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

Question

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.

Answer 1

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.

Answer 2

This is a partial answer.

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 $T_2$.

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I do not know how to show it is not $T_3$. What makes you think that this space is not regular?

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Hope this helps.