# Set of Rational Numbers Strictly between Zero and One has no Greatest or Least Element

## Theorem

Let $S \subseteq \Q$ be the subset of the set of rational numbers defined as:

$S = \set {r \in \Q: 0 < r < 1}$

Then $S$ has no greatest or smallest element.

However, $S$ has a supremum $1$ and an infimum $0$.

## Proof

We have that:

$\forall r \in S: 0 < r$

and:

$\forall r \in S: r < 1$

Hence $0$ and $1$ are lower and upper bounds of $S$ respectively.

Let $s \in S$.

Then $s \in \Q: 0 < s < 1$.

Aiming for a contradiction, suppose $s$ is the greatest element of $S$.

But then we have:

$0 < s < \dfrac {s + 1} 2 < 1$

and so $\dfrac {s + 1} 2 \in S$ but $s < \dfrac {s + 1} 2$.

This contradicts our assertion that $s$ is the greatest element of $S$.

Aiming for a contradiction, suppose $s$ is the smallest element of $S$.

But then we have:

$0 < \dfrac s 2 < s < 1$

and so $\dfrac s 2 \in S$ but $\dfrac s 2 < s$.

This contradicts our assertion that $s$ is the smallest element of $S$.

So $S$ cannot have either a greatest or smallest element.

Let $H$ be such that $0 < H < 1$.

Suppose $H$ is a smaller upper bound of $S$ than $1$.

Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $H < r < 1$.

But then $r$ is a rational number between $0$ and $1$ which is greater than $H$.

Hence $H$ cannot be an upper bound of $S$.

Thus, by definition, $1$ is the supremum of $S$

Suppose $H$ is a greater lower bound of $S$ than $0$.

Then from Between two Real Numbers exists Rational Number there exists a rational number $r$ such that $0 < r < H$.

But then $r$ is a rational number between $0$ and $1$ which is smaller than $H$.

Hence $H$ cannot be a lower bound of $S$.

Thus, by definition, $0$ is the infimum of $S$.

$\blacksquare$