Answer to Question #155676 in Calculus for Vishal

Question

Let a,b ∈ R with a < b. Let Ta,b ⊆ Q be the subset defined by Ta,b :=r ∈ Q : a<r<b. Prove that the set Ta,b is infinite.

Answer 1

Let's first prove that this set is non-empty. As $a<b$, there is such $n\in \mathbb{N}$ that $b-a>\frac{2}{n}$. Indeed, if that would not be the case, that would mean that $a-b \geq -\frac{2}{n}, \forall n\in\mathbb{N}$, then by the existence of a supremum, $a-b\geq \sup\{-2/n, n\in\mathbb{N} \}=0$, $a\geq b$ which is a contradiction. Now as $b-a>\frac{2}{n}$ for some fixed $n$, there is such $m\in \mathbb{Z}$ that $\frac{m-1}{n}< a\leq\frac{m}{n}$ ($m=\min\{x\in\mathbb{Z}, x\geq na \}$ and this set is non-empty and has a lower bound by the Archimedian property of real numbers). So $\frac{2m+1}{2n} \in \mathbb{Q}$, $\frac{2m+1}{2n}>\frac{m}{n}\geq a$. In addition, as $b-a>\frac{2}{n}$ and $\frac{m}{n}-a<\frac{1}{n}$(by construction of $m$) , we have $\frac{2m+1}{2n}<b$. Therefore there is a rational number $r, a<r<b$, so the set $T_{a,b}$ is non-empty. But it is enough to prove that it is infinite: as there is a rational number $r$ in $T_{a,b}$ for any real numbers $a,b$ there is a rational number $r'$ in $T_{a,r}$ which is also in $T_{a,b}$ ($a<r'<r<b$). THus there is at least two rational numbers $a<r'<r<b$. But therefore all the possible combinations of a form $q\in\mathbb{Q}, 0\leq q\leq1$ , $a<r'<q\cdot r' + (1-q)\cdot r<r<b$ are in $T_{a,b}$. As there is infinitely many such combinations, the set $T_{a,b}$ is infinite.