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 $S$.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (5)$