Ordering on Cuts is Compatible with Addition of Cuts

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Theorem

Let $\alpha$, $\beta$ and $\gamma$ be cuts.

Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.

Let $\beta < \gamma$ denotes the strict ordering on cuts defined as:

$\beta < \gamma \iff \exists p \in \Q: p \in \beta, p \notin \gamma$


Then:

$\beta < \gamma \implies \alpha + \beta < \alpha + \gamma$


Corollary

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


If:

$\alpha > 0^*$ and $\gamma > 0^*$

then:

$\alpha + \gamma > 0^*$


Proof

By definition of the strict ordering on cuts and addition of cuts:

$\alpha + \beta \le \alpha + \gamma$


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Suppose $\alpha + \beta = \alpha + \gamma$.

Then:

\(\ds \beta\) \(=\) \(\ds 0^* + \beta\) Identity Element for Addition of Cuts
\(\ds \) \(=\) \(\ds \paren {-\alpha} + \paren {\alpha + \beta}\) Definition of Negative of Cut
\(\ds \) \(=\) \(\ds \paren {-\alpha} + \paren {\alpha + \gamma}\)
\(\ds \) \(=\) \(\ds 0^* + \gamma\)
\(\ds \) \(=\) \(\ds \gamma\)


Thus:

$\beta = \gamma$

and the result follows.

$\blacksquare$


Sources

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