nLab long line

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 $B_{1/n}(t) \subseteq S$.

Related concepts

• line with two origins

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.

See also

• Wikipedia, Long line (topology)