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In Steen and Seebach's "Counterexamples in Topology", we see the definition of the Long Line (counterexample 45).

"The long line $L$ is constructed from the ordinal space $[0, \Omega)$ (where $\Omega$ is the least uncountable ordinal) by placing between each ordinal $\alpha$ and its successor $\alpha + 1$ a copy of the unit interval $I = (0,1)$. $L$ is then linearly ordered, and we give it the order topology."

Having given this a bit of thought, I need clarifying the following.

Are the ordinals $0, 1, 2, \ldots, \alpha, \alpha + 1, \ldots$ part of the space, or is $L$ just $\Omega$ instances of $(0,1)$ concatenated? If the latter, then it appears there may be a homeomorphism between $L$ and $[0,\Omega) \times (0,1)$ under the lexicographic ordering. If the former, then it is very much less simple.

So is $L$ like: $0, (0,1), 1, (0,1), 2, (0,1), \ldots, (0,1), \alpha, (0,1), \alpha + 1, (0,1), \ldots, (0,1), \Omega-1, (0,1)$

or is it like:

$(0,1), (0,1), (0,1), \ldots, (0,1), (0,1), (0,1), \ldots, (0,1), (0,1)$

with $\Omega$ instances of $(0,1)$?

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    $\begingroup$ The former but I don't know what your $\Omega-1$ is supposed to be. $\endgroup$
    – bof
    May 30, 2020 at 0:38
  • $\begingroup$ The former: it's a generalization of the real (semi-)line $\Bbb R_{\ge 0}=[0,\omega)$. $\endgroup$
    – Berci
    May 30, 2020 at 0:40
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    $\begingroup$ Seems to me $[0,\Omega)\times[0,1)$ is at least as simple as $[0,\Omega)\times(0,1)$. $\endgroup$
    – bof
    May 30, 2020 at 0:40
  • $\begingroup$ But $[0,1)$ is not the same thing as $(n, (0, 1) )$ where $n \ne 0$. Sorry, what am I missing? $\endgroup$ May 30, 2020 at 9:23
  • $\begingroup$ $L$ is order-isomorphic to the lexicographic order on $\Omega\times \Bbb R.$ $\endgroup$ May 30, 2020 at 11:12

2 Answers 2

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A better (IMO) description of the long line is $[0,\Omega) \times [0,1)$ ordered lexicographically: $$(\alpha, t) \le_L (\beta, u) \iff (\alpha < \beta) \lor \left(\alpha=\beta \land (t \le u)\right)$$ and then given the order topology (with basic elements all open intervals plus all right-open intervals of the form $[(0,0), (\alpha, t) \rangle$ (special case for the minimum, there is no maximum). This is also what Munkres does (he has more attention for ordered spaces, and it's one of his exercises (2nd edition, § 24, ex. 6) that a well-ordered set (like $[0,\Omega)$) times $[0,1)$ is a linear continuum (i.e. connected) in the lexicographic order topology.

So the minimum is $(0,0)$ and we start with a usual interval $[(0,0), (1,0)]\simeq [0,1]$, so no gaps or jumps. Up to $(\omega,0)$, it's just $[0,\infty)$, essentially, and there is no gap between that and $(\omega,0)$. Locally (in neighbourhoods of points) things look like $\Bbb R$. It only goes on for longer (it's no longer separable, or Lindelöf).

The S&S description is (I think) meant as $$X=[0,\Omega) \cup \bigcup_{\alpha < \Omega} I_\alpha$$

where each $I_\alpha$ is a disjoint copy of $(0,1)$ and the order within each $I_\alpha$ is the usual one, the order on $[0,\Omega)$ is the usual well-order among ordinals, and if $x \neq y$ belong to distinct intervals $I_\alpha, I_\beta$, the order of $\alpha$ and $beta$ alone determines which is smaller (so if $x \in I_\alpha$ and $\alpha < \beta$, then $x< y$. (Each $I_\alpha$ is the copy of $(0,1)$ between $\alpha < \alpha+1$ for each $\alpha$), so if $\alpha \in [0,\Omega)$ and $x \in I_\beta$ with $\beta > \alpha$, $x > \beta > \alpha$. So we can define all order relations for a linear order. The nice thing about the equivalent $\le_L$ is that general theory already implies this is a linear order, and we don't need to do case distinctions based on what kind of point (ordinal or interval point) we have, and the linear continuum fact is quite general. $\omega_1$ embeds as a closed subset in $X$ either way.

So it's like the first description you gave, not the second, long story short.

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  • $\begingroup$ "we glue together the 1 of an interval to the 0 of the next one," -- and in the S&S definition of this concept, the "gluing" process consists of replacing that $1$ point of overlap in successive unit intervals with successive ordinals? $\endgroup$ May 30, 2020 at 9:46
  • $\begingroup$ @PrimeMover There is no overlap in S&S. Any point of a copy of $(0,1)$ is strictly in-between its "neighbour" ordinals. In my model, the original ordinals are $(\alpha,0)$, and $(\alpha,1)$ does not occur too. In the S&S description at limits, the limit is still larger than all intervals and ordinals before it. $\endgroup$ May 30, 2020 at 9:53
  • $\begingroup$ By "overlap" I was trying to understand what you meant by "gluing". I was assuming you meant you were modelling $\ldots (0, 1), \alpha, (0, 1), \ldots$ by sticking successive copies of $[0,1]$ to $[0, 1]$ to [0,1]$ and "identifying" the upper end of one with the lower end of the other and replacing each with a single (ultimately arbitrary) ordinal. Your second comment went so far over my head I didn't even hear it whistle. I have trouble understanding complex aggregations of compound statements linked together as run-on sentences. $\endgroup$ May 30, 2020 at 10:07
  • $\begingroup$ I'm sorry Henno but every point you make confuses me even more. I said nothing about $[0,1], \alpha, [0,1], \alpha+1$ and I don't get what you mean by "$[0,1]$ copies". $\endgroup$ May 30, 2020 at 10:19
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    $\begingroup$ @PatrickR There is no end, so what would "at the end" even mean? Indeed, all $[0,\alpha)$ are homeomorphic to $[0,\infty)$ when $\alpha < \Omega$. It's quite a weird space that way. $\endgroup$ May 30, 2020 at 21:52
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It is like your first $$0, (0,1), 1, (0,1), 2, (0,1), \ldots, (0,1), \alpha, (0,1), \alpha + 1, (0,1), \ldots, (0,1),\ldots$$ except there is no $\Omega-1$ as $\Omega$ is a limit ordinal. which, in typography, is not too different from your second, because it is $$[0,1),[0,1),[0,1),[0,1),[0,1),[0,1),[0,1),[0,1),\ldots$$ the point is that there are uncountably many pieces. In your second there is only one piece, so it is order isomorphic to the real line.

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    $\begingroup$ As far as I can see, no point has an immediate predecessor or an immediate successor. Locally it's just like $\mathbb R$ except at the left endpoint. And what's that $\Omega-1$? $\endgroup$
    – bof
    May 30, 2020 at 4:03
  • $\begingroup$ @bof $\Omega - 1$ is the ordinal before $\Omega$. Such concepts are discussed in the various works I have on my shelf which discuss transfinite ordinals. I don't understand what it means, but then to paraphrase the words of Von Neumann you don't understand things in maths, you just get used to them $\endgroup$ May 30, 2020 at 9:26
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    $\begingroup$ @Ross Millikan Actually now I think about it, I don't understand about "uncountably many elements that do not have immediate predecessors". So which elements of $L$ do have immediate predecessors? I'm lost. $\endgroup$ May 30, 2020 at 9:39
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    $\begingroup$ no element of the long line has an immediate predecessor. It's a continuum. @PrimeMover is quite right in his objections. $\endgroup$ May 30, 2020 at 10:02
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    $\begingroup$ @PrimeMover I wondered if you were going to bring up Conway's surreal numbers. Sorry, Conway's $\Omega-1$ is not an ordinal. There is no ordinal number just before $\Omega$. $\endgroup$
    – bof
    May 30, 2020 at 10:43

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