What does order topology over Ordinal numbers look like, and how does it work? Space of ordinal numbers are one of the favorite examples of my professor in general topology. I quite understand the idea at the base of ordinal numbers (few things or nothing about the concept of ordinal itself in set theory). What we were told was just about constructing bigger and bigger ordinals like putting boxes into boxes into boxes...and that starting at a certain ordinal number, there are finite steps backwards and infinite forward.
So I still don't have a clear what a finite and what a successor ordinal are, I know the definitions and I have a vague idea, but I can't link it in a successful way with the proposition I reported at the end of the previous paragraph.
If somebody could explain these concepts in a better way, maybe then I could understand how to verify or disprove base-numerability, compactness and other important topological properties on the space. I'm not now interested in the ZF theory but in understanding how to manage and work on this space, in particular on $[0,\Omega]$ or $[0,\Omega)$, where $\Omega$ id the first uncountable ordinal.
 A: Ordinals are naturally ordered, and this ordering is a well-ordering. Namely, every non-empty set of ordinal has a least element. You can think about them as a continuation of the natural numbers. First we add $\omega$ which is the supremum of $\mathbb N$, then we add another point, $\omega+1$, etc. etc. 
Linear orders are particularly easy to define a topology from, we simply take the basic open sets to be intervals. Much like we would do in $\mathbb R$ or $\mathbb Q$.
There are generally two types of ordinals, successor ordinals and singular ordinals. Successor ordinals correspond to (linear) well-orders which have a maximal elements, for example if you add a point above all the natural numbers the order type is a successor ordinal, those would correspond to closed intervals, in some sense. Limit ordinals are, as the name suggests, limit of smaller ordinals and are not successor, to return to this example again the point which is the maximum in the aforementioned order (naturals+maximum) is itself a limit ordinal, often denoted $\omega$. Oh yes, there is also zero which corresponds to the empty set.
Finite ordinals are either zero or a natural number. In most modern mathematics natural numbers in set theory are often seen as the finite ordinals. Infinite ordinals are simply ordinals that have infinitely many points below them.
One of the qualities ordinal spaces offer is that every closed interval is compact, since being well-ordered is equivalent to the fact that there is no infinite decreasing sequence, so by covering with intervals the endpoints form a decreasing sequence which must be finite. They are also rich in isolated points, namely every successor ordinal is an isolated point. In fact the isolated points are exactly the successor ordinals and zero.
In the particular case of $[0,\Omega]$ and $[0,\Omega)$ the one thing to remember is that a countable sequence of countable ordinals have a countable limit (in ZFC), furthermore by their well-order every sequence of ordinals is either finite or contains an increasing sequence, and therefore has a limit point. Thus both the spaces are sequentially compact. The former space is a closed interval, and as remarked, is also compact; however the latter space is not compact since we can cover it with infinitely many disjoint intervals.
I feel that I cannot go much beyond that without getting into deeper parts of the theory of ordinals, and I feel that to truly understand these spaces you need to have at least some basic understanding on how ordinals work (order, arithmetics, etc.) as it is tied deeply with the properties of ordinal spaces.
Don't feel bad though. In my second year we were given the classical example of $\omega_1+1$ as a compact ordinal space. I didn't understand much at the time, and it took some set theory course later to understand the example in full.
A: I’m going to use the more standard notation $\omega_1$ for the first uncountable ordinal.
The most important thing to understand is that the linear order $\le$ on the set $[0,\omega_1]$ is a well-order: if $A$ is any non-empty subset of $[0,\omega_1]$, $A$ has a smallest element with respect to $\le$. That is, there is some $\alpha\in A$ such that $\alpha\le\beta$ for every $\beta\in A$. I’ll use this fact in a bit, but first let’s look a little closer at the elements of $[0,\omega_1]$.
The first countably infinitely many elements are the finite ordinals; you can think of these as being simply the non-negative integers, $0,1,2,3,\ldots$; this is, so to speak, the low end of the order $\le$. Now let $A=\{\alpha\in[0,\omega_1]:\alpha\text{ is not a finite ordinal}\}$. The set of finite ordinals is countable, and $[0,\omega_1]$ is uncountable, so $A\ne\varnothing$, and therefore $A$ has a least (or smallest) element; we call this element $\omega$. The set $\{0,1,2,\dots,\}\cup\{\omega\}$ is still countable, so the set $$[0,\omega_1]\setminus\Big(\{0,1,2,\dots,\}\cup\{\omega\}\Big)$$ is non-empty and therefore has a least element; we call this element $\omega+1$. This $\omega+1$ is the smallest ordinal after $\omega$: it comes right after $\omega$ in the order, so it’s the successor of $\omega$, just as $2$ is the successor of $1$. At this point we have a low end of $[0,\omega_1]$ that looks like this:
$$0,1,2,3,\dots,\omega,\omega+1$$
It should be intuitively clear that we can repeat this argument countably infinitely many times to produce $\omega+2,\omega+3,\dots\,$, and indeed $\omega+n$ for every finite ordinal $n$. Now we have an initial segment of $[0,\omega_1]$ that looks like this:
$$0,1,2,3,\dots,\omega,\omega+1,\omega+2,\omega+3,\dots$$
The only ordinals in this set that are not successors are $0$, since there’s nothing before it at all, and $\omega$, since there is nothing immediately before it: no matter what finite ordinal $n$ you consider, $n+1\ne\omega$.
But this set is still countable, so there is a smallest ordinal in
$$[0,\omega_1]\setminus\{0,1,2,\dots,\omega,\omega+1,\omega+2\dots\}\;;$$
this ordinal is denoted by $\omega\cdot 2$, and like $\omega$, it’s not a successor: it is not $\alpha+1$ for any $\alpha$. In other words, it’s a limit ordinal, as is $\omega$. ($0$ is a bit anomalous: it’s not a successor ordinal, but it’s also not a limit ordinal.)
After this the ordinals and their standard notations start getting a bit complicated, and we don’t really have to go into them to understand the topological properties of $[0,\omega_1)$ and $[0,\omega_1]$. It’s also true that the usual set-theoretic definition of the ordinals makes each ordinal equal to the set of its predecessors:
$$1=\{0\},2=\{0,1\},3=\{0,1,2\},\dots,\omega=\{0,1,2,\dots\},\omega+1=\{0,1,2,\dots,\omega\}\,,$$
and so on; this is the boxes within boxes that you mentioned. However, you don’t need to worry about this, either.
Now let’s see why every strictly decreasing sequence in $[0,\omega_1]$ is finite. Suppose that we had an infinite sequence $\langle\alpha_n:n\in\Bbb N\rangle$ such that $\alpha_0>\alpha_1>\alpha_2>\ldots\,$; then the set $A=\{\alpha_n:n\in\Bbb N\}$ would be a non-empty subset of $[0,\omega_1]$ with no least element, contradicting the fact that $[0,\omega_1]$ is well-ordered. Infinite increasing sequences are no problem at all, however: for each $\alpha\in[0,\omega_1)$, the set $[0,\alpha]$ is countable, so $[0,\omega_1)\setminus[0,\alpha]\ne\varnothing$, so there are elements of $[0,\omega_1)$ bigger than $\alpha$. The smallest of these is $\alpha+1$, the successor of $\alpha$. Thus, starting at any $\alpha\in[0,\omega_1$ I can form an infinite increasing sequence $\langle\alpha,\alpha+1,\alpha+2,\dots\rangle$ whose members are all still in $[0,\omega_1)$.
Next, let’s see why $[0,\omega_1)$ is first countable. Let $\alpha\in[0,\omega_1)$. Suppose first that $\alpha$ is a successor ordinal, say $\alpha=\beta+1$; then $(\beta,\alpha+1)=[\beta+1,\alpha+1)=[\alpha,\alpha+1)=\{\alpha\}$ is an open nbhd of $\alpha$ in the order topology, so $\alpha$ is an isolated point, and $\big\{\{\alpha\}\big\}$ is certainly a countable local base at $\alpha$! Note that $0$ behaves like a successor ordinal: $[0,1)=\{0\}$ is an open nbhd of $0$, so $0$ is also an isolated point.
Now suppose that $\alpha$ is a limit ordinal. For each $\beta<\alpha$ the set $(\beta,\alpha+1)=(\beta,\alpha]$ is an open nbhd of $\alpha$. Every open nbhd of $\alpha$ contains an open interval around $\alpha$, which in turn contains one of these intervals $(\beta,\alpha]$, so $$\mathscr{B}_\alpha=\Big\{(\beta,\alpha]:\beta<\alpha\Big\}$$ is a local base at $\alpha$. Finally, $\alpha<\omega_1$, and $\omega_1$ is the first ordinal with uncountably many predecessors, so there are only countably many $\beta<\alpha$, and $\mathscr{B}_\alpha$ is therefore countable. Thus, every point of $[0,\omega_1)$ has a countable local base, and $[0,\omega_1)$ is therefore first countable.
Note that $[0,\omega_1]$ is not first countable, because there is no countable local base at $\omega_1$: if $\big\{(\alpha_n,\omega_1]:n\in\Bbb N\big\}$ is any countable family of open intervals containing $\omega_1$, let $A=\bigcup_{n\in\Bbb N}[0,\alpha_n]$. Then $A$, being the union of countably many countable sets, is a countable subset of $[0,\omega_1)$, so $[0,\omega_1)\setminus A\ne\varnothing$. Pick any $\beta\in[0,\omega_1)\setminus A$; then $(\beta,\omega_1]$ is an open nbhd of $\omega_1$ that does not contain any of the sets $(\alpha_n,\omega_1]$, and therefore the family $\big\{(\alpha_n,\omega_1]:n\in\Bbb N\big\}$ is not a local base at $\omega_1$. That is, no countable family is a local base at $\omega_1$, so $[0,\omega_1]$ is not first countable at $\omega_1$.
Finally, let’s look at compactness. Suppose that $\mathscr{U}$ is an open cover of $[0,\omega_1]$. Then there is some $U_0\in\mathscr{U}$ such that $\omega_1\in U_0$. This $U_0$ must contain a basic open nbhd of $\omega_1$, so there must be an $\alpha_1<\omega_1$ such that $(\alpha_1,\omega_1]\subseteq U_0$. $\mathscr{U}$ covers $[0,\omega_1]$, so there is some $U_1\in\mathscr{U}$ such that $\alpha_1\in U_1$. This $U_1$ must contain a basic open nbhd of $\alpha_1$, so there is some $\alpha_2<\alpha_1$ such that $(\alpha_2,\alpha_1]\subseteq U_2$. Continuing in this fashion, we can construct a decreasing sequence $\alpha_1>\alpha_2>\alpha_3>\ldots\,$, which, as we saw before, must be finite. Thus, there must be some $n\in\Bbb Z^+$ such that $\alpha_n=0$, and at that point $\{U_0,\dots,U_n\}$ is a finite subcover of $\mathscr{U}$.
The space $[0,\omega_1)$, on the other hand, is not compact. It is countably compact, however. The easiest way to prove this is to show that $[0,\omega_1)$ has no infinite, closed, discrete subset. Suppose that $A$ is a countably infinite subset of $[0,\omega_1)$. Let $\beta$ be the smallest element of $[0,\omega_1)$ that is bigger than infinitely many elements of $A$. (You’ll have to explain why $\beta$ exists, using the fact that $A$ is countable and $[0,\omega_1)$ is well-ordered.) Finally, show that $\beta$ is a limit point of $A$. Then either $\beta\notin A$, in which case $A$ isn’t closed, or $\beta\in A$, in which case $A$ isn’t discrete.
