# Can the definition of "The Long Line" be clarified?

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 $$(0,1)$$?

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

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.

• "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? May 30, 2020 at 9:46
• @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. May 30, 2020 at 9:53
• 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]$and "identifying" the upper end of one with the lower end of the other and replacing each with a single (ultimately arbitrary) ordinal. Your second comment went so far over my head I didn't even hear it whistle. I have trouble understanding complex aggregations of compound statements linked together as run-on sentences. May 30, 2020 at 10:07 • I'm sorry Henno but every point you make confuses me even more. I said nothing about$[0,1], \alpha, [0,1], \alpha+1$and I don't get what you mean by "$[0,1]$copies". May 30, 2020 at 10:19 • @PatrickR There is no end, so what would "at the end" even mean? Indeed, all$[0,\alpha)$are homeomorphic to$[0,\infty)$when$\alpha < \Omega$. It's quite a weird space that way. May 30, 2020 at 21:52 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. • 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$\Omega-1$? – bof May 30, 2020 at 4:03 • @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 May 30, 2020 at 9:26 • @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. May 30, 2020 at 9:39 • no element of the long line has an immediate predecessor. It's a continuum. @PrimeMover is quite right in his objections. May 30, 2020 at 10:02 • @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\$.
– bof
May 30, 2020 at 10:43