Rational Number Not in Cut is Greater than Element of Cut
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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