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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.

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  • $\begingroup$ I guess $a,b$ are fixed. Your definition suggests that both vary in $\mathbb{N}$, in which case the topology would be discrete. $\endgroup$ – Paul Frost Mar 29 at 15:52
  • $\begingroup$ @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. $\endgroup$ – awllower Mar 29 at 15:57
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    $\begingroup$ @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. $\endgroup$ – Paul Frost Mar 29 at 16:03
  • $\begingroup$ @PaulFrost I see your point. Maybe I understood the definition wrongly. Thanks for pointing it out. $\endgroup$ – awllower Mar 29 at 16:20
  • $\begingroup$ @awllower I think the OP should clarify it. $\endgroup$ – Paul Frost Mar 29 at 16:35
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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.

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  • $\begingroup$ 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. $\endgroup$ – awllower Mar 30 at 1:56
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    $\begingroup$ @awllower I expanded that part somewhat. $\endgroup$ – Henno Brandsma Mar 30 at 6:30
  • $\begingroup$ It is quite an enlightening proof. Thanks! $\endgroup$ – awllower Mar 30 at 13:44
  • $\begingroup$ @awllower glad you like it! $\endgroup$ – Henno Brandsma Mar 30 at 13:46
  • $\begingroup$ 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 $\endgroup$ – The way of life Apr 3 at 20:47
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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 $T_2$.


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


Hope this helps.

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    $\begingroup$ Because it's a famous example of a Hausdorff connected countable space, and these cannot be regular. $\endgroup$ – Henno Brandsma Mar 30 at 8:12

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