Rational Number Not in Cut is Greater than Element of Cut

Theorem

Let $\alpha$ be a cut.

Let $p \in \alpha$.

Let $q \in \Q$ such that $q \notin \alpha$.

Then $q > p$.

Proof

We have been given that $p \in \alpha$ while $q \notin \alpha$.

Aiming for a contradiction, suppose $q \le p$.

If $q = p$ then $q \in \alpha$ immediately.

If $q < p$ then $q \in \alpha$ by property $(2)$ of the definition of cut.

In either case $q \in \alpha$ which contradicts $q \notin \alpha$.

Hence by Proof by Contradiction it follows that $q > p$.

$\blacksquare$

Sources

• 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.5$. Theorem