| Version 9 |
Version 8 |
| On the set $\Z^+$ of positive integers, let $a$ and $b$ be two different integers $b\neq0$ and consider the set |
On the set $\Z^+$ of positive integers, let $a$ and $b$ be two different integers $b\neq0$ and consider the set |
| \[ S(a,b)=\{a+kb\in Z^+ \colon k\in \Z\}\] |
\[ S(a,b)=\{a+kb\in Z^+ \colon k\in \Z\}\] |
| such set is the infinite arithmetic progression of positive integers with difference $b$ and containing $a$. |
such set is the infinite arithmetic progression of positive integers with difference $b$ and containing $a$. |
| The collection of all $S(a,b)$ sets is a basis for a topology on $\Z^+$. We will use a coarser topology induced by the following basis: |
The collection of all $S(a,b)$ sets is a basis for a topology on $\Z^+$. We will use a coarser topology induced by the following basis: |
| \[\mathbb{B}=\{S(a,b): \gcd(a,b)=1\}\] |
\[\mathbb{B}=\{S(a,b): \gcd(a,b)=1\}\] |
| \textbf{The collection $\B$ is basis for a topology on $\Z^+$}.\\ |
\textbf{The collection $\B$ is basis for a topology on $\Z^+$}.\\ |
| We first prove such collection is a basis. |
We first prove such collection is a basis. |
| Suppose $x\in S(a,b)\cap S(c,d)$. By Euclid's algorithm we have $S(a,b)=S(x,b)$ and $S(c,d)=S(x,d)$ and |
Suppose $x\in S(a,b)\cap S(c,d)$. By Euclid's algorithm we have $S(a,b)=S(x,b)$ and $S(c,d)=S(x,d)$ and |
| x\in S(x,bd)\subset S(x,d)\cap S(c,d) |
x\in S(x,bd)\subset S(x,d)\cap S(c,d) |
| besides, since $\gcd(x,b)=1$ and $\gcd(x,d)=1$ then $\gcd(x,bd)=1$ so $x$ and $bd$ are coprimes and $S(x,bd)\in \mathbb{B}$. This concludes the proof that $\B$ is indeed a basis for a topology on $\Z^+$. |
besides, since $\gcd(x,b)=1$ and $\gcd(x,d)=1$ then $\gcd(x,bd)=1$ so $x$ and $bd$ are coprimes and $S(x,bd)\in \mathbb{B}$. This concludes the proof that $\B$ is indeed a basis for a topology on $\Z^+$. |
| \textbf{The topology on $\Z^+$ induced by $\B$ is Hausdorff}.\\ |
\textbf{The topology on $\Z^+$ induced by $\B$ is Hausdorff}.\\ |
| Let $m,n$ integers two different integers, and let $d=|m-n|$. We can find an integer $t$ such that $t>d$ and such that |
Let $m,n$ integers two different integers, and let $d=|m-n|$. We can find an integer $t$ such that $t>d$ and such that |
| $\gcd(m,t)=\gcd(n,t)=1$. A way to accomplish this is to take any multiple of $mn$ greater than $d$ and add $1$. |
$\gcd(m,t)=\gcd(n,t)=1$. A way to accomplish this is to take any multiple of $mn$ greater than $d$ and add $1$. |
| Thus the sets $S(m,t)$ and $S(n,t)$ are disjoint since two arithmetic progression with the same common difference are either disjoint or the same, and $m\notin S(n,t)$ because the smallest difference between elementis in $S(n,t)$ is $t>|m-n|$. So the proposed sets are disjoint and $\Z^+$ becomes a Hausdorff space with the given topology. |
Thus the sets $S(m,t)$ and $S(n,t)$ are disjoint since two arithmetic progression with the same common difference are either disjoint or the same, and $m\notin S(n,t)$ because the smallest difference between elementis in $S(n,t)$ is $t>|m-n|$. So the proposed sets are disjoint and $\Z^+$ becomes a Hausdorff space with the given topology. |
| \textbf{The topology on $\Z^+$ induced by $\B$ is not completely Hausdorff}.\\ |
\textbf{The topology on $\Z^+$ induced by $\B$ is not completely Hausdorff}.\\ |
| We will use the closed-neighborhood sense for completely Hausdorff, which will also imply the topology is not completely Hausdorff in the functional sense. |
We will use the closed-neighborhood sense for completely Hausdorff, which will also imply the topology is not completely Hausdorff in the functional sense. |
| We need to determine first how do the $\overline{S(a,b)}$ look like in this topology. For sake of clarity, we will work with $S(3,5)$ first. Notice that if we had considered the former topology (where in $S(a,b)$, $a$ and $b$ didn't have to be coprime) the complement of $S(3,5)$ would have been $S(4,5)\cup S(5,5)\cup S(6,5)\cup S(7,5)$ and so $S(3,5)$ would have been closed. In general, in the finer topology, all basic sets were both open and closed. However, this is not true in our topology (for instance $S(5,5)$ is not open). |
We need to determine first how do the $\overline{S(a,b)}$ look like in this topology. For sake of clarity, we will work with $S(3,5)$ first. Notice that if we had considered the former topology (where in $S(a,b)$, $a$ and $b$ didn't have to be coprime) the complement of $S(3,5)$ would have been $S(4,5)\cup S(5,5)\cup S(6,5)\cup S(7,5)$ and so $S(3,5)$ would have been closed. In general, in the finer topology, all basic sets were both open and closed. However, this is not true in our topology (for instance $S(5,5)$ is not open). |
| First we show that $5\in \overline{S(3,5)}$. Indeed, any basic open set containing $5$ must be of the form $S(5,u)$ with $u$ coprime to $5$. |
First we show that $5\in \overline{S(3,5)}$. Indeed, any basic open set containing $5$ must be of the form $S(5,u)$ with $u$ coprime to $5$. |
| Now, the diophantine equation $ 3+5k = 5+ut $ can be written as |
Now, the diophantine equation $ 3+5k = 5+ut $ can be written as |
| \[ 5k+ut = 2\] |
\[ 5k+ut = 2\] |
| and it has solution because $\gcd(5,u)$ divides $2$. So |
and it has solution because $\gcd(5,u)$ divides $2$. So |
| \[\{3+5k \colon k\in \Z\} \cup \{ 5+ut \colon t\in \Z\}\neq \emptyset.\] |
\[\{3+5k \colon k\in \Z\} \cup \{ 5+ut \colon t\in \Z\}\neq \emptyset.\] |
| If $g$ is an element of such intersection, then we can find a positive element in such intersection (say, by adding multiples of $\mathrm{lcm}(5,u)$) and therefore |
If $g$ is an element of such intersection, then we can find a positive element in such intersection (say, by adding multiples of $\mathrm{lcm}(5,u)$) and therefore |
| \[S(3,5)\cap S(5,u)\neq \emptyset,\] |
\[S(3,5)\cap S(5,u)\neq \emptyset,\] |
| which means $5\in \overline{S(3,5)}$. |
which means $5\in \overline{S(3,5)}$. |
| So, in general $b\in \overline S(a,b)$ for any basic set $S(a,b)$ which shows that $S(a,b)\subset \overline{S(a,b)}$ (strictly). |
So, in general $b\in \overline S(a,b)$ for any basic set $S(a,b)$ which shows that $S(a,b)\subset \overline{S(a,b)}$ (strictly). |
| Now, back to the proof. If $m,n$ are two different positive integers, given two disjoint open neighborhoods $U_m$ and $U_n$ that contain them, we can always find basic sets $S(m,a)$ and $S(n,b)$ such that |
Now, back to the proof. If $m,n$ are two different positive integers, given two disjoint open neighborhoods $U_m$ and $U_n$ that contain them, we can always find basic sets $S(m,a)$ and $S(n,b)$ such that |
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The claim is that |
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g=\frac{ab}{ |
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\gcd(ab,mn) |
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\] is in both $\overline{S(m,a)}$ and $\overline{S(n,b)}$. |
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The proof is similar to what we did before. Any basic open set containing $g$ must be of the form $S(g,u)$ with $g,u$ coprime. |
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Then we consider the diophantine equation |
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$m+ax = g+uy$ which rewritten as |
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\[ ax+uy = m-g\] |
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shows that it has |
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\large To be finished |
