Ordinals and Countability I've been watching Prof Su's (Harvey Mudd College) incredible lecture on Ordinals and Transfinite Induction, and I have a few related questions. He starts by drawing and examining three sets of dots, strictly well-ordered by position ($> \implies \text{to the right of}$), where clustered dots represent carrying on adding dots forever:
(a) $\{\cdot \cdot \cdot \ \cdot \}$
(b) $\{ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdots$ } ($\mathbb{N}$)
(c) $\{ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdots \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdots \ \cdot \cdot \ \cdot \}$ (2 sets of $\mathbb{N}$ + 3 other things to the right)
He then notes that there are finitely many things in (a) and countably many things in both (b) and (c). Therefor (b) and (c) have the same cardinality. My first question is how can we say that (c) is countably infinite? Isn't the definition of a countable set a set that can be put in to bijection with some subset of (or the entirety of) $\mathbb{N}$? I get how that's true for (b), but what bijection could you use for (c) that would map every element of (c) to an element of the $\mathbb{N}$, and preserve distinctness? Wouldn't you run out of places to map things to after the first set of dots?
Next he constructs the ordinal numbers as a way to classify well-ordered sets. An ordinal is defined to be a set that is:
(i) transitive/complete - every member is a subset, and
(ii) strictly well ordered by membership
We start off with $\phi$ (the empty set, which is vacuously well-ordered) and define, the Successor function:
$$ S(\alpha) = \alpha \ \cup \ \{ \alpha \},  $$
which ensures that $S(\alpha)$ will satisfy (i) and (ii) if $\alpha$ does, and helps us to define the "next" few ordinals:
$$ \{ \phi \}, \ \{ \phi, \{\phi \} \}, \ \{ \phi, \{\phi \}, \{ \phi, \{\phi \} \} \}, \ \dots $$
We can also define the Supremum function for a set set of ordinals, A:
$$ \sup(A) = \bigcup\limits_{i} a_{i}, \ a_{i} \in A $$
If A is a collection of ordinals, then $\sup(A)$ is also an ordinal
We can then start assign labels to ordinals: 
$$ "0" \to \phi, \ "1" \to \{ \phi \}, \ "2" \to \{ \phi, \{\phi \} \}, $$ etc.
He then introduces a theorem by Von-Neumann:

Thm. Any well-ordered set is order isomorphic to one of the ordinals

e.g. (a) is order isomorphic to "4".
He then goes on to say that $\mathbb{N}$ is order isomorphic to the supremum of the successor of all the finite ordinals, which we call $\omega$:
$$ \omega \equiv \sup( \ S(\alpha): |\alpha|<\infty) $$
This is said to be a limit ordinal and the first infinite ordinal. It is also the first ordinal that is not a successor of any ordinal. From there, we can continue to take the Successor:
$$ S(\omega) = \omega \cup \{ \omega \} \to "\omega + 1" $$
and we're off to the races. We have, in order:
$$ 1, 2, 3, \dots, \omega, \omega + 1, \omega + 1, \dots \omega\cdot 2, \omega\cdot 2 + 1, \omega\cdot 2 + 2, \dots \omega\cdot 3, \dots \omega^2, \omega^2 + 1, \omega^2 + 2, \dots \omega^2 + \omega, \dots \omega^3, \dots \omega^4, \dots, \omega^{\omega}, \dots \omega^{{\omega}^{\omega}}, \dots,   $$
all of which are apparently countable, even as we move through $\epsilon_{0}$, which is defined to be the first ordinal that satisfies: $ \epsilon_{0} = \omega^{\epsilon_0} $. Then we have $\epsilon_{1}$ which also satisfies this (i.e. $\epsilon_{1} = \omega^{\epsilon_{1}}$) but is not the first to do so... and $\epsilon_2, \epsilon_3, $ etc. all of which are apparently countable, until we get to $$ \boxed{\omega_1}, $$ defined to be the supremum of all of the countable ordinals, which is the first uncountable ordinal.
My second set of questions is How can anything after $\omega$ by countable? isn't the set $\omega + 1$ too big to be put in bijection with $\mathbb{N}$? And how can the union of an infinite number of things (the Supremum of all the countable numbers) be said to be uncountable, if we've already had limit ordinals before, that we say are still countable, that are unions of infinite numbers infinite sets. Shouldn't they start to be uncountable at $\omega$?
What am I not understanding here?
 A: (c) $3,5,7,9,\dots$, $4,6,8,10,\dots$, $0,1,2$.
$\omega + 1 \longleftrightarrow 1,2,3,4,\dots,0$.
That's how all those sets are countable.
A: To answer your first question,
Define a function  $f(n)$ such that $f(0)$ maps to the first of the three finalmost elements, $f(1)$ maps to the second, and $f(2)$ to the third. For each positive integer greater than 2, let $f(n)$ map 3 as the first element of the first group, 4 as the first element of the second, 5 as the second element of the first, and so on, alternating between each of the two countable clusters of elements.
A: The other answers address your first question; let me tackle the second. 
Note that if $\alpha$ is countable, so is $\alpha+1$ (Hilbert's hotel). So $\omega_1$ (this is the standard notation for the supremum of the countable ordinals) can't be countable, since otherwise $\omega_1+1$ would also be countable and we would therefore have to have $\omega_1\ge\omega_1+1$, by definition of $\omega_1$. Basically, $\omega_1$ is the first thing you get after you've run out of countable ordinals, so it clearly can't be countable!
As far as intuition goes, I would argue that makes it different from countable limit ordinals - even really big ones like $\epsilon_0=\omega^{\omega^{\omega^{. . .}}}$ - is cofinality. Basically, each countable limit ordinal can be written as a countable limit of smaller countable ordinals. This is not true for $\omega_1$: the countable union of countable sets is countable (technically this uses a very small amount of choice), so if $\alpha_1<\alpha_2<\alpha_3< . . . <\omega_1$, then $\sup\{\alpha_i: i\in\mathbb{N}\}$ is also less than $\omega_1$.
A: To answer your first question as simply as possible, if we look at set c, take the the three elements at the end, and consider them mapped onto 1, 2 and 3 in the set of natural numbers. If we then map the first element of the first set of b within c and map that to 4, the first element of the second set of b within c onto 5, the second element of the first set of b within c onto 6 and so on and so forth, we have a bijection between c and the set of natural numbers.
In answering the first question we can solve the second problem easily, if you have the successor ordinal: ω + 1, we map the +1 onto 1, the next the each element of ω onto the set of the natural number set in turn. Any countable ordinal works this way, even ε0, we map ω onto 1, ω^ω onto 2 An day so on and so forth for each well ordered element within ε0, the fact that it is well ordered means it has a mapping function onto the set of natural numbers.
Hopefully this helps, I tried to explain this in a way that’s would make sense to people that aren’t very familiar with these concepts. I remember watching a good YouTube series on this topic once, so there are resources out there to help people starting to learn this section of set theory.
