By Seewoo Lee

2019-02-05 05:40:40 8 Comments

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).


@Hanul Jeon 2019-02-05 08:35:36

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.

@Dk-ium 2019-02-05 07:07:46

We are dealing with $S\times \mathbb{R}$ with the topology given by the base $\{a \times I : a \in S, I$ : open in $\mathbb{R}\}$. I claim that such $S_{1} \times \mathbb{R}$ and another $S_{2} \times \mathbb{R}$ if and only if $|S_{1}| = |S_{2}|$.

Indeed, consider a (set-theoretical) bjection $h : S_{1} \rightarrow S_{2}$, and gather all the homeomorephisms between $a \times \mathbb{R}$ and $h(a) \times \mathbb{R}$ for $a \in S_{1}$ in order to construct entire homeomorphism.

Definitely, it may disturb the "order" of the components, but it does not matter when you focus on topological properties. If one focuses on the equivalence by order-preserving maps, it becomes entirely different story.

@Eric Wofsey 2019-02-05 07:10:00

Note that saying each $a\times\mathbb{R}$ becomes a component doesn't actually determine the topology; you want to say more specifically that $S\times\mathbb{R}$ is topologized as the disjoint union of the subsets $a\times\mathbb{R}$.

@Dk-ium 2019-02-05 07:11:31

@EricWofsey Yes, precisely! Thanks for your comment, and I will elucidate more!

@Eric Wofsey 2019-02-05 07:15:15

My point wasn't that you have to say $a\times\mathbb{R}$ has the usual topology, but that you have to actually specify the entire topology of $S\times\mathbb{R}$. Consider for instance the product topology on $\mathbb{Q}\times\mathbb{R}$ whose components are $a\times\mathbb{R}$ but it is not the disjoint union of its components.

@Dk-ium 2019-02-05 07:20:51

@EricWofsey Now I got your point. My intuition was too naive, and there goes another amendment...

@SolveIt 2019-02-05 06:08:23

Edit: This answer doesn't apply to OP's question, which uses a nonstandard definition of long line. I had glossed over the definition assuming they would be equivalent. Mea culpa.

I think the answer to 1) is no. The largest ordinal you can embed into the long line is $\omega_1$ as per Asaf Karagila's answer in What uncountable ordinals live in the long line? , but you can easily embed larger ordinals in your long long line. Extending this line of thinking, we have a positive answer to 3).

@SolveIt 2019-02-05 08:25:58

And you're right of course. Never mind. I'm just nuking the whole thing and giving up on maths for the day.

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