero iff it is divisible by a large factorial. $\endgroup$ Mar 17, 2010 at 0:03
  • 4
    $\begingroup$ 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. $\endgroup$ Mar 17, 2010 at 7:10
4
$\begingroup$

There is a huge number of such topologies. Let $G$ be any discrete abelian group. By $\widehat G$ we denote the family (in fact, a group) of its characters, that is of homomorphisms from $G$ to the unit circle group $\Bbb T$. By Pontrjagin duality theory, the family $\widehat G$ separate points of $G$ (that is for each two distinct elements $g,h\in G$ there exists a character $\chi\in\widehat G$ such that $\chi(a)\ne\chi(b)$). Thus a diagonal product $\Delta\{\chi:\chi\in\widehat G\}:G\to\prod \{\Bbb T_\chi: \chi\in\widehat G\}$ is an injective homomorphism of $G$ into a compact group. In endows $G$ with a (totally bounded) group topology (called Bohr topology on the group $G$). In particular, if $G$ is infinite then its Bohr topology is indiscrete. On the other hand, according to this answer, $G$ admits continuum many (pairwise transversal) group topologies, none of which is totally bounded. The case of $G=\Bbb Z$ is espectially simple because it admits an injective homomorphism into each topological group with an element of infinite order. As a more concrete example, in this MSE answer I proposed for each $\kappa\le\frak c$ an injective homomorphism of the group $\Bbb Z$ onto a dense subroup of a compact group $\Bbb T^k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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