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This is in reference to the long ray(Alexandroff line) which can be extended in both directions to form the so-called Long Line defined by:

$$ L_{1} = \{ \omega _{1} \times[0,1) \}\cup \{ \omega _{1} \times[-1, 0) \} $$

So the long plane would be defined as:

$$L_{1} \times L_{1}$$

A similar object is mentioned in the wikipedia article on the bagpipe theorem.

As for the "longer line"(or maybe I should call it the long long line) would be defined as:

$$L_{2} = \{ \lambda _{2} \times[0,1) \}\cup \{ \lambda _{2} \times (-1, 0] \}$$ where $ \lambda_{2} $ is a limit ordinal whose cardinality is $\beth_ {2}$ (we know such a thing exists as it belongs to a class of transitive sets know as L). So the idea something that inherits the order topology of the real line but has cardinality larger than the continuum. Is there is another name for the longer line as I defined it what is it referred to?

For reference, I found this post by Seewoo Lee describing what he called the "long long line" but it's defined differently and still has the same cardinality of the long line.

Editing Notes: As others have pointed out, the cardinality of $ \omega _{2}$ is only guaranteed to be larger than the continuum if the GCH holds in the model of set theory we are using. However, using transitive models and the powerset operation we can construct ordinals that have cardinality larger than the continuum in all models of ZF set theory. And so $\lambda _{2} =\omega _{2} \leftrightarrow GCH \ holds$.

Mr X
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    The problem is that any such attempt will lose the homogeneity so important to the long line itself: the point corresponding to $\omega_1$ will not have a neighborhood isomorphic to $\mathbb{R}$ (more snappily, lots of ordinals below $\omega_2$ have uncountable cofinality, and so any long line built off of $\omega_2$ will fail to be first-countable). In a sense, the long line is already as long as it can be if we want it to look the way it does locally. – Noah Schweber Apr 10 '22 at 20:11
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    (Also, the author of the linked post was Seewoo Lee; Asaf was simply the most recent editor.) – Noah Schweber Apr 10 '22 at 20:12
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    See also here. – Noah Schweber Apr 10 '22 at 20:16
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    Your longer line is not necessarily larger than the continuum — that's independent, depends on the size of the continuum. If $2^{\aleph_0} \ge \aleph_2$ then both the long line and the longer line are of size continuum; they're different only if CH. – BrianO Apr 10 '22 at 23:50
  • @BrianO The cardinality of the continuum is $2^{\aleph_0}$ . Now there is a model of set theory( Gödel's constructible universe) where the continuum hypothesis is true. And in that model known as L, $ \omega {2} = 2^{2^{\aleph{0}}}$. So we indeed can construct an ordinal whose cardinality is larger than the continuum – Mr X Apr 11 '22 at 00:53
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    To keep nitpicking, the second uncountable ordinal is $\omega_1+1$, $\omega_2$ is the second uncountable cardinal – Alessandro Codenotti Apr 11 '22 at 08:17
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    @MrX Nobody said you couldn’t find/construct an ordinal larger than the continuum; you can in ZFC. In every model, the long line has size continuum, and your longer line has size $\text{max}(\aleph_2, 2^{\aleph_0})$. – BrianO Apr 11 '22 at 18:16
  • @NoahSchweber thanks. Your link basically answered my question as it had been asked before unbeknownst to me. – Mr X Apr 19 '22 at 22:54

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Consider what happens in the point around $\omega_1$, i.e. a "neighborhood" like $[\Gamma_0, \omega_1 + 0.5)$. In a certain sense, the part of this interval from $[\Gamma_0, \omega_1)$ is "denser" than that of $[\omega_1, \omega_1 + 0.5)$. That is, this interval is not uniformly "dense".

What do I mean by "more dense"? In particular, you can have an uncountably long and not-eventually constant sequence accumulating at $\omega_1$ from the left, but in the real numbers, any uncountably long sequence accumulating at a point must be eventually constant, i.e. it actually reaches the point. So there are like "lots more" points "crammed against" $\omega_1$ on the left that the sequence can keep going into to avoid actually having to get there.

That doesn't mean the space isn't "real", of course. It just means it breaks one of the usual motivations for talking about the original long line which is that around every point it still looks like a segment of $\mathbb{R}$. This one, though, doesn't look that way around every point, as I just showed.

The_Sympathizer
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    The Dedekind completion of an ordered field will not necessarily be homogeneous. Every point of the original ordered field looks the same, but some of the points you add to get the completion can look different. (More generally, it is quite hard for a complete linear order to be homogeneous--completeness forces there to exist points of all sufficiently small cofinalities from below (or above)). – Eric Wofsey Apr 10 '22 at 21:37
  • @Eric Wofsey: Hmm. I was aware this was the case "metrically", but not topologically - e.g. the point $\mathbf{\epsilon}$ defined as the supremum of the infinitesimals, i.e. the biggest infinitesimal. It is notoriously "sticky" in that in the natural extension of addition, you can add any infinitesimal to it and it just gets sucked up, so the resulting addition is not a field operation. But I figured that without worrying about the algebra, it did not so much affect the topology. I'd like to see how you'd find a topological property of that point, though, that shows it different. – The_Sympathizer Apr 10 '22 at 23:03
  • That said, your point does make a lot of sense - there is a sense where you can say topological spaces can describe structures with "fat points", something I realized when trying to wonder about how to visualize certain finite topological spaces intuitively. So it makes sense that could reoccur in an infinite case, too. – The_Sympathizer Apr 10 '22 at 23:05
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    For instance, suppose your ordered field has uncountable cofinality at infinity. Then every point of the ordered field has uncountable cofinality from above and below. Now just take any increasing sequence of length $\omega$ (say, the natural numbers) and its supremum gives a point of the completion that has countable cofinality from at least one side. – Eric Wofsey Apr 10 '22 at 23:05
  • @Eric Wofsey : WOW! :D I am curious, then, does this mean there aren't any other homogeneous complete orders than the reals? – The_Sympathizer Apr 10 '22 at 23:07
  • Well, the long line is homogeneous. There are also some homogeneous complete orders that are not dense (e.g., the integers). But this idea does show a homogeneous complete linear order cannot have any bounded increasing or decreasing sequence of length $\omega_1$. – Eric Wofsey Apr 10 '22 at 23:18
  • @Eric Wofsey : Yeah, so the reals are "as dense as it gets", while still being homogeneous and complete, right? – The_Sympathizer Apr 10 '22 at 23:35
  • Well, that might depend on what you mean by "dense". I think it may be possible to have a homogeneous complete linear order that does not even locally embed in $\mathbb{R}$, though I suspect this would be independent of ZFC. This is probably closely related to the existence of Suslin lines (though those are not required to be homogeneous). – Eric Wofsey Apr 11 '22 at 01:17
  • Might be this paper by Akin and Hrbacek is helpful? Concerning density, $\eta_\alpha$-sets might be worth considering. But they might fail to homogenous. – Ulli Apr 11 '22 at 08:40