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.

  • 3
    $\begingroup$ 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). $\endgroup$ Apr 17, 2010 at 16:34
  • 6
    $\begingroup$ 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". $\endgroup$ Apr 17, 2010 at 16:43
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 17, 2010 at 17:09
  • 5
    $\begingroup$ 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. $\endgroup$ Apr 17, 2010 at 17:17
  • $\begingroup$ @Qiaochu: there are non-isomorphic orders, whose order topologies are homeomorphic. @Robin: alright! $\endgroup$ Apr 17, 2010 at 17:46

1 Answer 1


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. Unless $Y$ is a circle, 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 closed subset $D \subset X$ 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 closed subset of $X$. This dictates what well-ordered closed subsets $D$ that are cofinal in $X$ look like.

  • If $\xi = \omega_1 + 1$, then any closed 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 closed 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.

  • $\begingroup$ 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 $\omega$. $\endgroup$ Sep 8, 2012 at 11:54
  • $\begingroup$ @Andreas: oops, my bad (and thanks). I hope it's fixed now. $\endgroup$
    – Todd Trimble
    Sep 8, 2012 at 14:04
  • 1
    $\begingroup$ @ToddTrimble: but how do you know that $S^1$ is the only 1-manifold which satisfies this property, that removing a point leaves one connected component? That seems like the most important part of the classification to me. $\endgroup$
    – ziggurism
    Nov 7, 2013 at 0:47
  • 1
    $\begingroup$ @ziggurism This is a sketch; I think Gale's article would have more details. There is a neighborhood of the point $y$ to be removed that is homeomorphic to a closed interval $I$. Let $x, z$ be the two endpoints of the interval. Now remove $y$; the result is connected, but manifolds are locally path connected, so the result is path connected. Hence there is a path $J$ connecting $x$ and $z$ in the complement of the point (and containing no interior point of $I$). I claim $I \cup J$ is the whole original manifold. But this union of two intervals is homeomorphic to $S^1$. $\endgroup$
    – Todd Trimble
    Nov 7, 2013 at 11:52
  • 1
    $\begingroup$ @ziggurism In any case, the point is that Gerald Edgar's point in his comment is applicable to this case: such a manifold has to be second-countable, and so we can just invoke the usual classification of second-countable Hausdorff connected topological 1-manifolds (as in Gale's article). $\endgroup$
    – Todd Trimble
    Nov 7, 2013 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.