# Contents

## Idea

An ordinary “line$ℝ$ is sometimes thought of as the result of gluing a countable number of copies of a half-open interval $\left[0,1\right)$ end-to-end in both directions. A long line is similarly obtained by gluing an uncountable number of copies of $\left[0,1\right)$ end-to-end in both directions, and is demonstrably “longer” than an ordinary line on account of a number of peculiar properties.

The long line is a source of many counterexamples in topology.

## Definition

###### Definition

Let ${\omega }_{1}$ be the first uncountable ordinal?, and consider $\left[0,1\right)$ as an ordered set. A long ray is the ordered set ${\omega }_{1}×\left[0,1\right)$ taken in the lexicographic order; as a space, it is given the order topology?. The long line is the space obtained by gluing two long rays together at their initial points.

The long line is a line in the sense of being a $1$-dimensional manifold (without boundary) that is not closed (so not a circle). However, it is not paracompact, so it is not homeomorphic to the real line (even though it is Hausdorff).

## Properties

Let $L$ denote the long line, and $R$ the long ray.

1. Every continuous function $f:L\to ℝ$ is eventually constant, i.e., there exists $x\in L$ and $c\in ℝ$ such that $f\left(y\right)=c$ whenever $y\ge x$ (and similarly $f$ is constant for all sufficiently small $x$).

2. $L$ is a normal (${T}_{4}$) space, but the Tychonoff product $L×\overline{L}$ with its one-point compactification is not normal. (See for example Munkres.)

3. Every continuous map $L\to L$ has a fixed point.

4. $R$ is sequentially compact but not compact.

5. The long line is not contractible. Proof sketch: Suppose $H:I×L\to L$ is a homotopy such that $H\left(0,-\right)$ is constant and $H\left(1,-\right)$ is the identity. For each $t\in \left[0,1\right]$ the image $imH\left(t,-\right)$ is an interval (either bounded or unbounded), since $L$ is connected. One may show the set

$\left\{t\in I:imH\left(t,-\right)\text{is bounded}\right\}$\{t \in I: \im H(t, -) \text{is bounded}\}

is both closed and open. It also contains $0$, hence is all of $I$. But it can’t contain $t=1$, contradiction.

## References

• Wikipedia

• Steen and Seebach, Counterexamples in Topology. Springer-Verlag, New York, 1978. Reprinted by Dover Publications, New York, 1995.

• J. Munkres, Topology (2nd edition). Prentice-Hall, 2000.

Revised on September 8, 2012 19:10:37 by Todd Trimble (67.81.93.25)