## Classification of 1-dimensional manifolds (not second-countable)

It is easy to see that every connected $1$-dimensional second-countable manifold (that is, what is often called just a manifold) is either homeomorphic to $\mathbb{R}$ or to $S^1$. Now let's drop the secound-countable-condition.

How do you prove that every connected $1$-dimensional manifold homeomorphic to $\mathbb{R}, S^1$, the long line or the long ray? And why are the long line and the long ray not homeomorphic?

A good survey about the latter spaces can be found in the wikipedia entry. Basically, a long ray is built up of $\omega_1$-many intervals pasted together, and the the long line consists of two long rays in both directions.

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I'm having trouble finding a reference for the long circle (it's not listed in the wikipedia entry, and I'm not sure how to interpret your last sentence). – Alison Miller Apr 17 2010 at 16:34
Compact and first-countable implies second-countable, right? So if you try a "long circle" the point at infinity won't have a local neighborhood such as the one required for a "manifold". – Gerald Edgar Apr 17 2010 at 16:43
thanks for the comments. I changed my question. – Martin Brandenburg Apr 17 2010 at 16:48
As the Wikipedia article states, an increasing sequence in the long ray converges; this isn't true of a decreasing sequence, so the two ends behave differently. – Qiaochu Yuan Apr 17 2010 at 17:09
Qiaochu, arguably a simpler way to see that the (1-sided) long ray is not homeomorphic to the (2-sided) long line is that removing a point from the former always creates one paracompact component, but removing a point from the latter never does. – Robin Chapman Apr 17 2010 at 17:17
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Here is a response to the first boxed question (the second was already answered by Robin Chapman). (Much belated of course, but I only just saw this question.)

Suppose that $Y$ is a connected (nonempty) topological 1-manifold without boundary; let $y$ be a point. As observed in comments, the complement $Y - \{y\}$ has two open connected components $U$ and $V$, and $Y$ can be reconstructed by gluing together $U \cup \{y\}$ and $V \cup \{y\}$, which are 1-manifolds with one boundary point each.

I found it technically easier to analyze the possibilities for such connected 1-manifolds with (at least) one boundary point. Recall that a 1-manifold with boundary is a topological space where every point has a neighborhood homeomorphic to an open subset of the interval $[0, 1]$. In conjunction with the gluing above, it suffices to establish the following result.

Theorem: Suppose $X$ is a connected 1-manifold with at least one boundary point. Then $X$ is homeomorphic to one of the following types of spaces:

• A closed interval $[0, 1]$.

• A half-open interval (homeomorphic to $\mathbb{R}_{\geq 0}$).

• A long half-open ray.

(I should say right away that a fully rigorous proof, with all i's dotted and t's crossed, would be somewhat lengthy. So I will content myself with a proof outline. See also reference [1], which should help fill most if not all the gaps.)

Proof: Observe that $X$ is path-connected, since it is connected and locally path-connected.

Let $0$ denote a boundary point, and order $X$ as follows: say $x \lt y$ if $x$ and $0$ belong to the same path component of $X - \{y\}$. It is not hard to show that $X$ is linearly ordered under $\lt$, with bottom element $0$. Every interval $[0, x]$ is a compact connected manifold with two endpoints (compact because there is a path from $0$ to $x$), and thus homeomorphic to the standard interval.

Suppose a subset $D \subset X$ (which we will assume is closed) is well-ordered under the order it inherits from $X$. The order type of such $D$ must be $\omega_1$ (the first uncountable ordinal) or less. For otherwise, there would be an initial segment $S$ of $D$ of order type $\omega_1 + 1$. In that case, if $s$ is the top element of $S$, the interval $[0, s)$, which is homeomorphic to $\mathbb{R}_{\geq 0}$, would contain $\omega_1$ as a suborder -- but this is absurd since $\mathbb{R}_{\geq 0}$ has a countable cofinal set.

We can now classify the possibilities for $X$, according to the smallest ordinal $\xi$ which does not occur as a well-ordered subset of $X$. This dictates what well-ordered subsets $D$ that are cofinal in $X$ look like.

• If $\xi = \omega_1 + 1$, then any well-ordered cofinal $D$ must be of type $\omega_1$, and $X$ is a topological union (a directed colimit) of open sets $[0, d)$ where $d$ ranges over $D$. This union is homeomorphic to a long half-open ray.

• If $\xi = \omega_1$, then any well-ordered cofinal $D$ is countable. This forces $X$ to be homeomorphic to $\mathbb{R}_{\geq 0}$.

(For an easy induction argument shows that for any countable ordinal $\alpha$, the lexicographically ordered set $\alpha \times [0, 1)$ with the order topology is homeomorphic to $\mathbb{R}_{\geq 0}$).

• If $\xi = \omega_0$, then $X$ is homeomorphic to $[0, 1]$.

(End of proof)

[1] David Gale, The Classification of 1-Manifolds: A Take-Home Exam, Amer. Math. Monthly, Vol. 94 No. 2 (February 1987), 170-175.

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 Thanks, Todd! – Martin Brandenburg Sep 8 at 10:12 I don't understand the argument showing the every discrete subset $D$ of your space is well-ordered. Even when $X$ is $[0,1]$, there is a decreasing, discrete $\omega$-sequence. Perhaps you intended $D$ to also be closed? Also, this argument (applied to the special case $S=D$) seems to say that each point $s$ in such a $D$ has only finitely many predecessors, which would mean that $D$ has order-type at most $\omega$. – Andreas Blass Sep 8 at 11:54 @Andreas: oops, my bad (and thanks). I hope it's fixed now. – Todd Trimble Sep 8 at 14:04