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The long line, and long ray, is defined with respect to $\omega_1$, the first uncountable ordinal. Is it possible to construct a long line with any of the larger uncountable ordinals, a very long line, if you will?

Would it still be a manifold, if so? From what I've seen, it is generally said that there are only 4 (Hausdorff) 1 dimensional manifolds. The line, circle, long ray and long line. But if a very long line existed, I would assume that it wouldn't be homeomorphic to the long line, since they would have a different cardinality.

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No, the long line is the longest possible line :). Or rather, you can define a long line $L_\lambda$

The issue is uncountable cofinality. The ordinal $\omega_1$ has the property that no countable sequence of countable ordinals has it as a limit (put another way: the countable union of countable sets is countable). This means that if we define the long line $L_\lambda$ of length $\lambda$ for $\lambda>\omega_1$ ($L_\lambda$ consists of $[0, 1)\times\lambda\times 2$ with the appropriate (non-product) topology), then no open neighborhood of the point $p=(0, \omega_1, 1)\in L_\lambda$ is homeomorphic to any open subset of $\mathbb{R}^1$, hence $L_\lambda$ is not a manifold. This is because $\mathbb{R}^1$ is separable; more specifically, because for any $q\in U\subseteq\mathbb{R}$ with $U$ open, there is a sequence $r_0, r_1, r_2, ...$ of points from $U$ which approach $q$. Identically, the open long ray of length $\lambda$ is not a manifold.

This is interesting, because it shows that the real obstacle isn't cardinality, but cofinality. A big theme in modern set theory is that cofinality is in many ways more fundamental than cardinality. (Note that there are uncountable ordinals of countable cofinality - $\omega_1+\omega$ is a silly example, and a more interesting one would be the cardinal $\omega_\omega$.)

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