There is a so-called long line in topology, which is a topological space with a base set $[0,1]\times \mathbb{R}$ with order topology given by lexicographic order: $(x_{1}, y_{1}) < (x_{2}, y_{2})$ if and only if $x_{1} < x_{2}$ or $x_{1} = x_{2}$ and $y_{1} < y_{2}$. Here are some properties of long line. (Actually, definition of long line in the link is more general than the definition above completely different, and we will use the above definition.)
Now let $L_{1}$ be the above long line. Define long long line $L_{2}$ as a topological space with a base set $[0, 1]\times L_{1}$ with order topology is given by lexicographic order as before. We can embed $L_{1}$ in $L_{2}$ as $L_{1} \simeq \{1/2\}\times L_{1}\hookrightarrow L_{2}$. My question is the following:
(1) Are $L_{1}$ and $L_{2}$ are homeomorphic?
(2) If not, we can also define a $long^{3}$ line $L_{3}$ in similar way, and even $L_{n}$ for any $n\geq 1$. What is a direct limit
$$
L_{\omega} = \lim_{\to}L_{n}?
$$
Is there any interesting property of $L_{\omega}$?
(3) If we proceed more, then we can also define $$L_{\omega + 1}, L_{\omega +2}, \dots, L_{2\omega}, \dots, L_{\omega^{2}}, \dots, L_{\omega^{\omega}}, \dots, L_{\epsilon_{0}}, \dots$$
for any given ordinals $\alpha$. Are they all different?
I think the most important question is (1). Thanks in advance.
Since the definition of my long line is different from the original one, here's the new version of questions for original definition.
First, we have the classical (original?) long ray $R_{1}$, which is a set $\omega_{1} \times [0, 1)$ with an order topology via lexicographic order.
Now we can define a long line $L_{1}$ by gluing two long rays $R_{1}$ with respect to their endpoints.
To define $L_{2}$, we define $R_{2}$ by $\omega_{1} \times R_{1}$ with an order topology (lexicographic order again) and glue two copies. By continuing this process, we can define $L_{\alpha}$ for any given ordinal $\alpha$ (I hope).
Best Answer
Let me redefine $L_2$ as a product of $\omega_1$ and the long line with lexicographical order, which seems more interesting. Note that $L_2$ is isomorphic to $\omega_1^2\times[0,1)$.
We can prove that both spaces are dense complete linear ordered set under the natural order. Completeness seems not trivial so I should sketch a proof of completeness of $I=\mu\times[0,1)$ for a limit ordinal $\mu$.
Divide $I$ into sets $I_{\alpha}=\{ \langle \alpha, r\rangle :r\in [0,1)\}$. Consider a subset $A$ of $I$ which is bounded above, so $A$ is contained in some $[(0,0), (\nu,0)]$ for some $\nu<\mu$.
Consider $B=\{\alpha\le\nu: I_\alpha\cap A\neq\varnothing\}$ and take $\beta=\sup B$. If $\beta\in B$, then finding a supremum of $A$ is reduced to finding a supremum of a set of $[0,1)$. If not, the point $(\beta,0)$ will be a supremum of $B$. (Added in Feb 10, 2019: we need to divide cases once more: because $\beta\in B$ does not guarantee $I_\beta$ has a supremum. If $r:=\sup I_\beta\in [0,1)$, $\langle\beta, r\rangle $ is a supremum of $A$. If $r=1$, $\langle\beta+1,0\rangle $ would be.)
Hence every connected subset of $I$ is an interval (see ยง24 of Munkres' Topology.) We can see that every compact interval in $L_1$ is separable. However $L_2$ is not.