# In an order topology, are connected sets convex, and are they intervals?

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.

• What's a convex subset of an ordered space?
– user1076376
Jun 12, 2023 at 7:25
• @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. Jun 12, 2023 at 15:27

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.
• 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. Jun 12, 2023 at 15:24
• (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. Jun 12, 2023 at 15:24