# 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 $$ and $z$ is an upper bound of $A$. Then the set $\{y\in X\mid y>z\}$ is open and contains $x$. Furthermore, all of its elements exceed $z$, so it is a subset of $U$. Thus, $U$ is open. $U$ contains $b$, so it is not empty. Let $x\in X\setminus U$. Then $x$ is not an upper bound of $A$, so there is an $a\in A$ such that $$. The set $$ 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 $$ so that there is no $c$ in $X$ with $$ (there is a gap between $a$ and $b$). Let $$ and let $V=\{x\in X\mid x>a\}$. Because there are no elements between $a$ and $b$, $U\cap V=\mathrm{\varnothing}$. 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 $$ (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 $$ or of the form $$, 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 $$ or of the form $$ for some $j$ and perhaps $k$. In either case, $$ and the set $$ 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. |
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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 |