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This is from Exercise 7 in p. 92 in Munkres's Topology.

Except for the trivial cases such as $Y$ is empty set or singleton, it seems if $Y$ is convex in an simply ordered set $X$ then $Y$ is interval or ray.

But I cannot start my proof because I cannot use $\sup$ or $\max$ functions.

What should I do?

le4m
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1 Answers1

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What about the set $\{p\in\Bbb Q:p^2<2\}$? It’s a non-trivial convex subset of $\Bbb Q$; does it meet Munkres’ definition of interval or ray?

Brian M. Scott
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    Thank you for your counterexample. It is neither interval or ray in $\mathbb{Q}$. – le4m Jun 29 '13 at 01:36
  • @julypraise: You’re welcome. – Brian M. Scott Jun 29 '13 at 01:39
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    @BrianM.Scott What about when $X$ is complete? – Alex Becker Jun 29 '13 at 01:43
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    @Alex: When $\langle X,\le\rangle$ is a Dedekind complete linear order, every order-convex set is an interval (open, closed, or half-open) or a ray (open or closed). – Brian M. Scott Jun 29 '13 at 01:49
  • @BrianM.Scott Thanks. – Alex Becker Jun 29 '13 at 01:56
  • @Alex: My pleasure. – Brian M. Scott Jun 29 '13 at 01:56
  • @BrianM.Scott example 3 on page 85 is the positive integers with order topology. Is this another example of a convex set with the property? – JEM Jun 24 '14 at 03:20
  • @BrianM.Scott So, to show that it is that ${p \in \mathbb{Q} ~|~ p^2 < 2 } = A$ can be neither a ray nor an interval, does not just apply of the density of $\mathbb{Q}$ several times over? E.g., if $A = (a,b)$ for positive $a,b \in \mathbb{Q}$, then $b < \sqrt{2}$, and the density of $\mathbb{Q}$ says $b < c < \sqrt{2}$, implying $c^2 < 2$, a contradiction. Of course this is just one case. One keeps repeating this same argument for each case? Does this sound right? – user193319 Oct 25 '16 at 18:37
  • @user193319: You don’t need to repeat. You just note that if $\sqrt2<q\in\Bbb Q$, then there is a $p\in(\sqrt2,q)$, so $p\in(\leftarrow,q)\setminus A$, while if $q<\sqrt2$, there is a $p\in(q,\sqrt2)$, so $p\in A\setminus(\leftarrow,q)$. – Brian M. Scott Oct 25 '16 at 18:57
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    @BrianM.Scott What does $(\leftarrow, q) \setminus A$ denote? – user193319 Oct 25 '16 at 19:25
  • @user193319: $(\leftarrow,q)$ is the open ray to the left from $q$; another notation for it is $(-\infty,q)$. $S\setminus A$ is the set difference: the set of things in $S$ but not in $A$. – Brian M. Scott Oct 25 '16 at 19:27