Product of Positive Cuts is Positive Cut

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Theorem

Let $0^*$ denote the rational cut associated with the (rational) number $0$.

Let $\alpha$ and $\beta$ be cuts such that $\alpha \ge 0^*$ and $\beta \ge 0^*$, where $\ge$ denotes the ordering on cuts.


Let $\gamma$ be 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 $q \ge 0$.


Then $\gamma$ is also a cut.


Thus the operation of multiplication on the set of positive cuts is closed.


Proof

By definition of $\gamma$, we have that $r < 0 \implies r \in \gamma$.

Hence $\gamma$ is not empty.


First suppose that either $\alpha = 0^*$ or $\beta = 0^*$.

Then by definition of cut:

$p \in \alpha \implies p < 0$
$q \in \beta \implies q < 0$

and so there exist no $r \in \gamma$ such that $r = p q$.

Thus $\gamma$ consists entirely of rational numbers $r$ such that $r < 0$.

That is:

$\gamma = \set {r \in \Q: r < 0}$

and so:

$\gamma = 0^*$

Hence $\gamma$ is a cut, as we needed to demonstrate.

$\Box$


Suppose that neither $\alpha$ nor $\beta$ are equal to $0^*$

That is:

$\alpha > 0$

and:

$\beta > 0$

Hence by definition of cut:

$0 \in \alpha$

and:

$0 \in \beta$

As a cut has no greatest element:

$\exists p \in \alpha: p > 0$
$\exists q \in \alpha: q > 0$

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 $\Q$.

Note that as $\alpha \ge 0^*$ and $\beta \ge 0^*$, it follows by definition of cut that, both $s \ge 0$ and $t \ge 0$.

We have:

\(\ds p\) \(<\) \(\ds s\) Definition of Cut: $s \notin \alpha$
\(\ds q\) \(<\) \(\ds t\) Definition of Cut: $t \notin \beta$
\(\ds \leadsto \ \ \) \(\ds \forall p \in \alpha, q \in \beta, p > 0, q > 0: \, \) \(\ds p q\) \(<\) \(\ds s t\) 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 $s < r$.

If $r < 0$, then $s < 0$ and so $s \in \gamma$ by definition of $\gamma$.

If $r > 0$ and $s < 0$, then $s \in \gamma$ by definition of $\gamma$.

Otherwise, we have the case where $r > 0$ and $s > 0$.

By definition of $\gamma$:

$r = p q$ for some $p \in \alpha, q \in \beta$ such that $p > 0$ and $q > 0$.

Let $t \in \Q$ such that $s = t q$.

Then $0 < t < p$.

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 $p \in \alpha, q \in \beta$ such that $p > 0$ and $q > 0$.

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

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