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.



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> 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?
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A good survey about the latter spaces can be found in the [wikipedia entry](https://en.wikipedia.org/wiki/Long_line_(topology)). 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.



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


