Fundamental Group of the Long Line Let $\omega_1$ be the first uncountable ordinal. The long
line $L$ is defined as the cartesian product of the first
uncountable ordinal $ \omega _{1}$ with the half-open interval
$ [0,1)$  equipped with the order topology that arises from
the lexicographical order on
$\omega _{1} \times [0,1)$. That means that of we
set by $"<_L"$ the lexicographic order on
$\omega_1 \times [0,1)$ then the topology on
$\omega_1 \times [0,1)$ is generated by "interval" base
$$ ((k_1,a), (k_2,b))=\{(k,x) \in \omega_1 \times [0,1) \vert \ \vert  
(k_1,a) <_L (k,x) <_L (k_2,b)$$
My question is if it's true that the fundamental group $\pi(L,l_0)$ of the long
line is trivial (let $l_0$ is any base point) or what's the reason? Although it seems to be a rather
natural question I nowhere found a serious discussion about this problem. My
intuition says that it should be trivial, but I'm not not sure.
My idea: I have to show that any loop $f: S^1 \to L$ is contractible.
My idea is to observe that one can find a finite number of "neighboured"
$a_1, a_2,..., a_n \in \omega_1$ and show that the image
of $f$ is contained in $\bigcup_{i=1}^n {a_i} \times [0,1) \subset L$ and
then because of finiteness of $a_1, a_2,..., a_n \in \omega_1$ one sees that
$\bigcup_{i=1}^n \{a_i\} \times [0,1) \cong [0,1)$ and this space is contractible.
Is the argument ok? My worries are based on my doubts if that's make
sense to say that we can fetch some "neighboured" $a_1, a_2,..., a_n \in \omega_1$.
What should "neighboured" here mean? If we believe in continuum hypothesis
then $\omega_1= \mathbb{R}$ and it is strage to say what does it mean that
two elements $a, b \in \mathbb{R}$ a "neighboured". If $a \neq b$ there is always
some other element between them, therefore I doubt that my argument above should
work, or not?
note that we were done if we could show that every loop $f: S^1 \to L$  must be contained in a subset of the form $\{r\} \times [0,1) \subset L$. This one is clearly contractible, so the loop would be also contractible. Is it possible to do it that way around?
Does somebody see how can I argue to conclude that the fundamental
group of $L$ is trivial (if that's really the case)?
 A: The key to understanding questions like this about the long line is the following observation.
Lemma: For any $x\in L$, the interval $I_x=\{y\in L:y\leq x\}$ is order-isomorphic to $[0,1]$.
Proof: Let $A$ be the set of $y\in L$ such that $(0,0)<y<x$ and the second coordinate of $y$ is rational.  Then $A$ is countable (since the first coordinate of $x$ is a countable ordinal so there are only countably many choices for the first coordinate of $y$) and densely ordered without endpoints, so it is order-isomorphic to $(0,1)\cap\mathbb{Q}$.  Also, $A$ is dense in $I_x$ and $I_x$ is Dedekind-complete (and has endpoints), so $I_x$ is the Dedekind-completion of $A$ (including endpoints).  So, $I_x$ is order-isomorphic to the Dedekind completion (including endpoints) of $(0,1)\cap\mathbb{Q}$, which is just $[0,1]$. $\blacksquare$
This Lemma is really the reason that $L$ is called the "long line".  If you cut it off at any point before the end, it really does just look like the ordinary real line.  To see that it is different, you have to go all the way "uncountably far" to the end.
In particular, it follows that $I_x$ is homeomorphic to $[0,1]$ and contractible.  But any continuous map from a separable space to $L$ has image contained in $I_x$ for some $x\in L$ (take $x$ to be an upper bound for the image of a countable dense subspace).  So, any continuous map from a separable space to $L$ is nullhomotopic.  In particular, this implies all the homotopy groups of $L$ are trivial.
(Note that the Lemma also implies that the image of a loop in $L$ does not have to be covered by finitely many sets of the form $\{a\}\times[0,1)$.  Indeed, for any infinite ordinal $\alpha<\omega_1$, the interval $I_{(\alpha,0)}$ contains infinitely many sets of the form $\{a\}\times[0,1)$, but can be covered by a single loop since it is homeomorphic to $[0,1]$.)
A: Yes, it suffices to show that the image of any continuous map $f:S^1\rightarrow L$ is bounded, i.e., contained in $\alpha\times[0,1)$ for some countable $\alpha$, and then use that $\alpha\times[0,1)\simeq[0,1)$. This is because $L$ has the so-called $\omega$-compactness property: the closure of any countable set is compact, and also $S^1$ is separable (actually even compact as pointed out in the comment).
