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

Question

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 $(0,1)$?

Answer 1

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.

Answer 2

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.