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I want to prove that the set of integers are Hausdorff.

Attempt : Suppose $a, b \in \mathbb{Z}$ where $a \neq b$. Then its pretty clear that if you put an open ball around each one, they are disjoint. One has to be careful though if the integers are consecutive. Consider $3$ and $4$. In a case like this, we would have to choose an open ball sufficiently small enough so that it did not intersect another open ball.

So I understand the idea as to why the integers are Hausdorff. I just feel uncomfortable with my argument as it is not very long and seems to get to the point a little to vaguely. Anyone have a good proof that ties in my ideas but is more concrete and detailed?

Thanks a lot!

user133183
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    Just choose a ball of radius $1/3$ if you want. – Najib Idrissi Mar 05 '14 at 21:08
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    It might help to make an inventory of the subsets of $\mathbb Z$ that are open, since the Hausdorff property involves being able to do something using open sets. – Dave L. Renfro Mar 05 '14 at 21:11
  • The answer does depend on the chosen topology. If you assume the topology that $\Bbb{Z}$ inherits from the usual metric on $\Bbb{R}$, then yours is the way to go, or, rather, the way nik refined it. The claim is actually false for some topologies. – Jyrki Lahtonen Mar 05 '14 at 21:15
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    If you're working with the relative topology on $\mathbb Z$ induced by the usual Euclidean topology on $\mathbb R$ (which is Hausdorff), then you can use the following general fact: If $(X,\tau)$ is a Hausdorff topological space and $Y\subseteq X$ is non-empty, then the relative topology on $Y$ induced by $\tau$ is Hausdorff. This is actually quite easy to prove and gives you a general result. – triple_sec Mar 05 '14 at 21:21

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It is in fact that easy, and you can even do better. You do not even have to take two points $m,n$ in $\Bbb Z$ and choose respective radii such that the balls are disjoint. You can also just take one number $n\in\Bbb Z$, and find a universal radius which works for every other interger. Consider the ball $B_1(n)$. Since every other integer has a distance of at least $1$ from $n$, this ball will be just the set $B_1(n)=\{n\}$.

Now for $k,l\in\Bbb Z$, the disjoint neighborhoods are $$k\in\{k\},\quad l\in\{l\},\quad \{k\}\cap\{l\}=\emptyset$$

What we actually did here is to show that the intergers $\Bbb Z$ have the discrete topology under the metric $d(k,l)=|k-l|$. And every discrete space is Hausdorff, because the singletons $\{x\}$ are open in the discrete topology.

Stefan Hamcke
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  • Thank you for your great advice as well as the other comments given. I think I can use a little bit of everyones help to make my proof sound better and still simplistic. – user133183 Mar 05 '14 at 21:56
  • @user133183: You are welcome. If you like an answer, you can upvote it by clicking on the arrow at the left. You can also accept the answer which you find most helpful, by clicking at the symbol at the left. I've seen at your profile that you haven't voted on nor accepted any answer to a question of yours so far, so I'd like to point you out to this feature, in case you didn't know that it exists. Sincerely, Stefan. – Stefan Hamcke Mar 14 '14 at 16:59