# Showing $\mathbb{N}$ with the topology generated from arithmetic progression is $T_2$ but not $T_3$

I'm trying to show that the natural numbers $$\mathbb{N}=\{1,2,...\}$$, with the topology that generated from the base $$\{ (a+nb)_{n=0}^{\infty} | a,b\in \mathbb{N} ,gcd(a,b)=1\}$$ is $$T_2$$ and not $$T_3$$.

I'm having problems showing both, I'll be happy for some help here.

Thanks.

• I guess $a,b$ are fixed. Your definition suggests that both vary in $\mathbb{N}$, in which case the topology would be discrete. – Paul Frost Mar 29 at 15:52
• @PaulFrost This topology does not seem to be discrete to me: $\{x\}$ is not open for every $x\in\mathbb N$, since finite sets are not open. Also, if $a$ and $b$ are fixed, how is $\mathbb N$ open in the generated topology? Those natural numbers less than $a$ and $b$ would not be covered in any open base then. – awllower Mar 29 at 15:57
• @awllower Mabe you are right - it depends on how you understand the definition. If you understand $(a+nb)_{n=0}^{\infty}$ as an infinite set, then I agree. If you understand it as a set of singletons (as I did), then the topology is discrete. So perhaps one should write $\{ a+nb \mid n \in \mathbb{N} \}$ instead. – Paul Frost Mar 29 at 16:03
• @PaulFrost I see your point. Maybe I understood the definition wrongly. Thanks for pointing it out. – awllower Mar 29 at 16:20
• @awllower I think the OP should clarify it. – Paul Frost Mar 29 at 16:35

To see $$\mathbb{N}$$ is Hausdorff: if $$x,y \in \mathbb{N}$$ and $$x \neq y$$, then pick $$p$$ a prime that is larger than both $$x$$ and $$y$$, and then $$U_p(x)= \{x+ap: a \in \mathbb{N}\}$$ and $$U_p(y)=\{y+ap: a \in \mathbb{N}\}$$ are basic open in $$\mathbb{N}$$ in this "relatively prime integer topology" and disjoint: If $$x+a_1p = y+a_2p$$ were a common point, then $$(a_2-a_1)p = x-y$$ and $$p$$ would divide $$x-y$$ which cannot be.

To see it is not regular try to separate the closed set $$E=\{2n: n \in \mathbb{N}\}$$ and $$1 \notin E$$ by disjoint open sets $$U$$ and $$V$$ respectively. As $$1 \in V$$ and $$V$$ is open, for some $$e$$: $$U_e(1) \subseteq V$$ with $$e$$ even (or else $$U_e(1)$$ already intersects $$E$$ and thus $$U$$), but then $$e \in E$$ so $$e \in U_a(b) \subseteq U$$ for some $$a,b$$ with $$\gcd(a,b)=1$$. So $$e=an_0+b$$ for some $$n_0 \in \mathbb{N}$$ and we know that $$\gcd(a,e)=1$$ as well. Then check that $$U_a(b)$$ and $$U_e(1)$$ intersect, which is a contradiction with the supposed disjointness of $$U$$ and $$V$$.

Alternatively, show that $$bd \in \overline{U_b(a)} \cap \overline{U_d(c)}$$ for a pair of basic open subsets. This shows that the space is also not Urysohn and is connected.

• I tried to show that $E$ and $1$ cannot be separated, but to no avail. Could you explain more why this is so? Thanks in advance. – awllower Mar 30 at 1:56
• @awllower I expanded that part somewhat. – Henno Brandsma Mar 30 at 6:30
• It is quite an enlightening proof. Thanks! – awllower Mar 30 at 13:44
• @awllower glad you like it! – Henno Brandsma Mar 30 at 13:46
• Is it trivial why $U_a(b)\cap U_e(1)\not=\emptyset$? I can't seem to prove it with Bezout Lemma or the likes of it. @HennoBrandsma – The way of life Apr 3 at 20:47

To show that this is $$T_2$$, take two points $$x\ne y\in\mathbb N$$. Take $$a\in\mathbb N$$ such that $$x\not\equiv y\pmod a$$ and co-prime to $$x$$, for example, some prime larger than $$x$$ and $$y$$, and then take $$b\in\mathbb N$$ such that $$\gcd(a,b)\not\mid(x-y)$$, so that $$\{x+na\mid n\in\mathbb N\}\cap\{y+mb\mid b\in\mathbb N\}=\emptyset$$; for example, we might take $$b=a$$, which is prime to $$y$$ as well. Then the two points $$x$$ and $$y$$ are separated by distinct neighbourhoods, which shows the space is $$T_2$$.
I do not know how to show it is not $$T_3$$. What makes you think that this space is not regular?