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The Long line topology is constructed from the ordinal space $[o,\omega_1)$ ( where $\omega_1$ is the least uncountable ordinal) by placing between each ordinal $\alpha$ and its successor $\alpha + 1$ a copy of the unit interval $I=(0,1)$. But in the Engelking's book, the Long Line topology defined on the set $V_{0}=W_{0}\times [0,1)$ where $ W_{0}$ is the set of all countable ordinal numbers. But why these are different ? Could you help me please?

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In the most usual formulation of set theory, $\omega_1$ is the set of all countable ordinals; so the two definitions of the long line are equivalent.

Squeezing an interval between each countable ordinal is equivalent to turning each countable ordinal into a half-open interval, and the set you get is the union of $\omega_1$-many disjoint half-open intervals, which is essentially the same as your set $V_0$.

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I think that we can not write this space as a disjoint union of two nonempty subsets. But, is there any dense subset of the long line topology? – ghb Jan 13 '14 at 14:13
Just commenting to add that Engelking's text is completely agnostic about what ordinal numbers actually are, whereas in set theory we have generally decided that ordinal numbers are von Neumann ordinals. Because of this agnosticism, Engelking must use different notations for the set of all countable ordinals and the first uncountable ordinal $\omega_1$. – inactive... for now Jan 13 '14 at 14:13
@ghb: You can't write it as the disjoint union of two nonempty subspaces; you can write any space (or set) with more than one element as a disjoint union of two nonempty subsets. – Clive Newstead Jan 13 '14 at 14:16
@ArthurFischer: Sure. What's important is that the cardinalities of the two sets are the same, and they both have a natural well-ordering of length $\omega_1$. – Clive Newstead Jan 13 '14 at 14:17
Is there any dense subsets? Can I take the real line as a uncountable ordered space? – ghb Jan 13 '14 at 14:20

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