Ordering on Cuts is Transitive
Jump to navigation
Jump to search
Theorem
Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let:
\(\text {(1)}: \quad\) | \(\ds \alpha\) | \(<\) | \(\ds \beta\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \beta\) | \(<\) | \(\ds \gamma\) |
where $<$ denotes the strict ordering of cuts:
- $\alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$
Then:
- $\alpha < \gamma$
Hence the ordering of cuts $\le$ is a transitive relation.
Proof
We have that:
- $\alpha < \beta$
By definition of strict ordering of cuts:
- $\exists p \in \Q: p \in \beta, p \notin \alpha$
Similarly, we have that:
- $\beta < \gamma$
and so:
- $\exists q \in \Q: q \in \gamma, q \notin \beta$
We have by definition of a cut that:
- $p \in \beta$ and $q \notin \beta$ implies that $p < q$
Together with $p \notin \alpha$, this implies that $q \notin \alpha$.
Thus we have:
- $q \in \gamma$ and $q \notin \alpha$
from which by definition of strict ordering of cuts:
- $\alpha < \gamma$
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.11$. Theorem