# If $T_n = \{\frac{a}{n} \mid a \in \mathbb{Z}\}$, then what are $\bigcup_{n\in\mathbb{N}}T_n$ and $\bigcap_{n\in\mathbb{N}}T_n$?

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 $$2a/2b$$.

• 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. May 8, 2020 at 7:03
• @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.
– user783401
May 8, 2020 at 7:08
• Eulerian, you can write it as an answer, yourself. May 8, 2020 at 7:10
• 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. May 9, 2020 at 15:26

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.

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

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

• 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.
– user783401
May 8, 2020 at 7:24
• @Eulerian ops, fixed. May 8, 2020 at 7:37
• 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). May 9, 2020 at 15:11