<?xml version="1.0" encoding="UTF-8"?>

<record version="14" id="4069">
 <title>long line</title>
 <name>LongLine</name>
 <created>2003-02-27 04:39:53</created>
 <modified>2007-07-18 21:29:29</modified>
 <type>Definition</type>
 <creator id="537" name="Dr_Absentius"/>
 <author id="3771" name="CWoo"/>
 <author id="537" name="Dr_Absentius"/>
 <author id="13753" name="Mathprof"/>
 <author id="2414" name="alozano"/>
 <author id="2760" name="yark"/>
 <author id="56" name="AxelBoldt"/>
 <author id="291" name="igor"/>
 <classification>
	<category scheme="msc" code="54G20"/>
 </classification>
 <preamble>\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}

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\newcommand{\FL}{\mathfrak{L}}
\newcommand{\FM}{\mathfrak{M}}
\newcommand{\Ga}{\alpha}
\newcommand{\Gb}{\beta}
\newcommand{\Gg}{\gamma}
\newcommand{\GG}{\Gamma}
\newcommand{\Gd}{\delta}
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\newcommand{\Ge}{\varepsilon}
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\newcommand{\GW}{\Omega}
\newcommand{\Gee}{\epsilon}
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\newcommand{\Grr}{\varrho}
\newcommand{\Gff}{\phi}
\newcommand{\Gss}{\varsigma}

\def\co{\colon\thinspace}</preamble>
 <content>\PMlinkescapeword{induced}
The \emph{long line} is a non-paracompact Hausdorff $1$-dimensional manifold
 constructed as
follows. Let $\GW$ be the first uncountable ordinal (viewed as an ordinal space) and consider the set 
$$L:=\GW\times [0,1)$$
endowed with the order topology induced by the 
lexicographical order, that is the order defined by 
$$(\Ga_1,t_1) &lt; (\Ga_2,t_2) \iff \Ga_1&lt;\Ga_2 \quad\text{or}\quad
(\Ga_1=\Ga_2 \quad\text{and}\quad t_1&lt;t_2)\,.$$
Intuitively $L$ is obtained by ``filling the gaps'' between consecutive
ordinals in $\GW$ with intervals, much the same way that 
nonnegative  reals are 
obtained by filling the gaps between consecutive natural numbers with intervals. 

Some of the  properties of the long line:
\begin{itemize}
\item $L$ is a chain.

\item $L$ is not compact; in fact $L$ is not Lindel\"of.

Indeed $\left\{\,[\,0,\Ga):\Ga&lt;\GW\right\}$ is an open cover of $L$ that has no
countable subcovering. To see this notice that 
$$\bigcup\left\{\, [\,0,\Ga_x):x\in X\right\}=\left[\,0,\sup\{\Ga_x:x\in X\}\right)\,$$
 and since the supremum of a countable
collection of countable ordinals is a countable ordinal such a union can
 never be $[\,0,\GW)$.

 \item However, $L$ is sequentially compact.

Indeed every sequence has a convergent subsequence. To see this notice that
given a sequence $a:=(a_n)$ of elements of $L$ there is an ordinal $\Ga$ such
that all the terms of  $a$ are in the subset $[\,0,\Ga\,]$. Such a subset is
compact since it is homeomorphic to $[\,0,1\,]$.

\item $L$ therefore is not metrizable.

\item $L$ is a $1$--dimensional locally Euclidean

\item $L$ therefore is not paracompact.

\item $L$ is first countable.

\item $L$ is not separable.

\item All homotopy groups of $L$ are trivial.

\item However, $L$ is not contractible.
 
\end{itemize}

\section*{Variants}
There are several variations of the above construction.

\begin{itemize}
\item Instead of $[\,0,\GW)$ one can use $(0,\GW)$ or $[\,0,\GW\,]$. The latter (obtained by adding a single point to $L$) is compact.

\item One can consider the ``double'' of the above construction. That is the
  space obtained by gluing two copies of $L$ along $0$. The resulting open
  manifold is not homeomorphic to $L\setminus \{0\}$.
\end{itemize}</content>
</record>
