Sum of Cuts is Cut
Theorem
Let $\alpha$ and $\beta$ be cuts.
Let $\gamma$ be the set of all rational numbers $r$ such that:
- $\exists p \in \alpha, q \in \beta: r = p + q$
Then $\gamma$ is also a cut.
Thus the operation of addition on the set of cuts is closed.
Proof
By definition of cut, neither $\alpha$ nor $\beta$ are empty.
Hence there exist $p \in \alpha$ and $q \in \beta$.
Hence there exists $r = p + q$ and so $\gamma$ is likewise not empty.
Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers.
Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$.
We have:
\(\ds p\) | \(<\) | \(\ds s\) | Definition of Cut: $s \notin \alpha$ | |||||||||||
\(\ds q\) | \(<\) | \(\ds t\) | Definition of Cut: $t \notin \beta$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall p \in \alpha, q \in \beta: \, \) | \(\ds p + q\) | \(<\) | \(\ds s + t\) | Rational Numbers form Totally Ordered Field | |||||||||
\(\ds \leadsto \ \ \) | \(\ds s + t\) | \(\notin\) | \(\ds \gamma\) | Definition of Cut |
Thus it is demonstrated that $\gamma$ does not contain every rational number.
Thus condition $(1)$ of the definition of a cut is fulfilled.
$\Box$
Let $r \in \gamma$.
Let $s \in \Q$ such that $s < r$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
Let $t \in \Q$ such that $s = t + q$.
Then $t < p$.
Hence $t \in \alpha$.
Hence by definition of $\gamma$, $t + q = s \in \gamma$.
Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$.
Thus condition $(2)$ of the definition of a cut is fulfilled.
$\Box$
Aiming for a contradiction, suppose $r \in \gamma$ is the greatest element of $\gamma$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
By definition of a cut, $\alpha$ has no greatest element:
Hence:
- $\exists s \in \Q: s > p: s \in \alpha$
But then $s + q \in \gamma$ while $s + q > r$.
This contradicts the supposition that $r$ is the greatest element of $\gamma$.
Hence $\gamma$ itself can have no greatest element.
Thus condition $(3)$ of the definition of a cut is fulfilled.
$\Box$
Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut.
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.12$. Theorem