Does the set ${\displaystyle A =\{q\in \mathbb {Q} |q<a\}} $ have a maximum element?

Question

Does the set A have a maximum element?

${\displaystyle A =\{q\in \mathbb {Q} |q<a\}} $

Thoughts: I don't think it does. We can suppose that it has a maximum element, q. Then q is a real number and a is also a real number (and they are not equal), so infinite rational numbers exist between them, let's say one of them is p. So, $ q<p<a$ which is absurd. If that's the correct way to think about it, can you help me prove that the rational numbers between a and q are infinite?

Answer 1

There are no maximum element in $A$. However, if you look for a supremum :

• If $a\in \mathbb Q$ and you want the supremum in $\mathbb Q$, then yes, and $a$ is the supremum.

• If $a\in \mathbb R\setminus \mathbb Q$ and you want the supremum in $\mathbb Q$, then no.

• If $a\in \mathbb R\setminus \mathbb Q$ and you want the supremum in $\mathbb R$, then yes, and the supremum is $a$.

Answer 2

Having an infinite number of rationals between any two real numbers is true, but is useless here: what matters is to find one.

But if $q = \sup(A)$, then $q\in A$ so $q\in \mathbb Q$ and $q<a$.

We know that for the sequence $u_n = \dfrac 1 {10^n}$ we have $\lim_{n\to\infty}{u_n}=0$. That means:

$$ \exists n\in \mathbb N, \forall m \ge n,\space{} u_m < a - q $$

We can conclude immediately:

$$\left\lbrace {q+\frac {10} n\in\mathbb Q\\ q<q+\frac{10} n < a}\right. \implies\left\lbrace{q+\frac{10} n\in A\\ q+\frac{10} n > \sup(A)}\right.$$

This is a contradiction, so $A$ has no maximum element.

You could have used any positive rational strictly greater than $1$ instead of $10$ to define the $u_n$ sequence, with the exact same result.

(BTW, the tail of the $u_n$ sequence exhibit an infinite numbers of rationals strictly between $q$ and $a$: $q+u_m)$