a space is connected under the ordered topology if and only if it is a linear continuum.

Let $X$ be totally ordered by the strict total order $<$ , and let it have the order topology.

Suppose $X$ is not a linear continuum. Then either $X$ is not bounded-complete, or the order on $X$ is not a dense total order.

Suppose $X$ is not bounded-complete. Let $A$ be a nonempty subset of $X$ that is bounded above by $b$ , but has no least upper bound. Let $U$ be the set of upper bounds of $A$ . If $x\in U$ , then $x$ is not a least upper bound of $A$ , so there is a $z\in X$ such that $z<x$ and $z$ is an upper bound of $A$ . Then the set $\{y\in X\mid y>z\}$ is open and contains $x$ . Furthermore, all of its elements exceed $z$ , so it is a subset of $U$ . Thus, $U$ is open. $U$ contains $b$ , so it is not empty. Let $x\in X\setminus U$ . Then $x$ is not an upper bound of $A$ , so there is an $a\in A$ such that $x<a$ . The set $\{y\in X\mid y<a\}$ is open, and contains no upper bounds of $A$ , so it is a subset of $X\setminus U$ . Thus $X\setminus U$ contains a neighborhood of each of its points, and is therefore open. Since $U$ and $X\setminus U$ are open, X is not connected.

Suppose the ordering of $X$ is not dense, so there are $a$ and $b$ in $X$ with $a<b$ so that there is no $c$ in $X$ with $a<c<b$ (there is a gap between $a$ and $b$ ). Let $U=\{x\in X\mid x<b\}$ and let $V=\{x\in X\mid x>a\}$ . Because there are no elements between $a$ and $b$ , $U\cap V=\emptyset$ . By transitivity and trichotomy, $U\cup V=X$ . $U$ and $V$ are both open. $a\in U$ and $b\in V$ , so neither $U$ nor $V$ is empty. Thus, $U$ and $V$ separate $X$ , so $X$ is not connected.

Therefore, if $X$ is connected, then $X$ is a linear continuum.

Now suppose that $X$ is disconnected and bounded-complete, and that $U$ and $V$ are (nonempty, open and closed) sets separating $X$ . Suppose that $a\in U$ , and suppose also that there is an element $b\in V$ such that $a<b$ (if there is none, swap the names of $U$ and $V$ , or reverse the ordering). $Z=\{x\in V\mid x>a\}$ is open (it is the intersection of two open sets), and contains $b$ (so it is not empty). $Z$ is bounded below by $a$ , so it has a greatest lower bound $z$ .

If $z\in U$ , then, since $U$ is open, there is an interval in $U$ containing $z$ , which must, to exclude $b$ , be of the form $\{x\mid x<k\}$ or of the form $\{x\mid j<x<k\}$ , for some $k$ and perhaps $j$ . But then $k$ would be a lower bound of $Z$ , contradicting the fact that $z$ is the infimum of $Z$ .

If $z\in V$ , then, since $V$ is open, there is an interval in $V$ containing $z$ , which must, to exclude $a$ , be of the form $\{x\mid j<x\}$ or of the form $\{x\mid j<x<k\}$ for some $j$ and perhaps $k$ . In either case, $a<j<z$ and the set $W=\{x\mid j<x<z\}$ is a subset of $V$ . If $x\in W$ , then $x\in V$ and $x>a$ , so $x\in Z$ , contradicting the fact that $z$ is the infimum of $Z$ . Thus, there are no elements of $X$ between $j$ and $z$ , so the order on $X$ is not dense. This proves that if $X$ is a linear continuum, then $X$ is connected.

Title	a space is connected under the ordered topology if and only if it is a linear continuum.
Canonical name	ASpaceIsConnectedUnderTheOrderedTopologyIfAndOnlyIfItIsALinearContinuum
Date of creation	2013-03-22 17:38:38
Last modified on	2013-03-22 17:38:38
Owner	dfeuer (18434)
Last modified by	dfeuer (18434)
Numerical id	7
Author	dfeuer (18434)
Entry type	Result
Classification	msc 54B99
Classification	msc 06F30
Related topic	LinearContinuum
Related topic	OrderTopology