Existence of Unique Inverse Element for Addition of Cuts
Theorem
Let $\alpha$ be a cut.
Let $0^*$ be the rational cut associated with the (rational) number $0$:
- $0^* = \set {r \in \Q: r < 0}$
Then there exists a unique cut $\beta$ such that:
- $\alpha + \beta = 0^*$
where $+$ denotes the operation of addition of cuts.
Proof
Proof of Uniqueness
Suppose $\alpha + \beta_1 = \alpha + \beta_2 = 0^*$.
We have:
\(\ds \beta_2\) | \(=\) | \(\ds 0^* + \beta_2\) | Identity Element for Addition of Cuts | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\alpha + \beta_1} + \beta_2\) | Definition of $\beta_1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\alpha + \beta_2} + \beta_1\) | Addition of Cuts is Commutative and Addition of Cuts is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds 0^* + \beta_1\) | Definition of $\beta_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \beta_1\) | Identity Element for Addition of Cuts |
So if such a $\beta$ exists, it is indeed unique.
$\Box$
Proof of Existence
Let $\beta$ be the set of all rational numbers $p$ such that $-p$ is an upper number of $\alpha$, but not the smallest upper number, were it to exist.
We verify that the conditions are satisfied for $\beta$ to be a cut.
It follows directly from the definition that $\beta \ne \O$ and $\beta \ne \Q$.
Thus condition $(1)$ is satisfied.
$\Box$
Let $p \in \beta$.
Let $q \in \Q$ be a rational number such that $q < p$.
Then:
- $-p \notin \alpha$
and:
- $-q > -p$
Thus $-q$ is an upper number of $\alpha$, but not the smallest.
Thus $q \in \beta$.
Thus condition $(2)$ is satisfied.
$\Box$
Let $p \in \beta$.
Then by definition $-p$ is an upper number of $\alpha$, but not the smallest.
Thus there exists a rational number $q$ such that $-q < -p$ and such that $-q \notin \alpha$.
Let:
- $r = \dfrac {p + q} 2$
- $-q < -r < -p$
so $-r$ is also an upper number of $\alpha$, but not the smallest.
So we have found a rational number $r$ such that $r > p$ where $r \in \beta$.
Thus condition $(3)$ is satisfied.
$\Box$
We have demonstrated that $\beta$ is a cut.
We now verify that $\alpha + \beta = 0^*$.
Suppose $p \in \alpha + \beta$.
Then:
- $p = q + r$
for some $q \in \alpha, r \in \beta$.
Hence:
- $-r \notin \alpha$
- $-r < q$
- $q + r < 0$
and:
- $p \in 0^*$
Suppose $p \in 0^*$.
Then:
- $p < 0$
By Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational, there exist $q \in \alpha, r \notin \alpha$ such that:
- $r$ is not the smallest upper number of $\alpha$
$r - q = -p$
Because $-r \in \beta$:
\(\ds p\) | \(=\) | \(\ds q - r\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds q + \paren {-r}\) | ||||||||||||
\(\ds \) | \(\in\) | \(\ds \alpha + \beta\) |
Thus by definition of set equality:
- $\alpha + \beta = 0^*$
and the proof is complete.
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.16$. Theorem