Convex subsets in ordered sets: intervals or half rays?

[nonequilibrium]

Homework Statement

It might not be a real topology question, but it's an exercise question in the topology course I'm taking. The question is not too hard, but I'm mainly doubting about the terminology:

Let X be an ordered set and [itex]Y \subsetneq X[/itex] convex. Does it follow that Y is an interval or half ray?

Homework Equations

N.A.

The Attempt at a Solution

I would think not, unless I'm misunderstanding the terminology. Take the rational numbers and the subset denoted in [itex]\mathbb R[/itex] as [itex][\sqrt{2},2] \cap \mathbb Q[/itex]. It is indeed convex in [itex]\mathbb Q[/itex], but it's not an interval, cause I can't write it as [itex][q_1,q_2][/itex] or [itex]]q_1,q_2][/itex] with [itex]q_i \in \mathbb Q[/itex], or is my notion of interval too narrow?

 

[tiny-tim]

hi mr. vodka!

doesn't "convex" mean that, between any two elements of Y, there's no element in X that isn't in Y ?

 

[nonequilibrium]

Uhu. Which is true if you view that set as a part of the rational numbers, right?

 

[nonequilibrium]

How is that in Q?

 

[tiny-tim]

oops! misread the question!

let me start again …

√2 isn't in Q, so what's the meaning of [itex][\sqrt{2},2] \cap \mathbb Q[/itex] ?

 

[nonequilibrium]

Well you know what it means in R, right? And then you can interpret it as a subset of Q.

 

[HallsofIvy]

Well, [itex][-\sqrt{2},\sqrt{2}][/itex] as a subset of R is not the same set as [itex][-\sqrt{2}, \sqrt{2}][/itex] as a subset of Q. The first contains many points not in the second.

 

[nonequilibrium]

but that's not the subset I regarded, I regarded the intersection with Q