Are ordinal numbers produced by the power set axiom? I am a nube just getting into mathematics and set theory.
I am learning about how we can produce the list of ordinal numbers by purely using the null set, with 0 standing for Ø, 1 standing for {Ø}, 2 standing for {Ø, {Ø}} and so forth. What I am confused about is the operation at play here to produce the larger sets with more elements. It seems to me to be the power set axiom being applied to create a new larger set. But elsewhere I have seen this called the axiom of subsets. Is this the same thing? Or am I confused?
Thanks so much :)
A
 A: 
What I am confused about is the operation at play here to produce the larger sets with more elements.

Every natural number is defined as the set of all preceding natural numbers. The successor of a natural number is the union of that number with the singleton set of that number.
$S(n) := n\cup \{n\}$
Since if for all $n\in\Bbb N$, $n=\{m\in\Bbb N: m<n\}$, then $n+1=\{m\in\Bbb N:m<n\vee m=n\}$
So $$\small\begin{array}{lll}0&&=\{\}&=\emptyset&=\{\}\\1&=0\cup\{0\} &= \{\{\}\}&=\{\emptyset\}&=\{0\}\\2&=1\cup \{1\} &= \{\{\},\{\{\}\}\}&=\{\emptyset,\{\emptyset\}\}&=\{0,1\}\\3&=2\cup\{2\}&=\big\{\{\},\{\{\}\}, \{\{\},\{\{\}\}\}\big\}&=\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}&=\{0,1,2\}\\\vdots&\vdots&\vdots&\vdots&\vdots\\n+1&= n\cup \{n\}&\cdots&\cdots&=\{m:m<n+1\}\end{array}$$
A: The answer is no for small ordinals, and yes for large ordinals.
Producing finite ordinals does not require the axiom of power set. What we only need is the union and pairing. In fact, generating hereditarily finite sets needs no power set.
(Hereditarily finite sets are finite sets whose elements are also finite, elements of elements are also finite, and so on.)
In fact, we can produce all hereditarily finite sets by using adjunction operator $x;y:= x\cup\{y\}$ from the empty set.
For countable ordinals, the answer seems subtle. Sufficiently small countable ordinals, like $\omega$, $\omega^\omega$ or $\epsilon_0$, do not require the axiom of power set to be defined. However, more axioms sometimes result in the description of larger countable ordinals, although there is no easy example in my mind.
We need the axiom of power set to produce uncountable ordinals. The usual way to produce uncountable ordinal is using Hartogs number, although I think there is a more direct way: for example, consider the map $F$ from $\mathcal{P}(\omega\times\omega)$ to the class of all ordinals, defined by
$$F(r)=\begin{cases}\text{the ordertype of $r$} & \text{if $r$ is a well-order over $\omega$},\\ 0 & \text{otherwise.}\end{cases}$$
Then we can see that the image of $F$ is an ordinal, and is equal to $\omega_1$.
The axiom of power set is necessary to produce uncountable ordinals. It is known that the set of all hereditary countable sets $H_{\omega_1}$ satisfies $\mathsf{ZFC}$ without the axiom of power set, and it thinks every set (especially, every ordinal) is countable. Hence $H_{\omega_1}$ thinks there is no uncountable ordinal.
