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?
[Math] The construction of the Long Line topology
general-topology
Related Solutions
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.
The key to understanding questions like this about the long line is the following observation.
Lemma: For any $x\in L$, the interval $I_x=\{y\in L:y\leq x\}$ is order-isomorphic to $[0,1]$.
Proof: Let $A$ be the set of $y\in L$ such that $(0,0)<y<x$ and the second coordinate of $y$ is rational. Then $A$ is countable (since the first coordinate of $x$ is a countable ordinal so there are only countably many choices for the first coordinate of $y$) and densely ordered without endpoints, so it is order-isomorphic to $(0,1)\cap\mathbb{Q}$. Also, $A$ is dense in $I_x$ and $I_x$ is Dedekind-complete (and has endpoints), so $I_x$ is the Dedekind-completion of $A$ (including endpoints). So, $I_x$ is order-isomorphic to the Dedekind completion (including endpoints) of $(0,1)\cap\mathbb{Q}$, which is just $[0,1]$. $\blacksquare$
This Lemma is really the reason that $L$ is called the "long line". If you cut it off at any point before the end, it really does just look like the ordinary real line. To see that it is different, you have to go all the way "uncountably far" to the end.
In particular, it follows that $I_x$ is homeomorphic to $[0,1]$ and contractible. But any continuous map from a separable space to $L$ has image contained in $I_x$ for some $x\in L$ (take $x$ to be an upper bound for the image of a countable dense subspace). So, any continuous map from a separable space to $L$ is nullhomotopic. In particular, this implies all the homotopy groups of $L$ are trivial.
(Note that the Lemma also implies that the image of a loop in $L$ does not have to be covered by finitely many sets of the form $\{a\}\times[0,1)$. Indeed, for any infinite ordinal $\alpha<\omega_1$, the interval $I_{(\alpha,0)}$ contains infinitely many sets of the form $\{a\}\times[0,1)$, but can be covered by a single loop since it is homeomorphic to $[0,1]$.)
Best Answer
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 $V_0$.