Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational
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Theorem
Let $\alpha$ be a cut.
Let $r \in \Q_{>0}$ be a (strictly) positive rational number.
Then there exist rational numbers $p$ and $q$ such that:
- $p \in \alpha, q \notin \alpha$
- $q - p = r$
such that $q$ is not the smallest upper number of $\alpha$.
Proof
Let $s \in \alpha$ be a rational number.
For $n = 0, 1, 2, \ldots$ let $s_n = s + n r$.
Then there exists a unique integer $m$ such that:
- $s_m \in \alpha$
and:
- $s_{m + 1} \notin \alpha$
If $s_{m + 1}$ is not the smallest upper number of $\alpha$, take:
- $p = s_m$
- $q = s_{m + 1}$
If $s_{m + 1}$ is the smallest upper number of $\alpha$, take:
- $p = s_m + \dfrac r 2$
- $q = s_{m + 1} + \dfrac r 2$
The result follows.
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.15$. Theorem