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Problem: $X$ is an ordered set with order topology. Is it true that (1) $A\subseteq X$ is connected $\implies$ $A$ is convex (2) $A\subseteq X$ is connected $\implies$ $A$ is an interval ? (Here interval can be open, closed, half-open half-closed, and boundary can be $(-\infty$ or $+\infty)$, or has one element since $\{a\}=\{x\ :\ a\le x\le a \}$).

Motivation: In the order topology of $\mathbb{R}$ (which is the same as usual topology on $\mathbb{R}$), it is easy to see that $A\subseteq \mathbb{R}$ is connected $\iff$ $A$ is convex $\iff$ $A$ is an interval. I want to know if this holds for general order topology. Now I have worked out four arrows:

(1) Convex does not imply connectedness. counterexample: the order topology on $\mathbb{Z}_{\ge 0}$ with subset $\{2,3,4\}$. $\{2,3,4\}=\left(\{2,3,4\}\cap\{x\in\mathbb{Z}_{\ge 0}\ :\ 1< x<3\} \right) \cup\left( \{2,3,4\}\cap\{x\in\mathbb{Z}_{\ge 0}\ :\ x>2\}\right) $ and is hence not connected.

(2) Convex does not imply being an interval, counterexample: the order topology on $\mathbb{Q}$ and the subset $\mathbb{Q}\cap (\alpha, \beta)$, where $\alpha$, $\beta\in\mathbb{R}-\mathbb{Q}$.

(3) Being an interval implies convex: obvious.

(4) Being interval does not imply connectedness: the counterexample in (1) again works.

I did not work out the two arrows in my question.

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    $\begingroup$ What's a convex subset of an ordered space? $\endgroup$
    – user1076376
    Jun 12, 2023 at 7:25
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    $\begingroup$ @DumperDGarb $A$ is convex, provided $a,\ b\in A$, $a<b$ $\implies$ $\{x\ :\ a<x<b\}\subseteq A$. This term is used in Munkres's Topology, for example. $\endgroup$
    – Asigan
    Jun 12, 2023 at 15:27

1 Answer 1

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With $(X, \leq)$ in the order topology, $A \subseteq X$ connected does imply convexity. For if $A$ is not convex, there are $a < b < c$ of $X$ s.t. $a, c \in A$ but $b \not\in A$. So $A \cap (-\infty, b), A \cap (b, +\infty)$ form a disconnection of $A$.

However, $A$ connected does not imply $A$ is necessarily an interval. Indeed for a counterexample consider $X = (\mathbb R^2, \leq_{\text{dict}})$ in the dictionary order and $A = \{0\} \times \mathbb R$. The latter is connected but you can't write it as an interval since "it has no endpoints in $(\mathbb R^2, \leq_{\text{dict}})$" so to speak.

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    $\begingroup$ Thanks. For others' reference I supplement some details on the example $A = \{0\} \times \mathbb {R}$. (1) $A$ is connected. proof: $A$ is convex $\implies$ the subspace topology on $A$ is the same as the order topology restricted on $A$, which is homeomorphic to the order topology on $\mathbb{R}$ (consider $0\times a\mapsto a$), and the same with the usual topology on $\mathbb{R}$, and hence connected. $\endgroup$
    – Asigan
    Jun 12, 2023 at 15:24
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    $\begingroup$ (2) $A$ is not an interval. Assume it is $\{x\ : \ a<x<b\}$, $a=x_a\times y_a$, $b=x_b\times y_b$. Then $x_b<0$, $x_b=0$ and $x_b>0$ are all impossible. The same argument implies $A$ is not an interval of other form. $\endgroup$
    – Asigan
    Jun 12, 2023 at 15:24

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