What are the charts that make up an atlas for the long line? This question is prompted while I was working through the MIT OCW (Massachusetts Institute of Technology, Open CourseWare) for 18.965, ``Geometry of Manifolds,'' in its Lecture 2, 
http://ocw.mit.edu/courses/mathematics/18-965-geometry-of-manifolds-fall-2004/lecture-notes/lecture2.pdf
Near verbatim, the setup for the example of a long line from there is this:
Let $S_{\Omega}$ denote the smallest uncountable totally ordered set.
Consider the product $X = S_{\Omega} \times (0,1]$ with dictionary order topology.
Give $X$ charts as follows.
$\forall \, (\omega, t) \in X$, \
if $t\neq 1$, let $U_{(\omega,t)} = \lbrace \omega \rbrace \times (0,1)$ and
$\begin{aligned}                                                                                              
  & \phi_{(\omega,t)}: U \to \mathbb{R} \\                                                                    
& \phi_{(\omega,t)}(\omega,t) = t                                                                             
\end{aligned}$
If $t=1$, ``let $S(\omega)$ denote the successor of $\omega$.''
Set $U_{(\omega,1)} = \lbrace \omega \rbrace \times (0,1]\text{sup}\lbrace S(\omega) \rbrace \times (0,1)$ and
$
\phi_{(\omega,t)}(\eta, t) = \begin{cases} t & \text{ if } \eta = \omega \\
  t + 1 & \text{ if } \eta = S(\omega) \end{cases}
$
My questions are the following: what is the domain for that second chart, $(\phi_{(\omega,t)}, U_{(\omega,1)}\
)$? It's unclear to me if this is a union or what's going on with the supremum for $S(\omega)$.
Also, in this context, could someone give an example of what it means to be a successor for $\omega$, to clarify what it means, and in general?
Thanks, and for those working through MIT OCW 18.965 or are seriously interested in learning differential topology online, I'm keeping a blog with my progress and some, hopefully helpful, thoughts, discussions, and solutions at ernestyalumni.wordpress.com.    
Mrowka, Tomasz. 18.965 Geometry of Manifolds, Fall 2004. (MIT OpenCourseWare: Massachusetts Institute of Technology), http://ocw.mit.edu/courses/mathematics/18-965-geometry-of-manifolds-fall-2004} (Accessed 29 Nov, 2014). License: Creative Commons BY-NC-SA
 A: As indicated in the first comment above, the set $S_\Omega$ is well ordered. Modern notation usually has $\omega$ as the first infinite ordinal, and $\omega_1$ as the first uncountable ordinal (same as $S_\Omega$), but I will stick with the notation as used in the above question (in which $\omega$ may be an arbitrary element of $S_\Omega$). 
For any $\omega\in S_\Omega$ consider the set $\{\xi\in S_\Omega : \xi > \omega \}$. Since this set is non-empty (since $S_\Omega$ has no last element), and since $S_\Omega$ is well-ordered, the set $\{\xi\in S_\Omega : \xi > \omega \}$ must have a minimal element, and this is what is denoted by $S(\omega)$ in your question, the successor of $\omega$. So, $S(\omega)=\min\{\xi\in S_\Omega : \xi > \omega \}$. 
For the case when $t=1$ on one hand you have the half-open interval $I=\lbrace \omega \rbrace \times (0,1]$ and following it you have the open interval $J=\lbrace S(\omega) \rbrace \times (0,1)$. Note that in the dictionary order if $U=I\cup J$, then $U$ itself may be thought of as an interval, namely if $a=\inf I$, $b=\sup I = \max I = \langle \omega,1 \rangle$, $c=\inf J$, and $d=\sup J$, then $U=(a,d)$, but also $U=(a,b]\cup (c,d)$. In fact if you unravel the definition of $\inf$ you would realize that $c=b$, so when you look at $U=(a,b]\cup (c,d)$ there is no gap in between $b$ and $c$, so you could really think of it as just one interval $U=(a,d)$ where 
$a=\inf\Bigl(\lbrace \omega \rbrace \times (0,1]\Bigl)$, $d=\sup\Bigl(\lbrace S(\omega) \rbrace \times (0,1)\Bigr)=\langle S(S(\omega)),0 \rangle$ (think of the latter equality), and this interval is a "usual" neighborhood of $b=c=\max \Bigl(\lbrace \omega \rbrace \times (0,1]\Bigl)= \langle S(\omega),0 \rangle= \inf\Bigl(\lbrace S(\omega) \rbrace \times (0,1)\Bigl)$ which is "in the middle" if the interval. (If some part of this is unclear please keep reading.) 
What $\phi_{(\omega,t)}$ does is that it sends the interval $I=\lbrace \omega \rbrace \times (0,1]$ onto the interval $(0,1]$ in the real line, and it sends 
the interval $J=\lbrace S(\omega) \rbrace \times (0,1)$ onto $(1,2)$ in the real line. Of course $(0,1]\cup (1,2) = (0,2)$ is a usual neighborhood of the number $1$ in the real line. So the map $\phi_{(\omega,t)}$ makes it so that $(a,d)$ (in the notation introduced above) is a usual neighborhood of $b=c=
\langle \omega, 1 \rangle = \langle S(\omega), 0 \rangle$. I am abusing notation since $\langle S(\omega), 0 \rangle$ was never introduced formally in the definition in your question, since $\langle S(\omega), 0 \rangle \not\in \lbrace S(\omega) \rbrace \times (0,1]$,
but you could think of $\langle \omega, 1 \rangle$ as what 
$\langle S(\omega), 0 \rangle$ "should be". 
