9
$\begingroup$

For each positive integer $n$, let $T_n = \{\frac{a}{n} \mid a \in \mathbb{Z}\}$.

What are $\bigcup_{n \in \mathbb{N}} T_n$ and $\bigcap_{n \in \mathbb{N}} T_n$?

I'm pretty sure the first one is just $\mathbb{Q}$, the rationals. Since the set will have all possible numerators over all possible denominators. The second I'm not so sure of. It's certainly not empty, since $1/2 = 2/4$ so $T_2$ and $T_4$ have non-empty intersection. I am currently leaning towards this set also being $\mathbb{Q}$, since we will get every fraction here as well: if $a/b$ is in $T_b$, then it is also in $T_{2b}$ as $2a/2b$.

$\endgroup$
4
  • 2
    $\begingroup$ You may think about questions like is 1/3 in T_2? Note that the intersection is over all n, checking T_2 and T_4 have nonempty intersection is not enough to make sure the intersection over all n is also nonempty. $\endgroup$
    – Ken Leung
    May 8, 2020 at 7:03
  • 1
    $\begingroup$ @KenLeung Then it must be the set of integers, right? In each $T_n$ we will have all the integers, but as you noted we cannot have all rationals since the relatively prime denominators will have empty intersection except at the integers. $\endgroup$
    – user783401
    May 8, 2020 at 7:08
  • 1
    $\begingroup$ Eulerian, you can write it as an answer, yourself. $\endgroup$
    – Aravind
    May 8, 2020 at 7:10
  • 5
    $\begingroup$ Hint $ $ A fraction writable with coprime denominators $\,p,q\,$ is an integer, since its least denominator $d$ ("order") divides the coprimes $\,p,q\,$ so $\,d=1.\,$ It is an additive form of $\,a^p = 1 = a^q\,\Rightarrow\, a = 1\,$ by $\,a\,$ has order $\,d=1,$ since $\,d\,$ divides coprimes $\,p,q.\,$ The least denominator is the fraction's order in $\Bbb Q/\Bbb Z.\,$ See here for further discussion of this basic result and generalizations. $\endgroup$ May 9, 2020 at 15:26

2 Answers 2

10
$\begingroup$

You are correct that the union is just $\mathbb{Q}$. As you noted, every pair $(a,n)$ appears in the union and so we have all the rationals. On the other hand, as noted in a comment, the intersection cannot be all the rationals. Clearly the integers are a subset the intersection since every $(mn)/n$ will appear in each $T_n$. Now suppose some rational $p/q$ with $gcd(p,q)=1$ and $q \neq 1$ is in the intersection. Then $p/q$ is not in $T_p$, since $p$ and $q$ are relatively prime (there is no integer $x$ so that $p^2 = xq$ is $p$ and $q$ are relatively prime). Thus, the intersection is simply the integers.

$\endgroup$
4
  • 2
    $\begingroup$ More conceptually, a fraction writable with coprime denominators is an integer - see my comment on the question. $\endgroup$ May 9, 2020 at 15:10
  • $\begingroup$ About the union, 0/0 and 1/0 are in the union but not in Q. $\endgroup$ May 21, 2020 at 8:14
  • $\begingroup$ @AlbertHendriks no that is not true, the union is only over natural numbers $\endgroup$
    – SescoMath
    May 21, 2020 at 8:26
  • $\begingroup$ @PhysMath Ah, I thought consensus was to include 0 in N, but Google shows me it's not. $\endgroup$ May 21, 2020 at 9:10
7
$\begingroup$

You are correct that $\bigcup_{n\in\mathbb{N}}T_n=\mathbb{Q}$. That's simply because any $q\in\mathbb{Q}$ is of the form $q=\frac{a}{b}$ for some $a\in\mathbb{Z}$ and $b\in\mathbb{N}$ and so $q\in T_b$.

You are wrong that $\bigcap_{n\in\mathbb{N}}T_n=\mathbb{Q}$. This has no chance of happening since $\bigcap_{n\in\mathbb{N}}T_n\subseteq T_m$ for any $m$ and each $T_m$ is a proper subset of $\mathbb{Q}$.

So first note that if $n\in\mathbb{Z}$ then $n\in T_b$ for any $b\in\mathbb{N}$. That's because $n=\frac{bn}{b}$ regardless of $b$. Meaning $\mathbb{Z}\subseteq T_b$ for any $b\in\mathbb{N}$ and so $\mathbb{Z}\subseteq\bigcap_{n\in\mathbb{N}}T_n$.

We will show that $\bigcap_{n\in\mathbb{N}}T_n\subseteq\mathbb{Z}$. So assume that $q\in\mathbb{Q}\backslash\mathbb{Z}$, i.e. $q=\frac{a}{b}$ for some $a\in\mathbb{Z}\backslash\{0\}$ and $b>1$ relatively prime. Obviously $q\in T_b$. Take a prime number $p$ not dividing $b$. It is enough to show that $q\not\in T_p$. Indeed, $\frac{a}{b}=\frac{x}{p}$ implies $ap=xb$ which cannot hold since $b>1$ is relatively prime with $a$ and with $p$.

This shows that $\bigcap_{n\in\mathbb{N}}T_n =\mathbb{Z}$.

$\endgroup$
3
  • 2
    $\begingroup$ If $a$ is $2b$ then $a/b = 2$ now just take $a' = 2b'$. In your example, instead of considering $1/2$, notice $4/2 = 6/3$. Your answer is wrong. $\endgroup$
    – user783401
    May 8, 2020 at 7:24
  • 3
    $\begingroup$ @Eulerian ops, fixed. $\endgroup$
    – freakish
    May 8, 2020 at 7:37
  • 5
    $\begingroup$ More conceptually, a fraction writable with coprime denominators is an integer - see my comment on the question. fyi: there is a meta question about this answer (thta's how I found the question). $\endgroup$ May 9, 2020 at 15:11

You must log in to answer this question.