“The long line is a topological space somewhat similar to the real line, but in a certain sense ›longer‹.”¹

To understand this statement, we will firstly do a short recap of topology. A lot of statements will just be given, but the reader is encouraged to verify that these hold.

Topology-Basics

A topological space is a tuple $(X,\mathcal{T})$ consisting of a non-empty set $X$ and a topology $\mathcal{T}$ on $X$, i.e. $\mathcal{T}$ is a collection of subsets of $X$, that contains $\emptyset$ and $X$ and is closed under binary intersections and arbitrary unions. We call the elements of $\mathcal T$ the open sets in $X$. When $\mathcal{T}$ is understood, we sometimes refer to the whole space just by $X$. We call a collection $\mathcal{B} \subseteq \mathcal{T}$ a basis for $\mathcal{T}$ if every element of $\mathcal{T}$ is the union of elements in $\mathcal{B}$. In this case, we say that $\mathcal{B}$ generates the topology. Similarly, every set $\mathcal{B} \subseteq \mathcal{P}(X)$, which is closed under intersections and covers $X$ (i.e. $\bigcup\mathcal{B} = X$) forms a basis for some topology. For example, if we have a linearly ordered space $(X,<)$ with at least two elements, then we can define an interval as the set of all elements between two given points. The collection of all intervals forms a basis for the order topology on $X$. This topology will play an important role for the long line.

Revisiting $\mathbb{R}_+$

Before we look at the long line, we will construct the long ray as a preliminary step. To make the construction easier to understand, we will first look at a different way to view the topological space of non-negative real numbers $\mathbb{R}_+$ (including $0$). Note that the usual topology on $\mathbb{R}_+$ is just the order topology given by the usual ordering. One can easily see that this basis as described earlier is uncountable. However, since the rationals are dense in the reals, we can restrict $a$ and $b$ to be rationals and would obtain the same topology. This means that $\mathbb{R}_+$ is second countable, i.e. has a countable basis. Furthermore, this makes it first countable: Every point has a countable neighborhood basis. Given a point $x \in X$, we call a collection $\mathcal{B}_x \subseteq \mathcal{T}$ of opens containing $x$ a neighborhood basis if for every open set $U$ containing $x$, there is a $V \in \mathcal{B}_x$, such that $V \subseteq U$. So by considering

we see that second countable spaces are also first countable. Instead of viewing $\mathbb{R}_+$ as the subset of $\mathbb{R}$ where all values are greater or equal to $0$, we can also construct it via

This has a nice intuition: Since every element is a pair $(n,r)$ with a natural number $n$ and a decimal $r$, we can view it as the number $n+r$. Moreover, if we give it the lexicographical ordering by setting $(n,r)<(m,q)$ iff $nhomeomorphic to the original space, but since homeomorphisms are the isomorphisms in topology, we will consider them to be the same. You might notice that instead of $\mathbb{N}$, we could have written the ordinal $\omega$. So what if we used different ordinals? For finite ordinals (apart from $0$), we essentially get the intervals $[0,n)$, which are homeomorphic to $[0,1)$ by division by $n$. Using the fact that $[0,1) \cong \mathbb{R}_+$, one can inductively show that for all countable ordinals the result is still homeomorphic to $\mathbb{R}_+$. But what happens after that?

Constructing $L_+$

The smallest uncountable ordinal is $\omega_1$. If you are unfamiliar with ordinals, you can imagine the natural numbers, but there are uncountably many of them. By saying that they are like natural numbers, we want to emphasize that every number has a unique successor. For a deeper digression, we refer to (Levy 2012). This way, we can define

Giving it the lexicographical ordering and the order topology yields the long ray. Firstly, let us look at why the long ray is a line, i.e. in what sense is it similar to $\mathbb{R}_+$. As we will see it does not globally look like $\mathbb{R}_+$, so it is not homeomorphic to it. However, it is locally homeomorphic to it. Every point has a neighborhood which is homeomorphic to $\mathbb{R}_+$. Indeed, if $(\alpha,r) \in L_+$, then $\alpha$ must be a countable ordinal. So $\alpha +1$ is still countable and

Since $\alpha+1$ is also countable, we already discussed that $U$ must be homeomorphic to $\mathbb{R}_+$, so it is the neighborhood we were looking for. Because local homeomorphisms preserve first countability, we get that $L_+$ is first countable.

Properties of $L_+$

In $\mathbb{R}_+$, there are sequences like $a_n:=n$ that do not converge. This can almost not happen in $L_+$: While alternating sequences still do not converge, every sequence has a convergent subsequence, which topologists call sequential compactness. For any sequence $(a_n)_{n \in \mathbb{N}}$ in $L_+$, we know that

for all $n$ with all $\alpha_n$ countable. Thus the supremum of them must be a countable ordinal $\alpha$. So all $a_n$ are smaller than $a:= \underline{\alpha+1}$. (I use the notation $\underline\beta:= (\beta,0)$ to improve readability.) Thus $a_n \in [\underline0,a]$ for all $n$. We have already seen that $[\underline0,a) := (\alpha +1) \times [0,1) \cong [0,1)$, so it should not be hard to convince yourself that $[\underline0,a] \cong [0,1]$. Since the latter is sequentially compact, our sequence must converge. This means that the long ray is too long for sequences to diverge. Now we can also conclude that $L_+$ cannot be homeomorphic to $\mathbb{R}_+$: If we had

then instead of calculating the limit of $a_n:= n$ in $\mathbb{R}_+$, we could calculate the limit of $b_n:=f(a_n)$ in $L_+$. If $b:= \lim b_n$ in $L_+$, then $a:=f^{-1}(b)$ would be the limit of $a_n$ in $\mathbb{R}_+$. However, we know that this sequence does not converge in $\mathbb{R}_+$. Thus they cannot be homeomorphic.

I already hinted at the fact that the long line is in a sense the “longest” a line can be. What I mean by that is that spaces constructed via

with an ordinal $\kappa > \omega_1$ fail to be locally homeomorphic to $\mathbb{R}_+$. And if they do not even locally look like $\mathbb{R}_+$, in what sense is that space still a line? If $L_+^\kappa$ were locally homeomorphic to $\mathbb{R}_+$, then there must be a neighborhood $U$ of $\underline{\omega_1} \in L_+^\kappa$ and a homeomorphism $f\colon U \cong \mathbb{R}_+$. Since $\mathbb{R}_+$ is path-connected and $\underline0 \in U$, there is a path $\gamma$ from $f(\underline 0)$ to $f(\underline{\omega_1})$. Since $f^{-1} \circ \gamma$ is continuous, $[0,1]$ is compact, $L_+^\kappa$ is Hausdorff and we can assume that $\gamma$ is bijective, by (Munkres 2000, Th. 26.6) this would mean that $f^{-1}\circ \gamma$ is a homeomorphism. But since

the long ray would be homeomorphic to some subset of $[0,1]$. This must be of the form $[0,p)$ for $p \in (0,1]$. But since $[0,p)$ is homeomorphic to $\mathbb{R}_+$, we would get $L_+ \cong \mathbb{R}_+$. But we concluded earlier this cannot happen, so there cannot be any path from $\underline0$ to $\underline{\omega_1}$. After adding just one more copy of $[0,1)$ to $L_+$, it fails to be locally homeomorphic to $\mathbb{R}_+$. So the long ray is indeed the longest a ray can be.

The long line

So far, we have only talked about the long ray, although the article is supposed to be about the long line. It is rather easy to get the long line from two long rays: We can just glue them together at $\underline0$. Formally we can define this as

where $\sim$ is the smallest equivalence relation identifying both zeros, but nothing more. The intuition for this is literally taking two long rays and glueing them together at $\underline0$ with one pointing into the “positive” direction and the other in the “negative” direction.

We have seen many interesting properties of $L_+$, so which ones does the long line have? Basically all of them! This is because the subspace topology on $L_+$ given by $L$ is just the topology we were considering the whole time on it. In a similar way as for the long ray, we can see that the long line is locally homeomorphic to $\mathbb{R}$ (since it now does not have a smallest element). But it cannot be homeomorphic to $\mathbb{R}$, because then the long ray would have to be homeomorphic to $\mathbb{R}_+$. So it is first countable, but cannot be second countable (otherwise the long ray would have to be second countable as well). It is also sequentially compact by a very similar argument. So we can just claim all these results now because we did the work earlier.

A modified classification theorem

The topology-minded people might remember the classification theorem of connected $1$-manifolds. It states that up to homeomorphism the only connected $1$-manifolds are $\mathbb{R}$ and the circle $S^1$. Intuitively, connected $1$-manifolds are topological spaces that are locally homeomorphic to $\mathbb{R}$ and satisfy some additional criteria. You might notice that the long line is locally homeomorphic to $\mathbb{R}$, but not homeomorphic to either $\mathbb{R}$ or $S^1$, so it is not covered by this theorem. This is because manifolds are required to be second countable. After dropping this condition, there are up to homeomorphism four connected $1$-manifolds: $\mathbb{R}$, $S^1$, $L$ and the half-long line, which is $L_+$ glued together with $\mathbb{R}_+$ (Frolík 1962). So the long line does not only serve as a funny counterexample, but plays an important role in the classification of topological spaces.

Bibliography

Frolík, Zdeněk. 1962. “On the Classification of 1-Dimensional Manifolds.” Acta Universitatis Carolinae. Mathematica Et Physica 3 (1): 1–4. http://hdl.handle.net/10338.dmlcz/142137 .

Levy, Azriel. 2012. Basic Set Theory. Dover Books on Mathematics. Dover Publications.

Munkres, James R. 2000. Topology. 2nd ed. Prentice Hall, Inc.