It's late and I only read the first few lines. Just some quick comments before I take a look at the rest of this tomorrow.

Let $I = \{ x : x \text{ is an infinite binary string} \}$.

Ok, I is the set of infinite binary sequences. In the context of formal set theory we shouldn't write $\{x : \text{anything}\}$ because that's dangerously close to Russell's paradox. And it sounds awkward to me. Just say, "Consider the set of foobars," rather than, "Consider the set of all x where x is a foobar." The latter is redundant and the notation is imprecise.

Let's take a moment to reflect on $I$. A

*binary sequence* or

*bitstring* is a function $b : \mathbb N \to 2$, where $2$ is the symbol that stands for the set $\{0, 1\}$. The set of all such functions is called $2^\mathbb N$ and it has cardinality $2^{\aleph_0}$. We usually notate the sequence $b_1, b_2, b_3, \dots$ as $b_1 b_2 b_3 \dots$. We can map bitstrings $b$ to the real interval $[0, 1)$ via the formula $b = \sum \frac{b_i}{10^i}$ and when we do that we notate $b$ by putting a decimal point in front of the string. The map is bijective if we wave our hands at the pesky trailing $9$'s, of which there are only countably many. We can always patch this up.

Let $F = \{ x : x \text{ is a finite binary string} \}$.

Likewise I'd just say, "Let $F$ be the set of finite bitstrings." We know that this is a countable set, since there are finitely many strings of length 1, finitely many of length 2, dot dot dot, and a countable union of finite sets is countable.

For each $a \in I$, $a = a_{1}a_{2}a_{3}\dots$ where each $a_i$ equals $0$ or $1$.

Yes.

For each $a \in I$, let $f(a) = \{ a_{1}, a_{1}a_{2}, a_{1}a_{2}a_{3}, \dots \}$.

So $f$ inputs a bitstring and outputs the set of its initial segments. Ok.

Let $\prec$ be a well ordering of $I$ (assumes the axiom of choice)

Yes, you may well-order $I$.

: $a^1, a^2, a^3, \dots$.

Aha. No. No. Not so simple. Well orders can be very wild. Consider a few well orders on the naturals:

* 2, 3, 4, 5, ..., 1

That has a first and last element.

* 2, 4, 6, ..., 1, 3, 5, ...

Odds before evens. That's $\omega + \omega = \omega \cdot 2$. Note that this is not the same as $2 \cdot \omega$, which is $\omega$ copies of a 2-element set, an entirely different order type. Ordinals are very weird.

* $2 \omega, 3 \omega, \dots, \omega \omega = \omega^2$.

* $\omega, \omega^2, \omega^3, \dots, \omega^\omega$.

* $\omega, \omega^\omega, \omega^{\omega^\omega}, \dots$. The ordinal number at the end of this process, a countably infinite exponential tower of $\omega$'s, is a magic ordinal named $\epsilon_0$.

https://en.wikipedia.org/wiki/Epsilon_numbers_(mathematics)

* After that the $\epsilon$ numbers keep going.

I'm only mentioning all this to show you how absolutely wild the ordinals are. And every ordinal I mentioned so far is a

*countable* ordinal. They're all just different ways of rearranging the natural numbers.

Now imagine how incomprehensible must be a well-order of an uncountable set!

So no, you may not assume that it is a

*list*, which is defined as the particular countable well-order that we call the "usual order on the naturals," namely 1, 2, 3, 4, ...

That's only one particular well-order of a countable set, and it's the simplest one. There are many others and they are one of the most mind-boggling objects in mathematics that I know. The contemplation of the countable ordinals makes me dizzy.

And now you are attempting to contemplate an uncountable ordinal.

I'm not saying we can't do that. It will be fun to work with this. But let's be aware of the pitfalls and complications.

The first uncountable ordinal is $\aleph_1$. In ordinal notation it's $\omega_1$, and $\aleph_1$ refers to the same number as a cardinal.

The best (probably the only) way to visualize $\omega_1$ is that it's the set of countable ordinals. In other words say you continue developing the countable ordinals as I did earlier, and you consider every possible way that you could well-order the naturals. Any set of ordinals is well ordered, in fact every ordinal is an initial segment of all the ones above it. So one, the set of all countable ordinals is in fact a set [needs proof]; and two,

*it's an ordinal* [ditto]. It can't be a countable ordinal, because a set can't be a member of itself. So the set of all countable ordinals must be the first uncountable ordinal.

I've just sketched a proof of the existence of an uncountable ordinal that did not depend on the axiom of choice. In fact the existence of $\omega_1$ is a theorem of ZF. Isn't that interesting? We need Choice to well-order an arbitrary set. But we do not need Choice to show that there is

*some* uncountable well-ordered set.

But now here is a problem you might have. We just described the uncountable ordinal $\omega_1$. But when you well order $2^{\aleph_0}$, you have no conceivable idea what that ordinal looks like. As wild as the countable ordinals are, how wild must the uncountable ordinals be? And what if the Continuum hypothesis is false and the $2^{\aleph_0} = \aleph_{47}$ or worse. How wild might your well-order of $I$ be?

I will end for tonight and take up the rest of this tomorrow. I hope this review of ordinals has been helpful. When you write $a^n$ are you thinking of wild uncountable ordinals but just writing them like the usual order on the naturals? Or are you intuitively thinking of the usual order and possibly confusing yourself? I couldn't tell from your exposition but I didn't read it closely yet.

ps -- When you ask if there's a least ordinal such that such and so, the answer is always yes. That's the thing about ordinals. If there's ANY ordinal such that X, there's a least one such that X, where X is any property.