Product of Cuts is Zero Cut iff Either Factor equals Zero Cut
Theorem
Let $\alpha$ and $\beta$ be cuts.
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Then:
- $\alpha \beta = 0^*$
- $\alpha = 0^*$ or $\beta = 0^*$
where $\alpha \beta$ denotes the product of $\alpha$ and $\beta$.
Proof
By definition, we have that:
- $\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\ -\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\ -\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\ \size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^* \end {cases}$ where:
- $\size \alpha$ denotes the absolute value of $\alpha$
- $\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
- $\ge$ denotes the ordering on cuts.
First we note that from Absolute Value of Cut is Zero iff Cut is Zero:
- $\size {0^*} = 0^*$
Necessary Condition
Let $\alpha = 0^*$ or $\beta = 0^*$.
Then from Product of Cut with Zero Cut equals Zero Cut it follows directly that:
- $\alpha \beta = 0^*$
$\Box$
Sufficient Condition
Let $\alpha \beta = 0^*$.
Suppose $\alpha > 0^*$ and $\beta > 0^*$.
By definition of multiplication of positive cuts:
$\alpha \beta$ is the set of all rational numbers $r$ such that either:
- $r < 0$
or
- $\exists p \in \alpha, q \in \beta: r = p q$
where $p \ge 0$ and $q \ge 0$.
We have that $\alpha \beta = 0^*$.
Aiming for a contradiction, suppose $r = p q$ where $p \ge 0$ and $q \ge 0$.
Then $r \ge 0$.
But then $r \notin 0^*$ by definition of $0^*$.
Thus there is no $p \in \alpha, q \in \beta$ such that $r = p q$ where $p \ge 0$ and $q \ge 0$.
It follows that $\alpha > 0^*$ and $\beta > 0^*$ cannot both be satisfied.
It follows further that for $\size \alpha \, \size \beta = 0^*$ it is not possible for $\size \alpha > 0^*$ and $\size \beta > 0^*$.
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: $1.26$. Theorem $\text {(e)}$