Interval of Ordered Set is Convex
Jump to navigation
Jump to search
This article, or a section of it, needs explaining, namely:
Obvious though it is, the above needs to be stated as a theorem in its own right.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
Theorem
Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $I$ be an interval: be it open, closed, or half-open in $S$.
Then $I$ is convex in $S$.
Proof
Any interval can be represented as the intersection of two rays.
Obvious though it is, the above needs to be stated as a theorem in its own right.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.
Thus by Ray is Convex and Intersection of Convex Sets is Convex Set (Order Theory), $I$ is convex.
$\blacksquare$
