Contents

# Contents

## Idea

The ordinary “line$\mathbb{R}$ (i.e. the real numbers equipped with their Euclidean metric topology) may be thought of as the result of gluing a countable set of copies of the half-open interval $[0, 1)$ end-to-end in both directions. A long line is similarly obtained by gluing an uncountable set of copies of $[0, 1)$ 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 the half-open interval $[0, 1)$ as an totally ordered set. A long ray is the ordered set $\omega_1 \times [0, 1)$ 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 locally Euclidean space (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

All the assertions below apply to both the long line and the long ray. We write $L$ to cover both cases even if we only treat one.

1. Every continuous function $f\colon L \to \mathbb{R}$ is eventually constant, i.e., there exists $x \in L$ and $c \in \mathbb{R}$ such that $f(y) = c$ whenever $y \geq x$ (and similarly $f$ is constant for all sufficiently small $x$).

2. $L$ is a normal ($T_4$) space, but the Tychonoff product $L \times \bar{L}$ with its one-point compactification is not normal. (See for example Munkres.)

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

4. $L$ is sequentially compact but not compact. (Being sequentially compact, they are also countably compact spaces.) Thus, the image $f(L)$ under a continuous map $f: L \to \mathbb{R}$, being sequentially compact, is a compact subspace of $\mathbb{R}$.

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

$S = \{t \in I: \im H(t, -)\; \text{is bounded}\}$

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

Let us flesh out this sketched proof. First we show $S$ is closed. Denote the bottom element of $L$ by $\bot$. If $t_n$ is a sequence in $S$ with limit point $t \in I$, then $H(\{t_n\} \times L) \subseteq [\bot, b_n]$ for some sequence of bounds $b_n \in L$; this sequence has an upper bound $b \in L$. Then $H(\{t_n\} \times L) \subseteq [\bot, b]$ for all $n \in \mathbb{N}$, which is the same as saying $\bigcup_n \{t_n\} \times L \subseteq H^{-1}([\bot, b])$. Now $\{t\} \times L$ is included within the closure of the union, which is included within the closed set $H^{-1}([\bot, b])$. But $\{t\} \times L \subseteq H^{-1}([\bot, b])$ means $H(\{t\} \times L) \subseteq [\bot, b]$; hence $t \in S$.

Now we show $S$ is open. This uses a countably compact version of the tube lemma. If $t \in S$, then we can find $b \in L$ such that $\{t\} \times L \subseteq H^{-1}([\bot, b))$, where the right side is open in $I \times L$. Let $B_r(t) \subseteq I$ denote the open ball of radius $r$ centered at $t$, and for $c \lt d$ in $L$ let $(c, d)$ denote the open interval between $c$ and $d$. Consider the collection $T$ of all triples $(n, c, d) \in \mathbb{N} \times L \times L$ such that

$U_{n, c, d} \coloneqq B_{1/n}(t) \times (c, d) \subseteq H^{-1}([\bot, b)).$

For each fixed $n \in \mathbb{N}$, put $U_n = \bigcup_{(n, c, d) \in T} (c, d)$. The $U_n$ form a countable open cover of $L$, since for every $x \in L$ there is a $U_{n, c, d}$ containing the pair $(t, x)$. Since $L$ is countably compact, there is a finite subcover $U_{n_1}, \ldots, U_{n_k}$. Putting $n = \max \{n_1, \ldots, n_k\}$, we see

$B_{1/n}(t) \times L = B_{1/n}(t) \times \bigcup_{i=1}^k U_{n_i} = \bigcup_{i=1}^k B_{1/n}(t) \times U_{n_i} \subseteq \bigcup_{i=1}^k B_{1/n_i}(t) \times U_{n_i} \subseteq H^{-1}([\bot, b))$

from which it immediately follows that $B_{1/n}(t) \subseteq S$.

## References

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

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