How explicitly can we find the cantor diagonal set to directly prove that $\omega_1$ is uncountable?

Question

Recall that $\omega_1$ is defined as:

$$\omega_1 = \{\text{Set of all countable ordinals}\}$$

and the cantor diagonal set is defined as:

$$T = \{ s \in S: s \not\in f(s) \}$$

where $f : S \to \mathcal{P}(S)$ not surjective

We note from this proof due to Carl Mummert that $\omega_1$ is uncountable and that $\omega_1 \subset \mathcal{P}(S)$ for some $S$ that is the set of all possible partial orderings of $\Bbb{N}$ (thus well orderings will be a subset of this)

However, in order to initiate a cantor diagonal style proof of $\omega_1$ will need to be able to define $f$. Now, we knew that from this, $\omega_1$ is impredicative thus $f$ is not a binary function.

But how explicit and/or constructive can $f$ be? Is there known bounds in $ZF\neg C$ (because the answer is trivial in $ZFC$ by defining $f$ to be a choice function) that limits how explicit we can specify $f$?

The motivation of this question, in case it is unclear, is to try the best to obtain a description of the uncountable number of uncomputable countable ordinals in $\omega_1$ to better understood them.

Answer 1

You write:

It will be cool if we can show how exactly when we try to count, $\omega_1$ with a bijection, how we ran out of things to pair up.

Here is a repackaging of the usual proof of the uncountability of $\omega_1$ which does exactly this (it should be noted that there's nothing new here mathematically, I'm just tweaking the language):

Recall that $\omega_1$ is the set of countable ordinals (this is not how you defined it). Suppose I have some function $f:\mathbb{N}\rightarrow\omega_1$; I want to construct an element $\alpha_f$ of $\omega_1$ - that is, a countable ordinal - not in $ran(f)$.

The definition of $\alpha_f$ is quite simple:

Let $\alpha_f=\bigcup\{f(i)\cup\{f(i)\}: i\in\mathbb{N}\}$.

The ordinal $\alpha_f$ is the smallest ordinal bigger than every ordinal in $ran(f)$.

Now we turn to the verification: showing that $\alpha_f\in\omega_1\setminus ran(f)$. It's easy to show that $\alpha_f$ is an ordinal: $\bigcup ran(f)$ is an ordinal since the union of any set of ordinals is an ordinal, and for any ordinal $\beta$ the set $\beta\cup\{\beta\}$ is also an ordinal (namely $\beta+1$). It's also easy to show that $\alpha_f$ is countable, since it's a countable union of countable sets. So we just need to show $\alpha_f\not\in ran(f)$. To do this, we first observe:

For every $\beta\in ran(f)$, there is an injective, order-preserving, non-surjective map from $\beta$ to $\alpha_f$.

This is a good exercise. We now use the following crucial fact about ordinals:

If $\eta,\theta$ are ordinals and there is an injective, order-preserving, non-surjective map from $\eta$ to $\theta$, then there is no order-preserving injection from $\theta$ to $\eta$. In particular, $\eta\not=\theta$.

This is an instance of a more general fact about well-orderings: if $L_1, L_2$ are well-orderings and there is an injective, order-preserving, non-surjective map from $L_1$ into $L_2$, then there is no injective order-preserving map from $L_2$ into $L_1$.

So we have $\alpha_f\not\in ran(f)$, and we're done.

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Note that this construction is actually much better-behaved than the diagonal argument! The ordinal $\alpha_f$ depends only on $ran(f)$: if we compose $f$ with a permutation of $\mathbb{N}$, the resulting ordinal is unchanged. This is in contrast to the diagonal argument on $\mathbb{R}$, where the real so produced depends on the order of the list of reals we begin with (and indeed it is consistent with ZF that this must be the case).

There are other "order-independent" ways to build an ordinal not in $ran(f)$. The most natural alternative to the construction of $\alpha_f$ given above, in my opinion, is:

Let $\zeta_f=\bigcup\{f(i)\cup\{f(i)\}: \forall \eta\le f(i)\exists j\in\mathbb{N}(f(j)=\eta)\}$.

This $\zeta_f$ is the least ordinal not in $ran(f)$.

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Now it's worth noting that there are many results in set theory which indicate that there can be no "concrete" picture of $\omega_1$, in various senses. For example, it is consistent with $\mathsf{ZF}$ that there is no injection from $\omega_1$ to $\mathbb{R}$ or from $\mathbb{R}$ to $\omega_1$. In $\mathsf{ZFC}$ of course there will be an injection from $\omega_1$ to $\mathbb{R}$, but it can still be very complicated: any such injection $i$ induces a relation $\trianglelefteq_i$ on a subset $S_i$ of $\mathbb{R}$, and any such pair $(S_i, \trianglelefteq_i)$ must be extremely complicated - certainly not Borel, and under additional ("large cardinal") hypotheses not even projective.

In a different direction, if we add an axiom like $\mathsf{MA+\neg CH}$ to $\mathsf{ZFC}$ (which we can consistently do if $\mathsf{ZFC}$ itself is consistent) then we can show roughly that $\omega_1$ can't be described in a concrete way as the smallest size of a set of reals which is "big enough" in some combinatorial sense. E.g. under $\mathsf{MA}$ each of the following descriptions:

• The smallest size of a set of functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that every $g:\mathbb{N}\rightarrow\mathbb{N}$ is dominated by some $f$ in the set

• The smallest size of a set of functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that every $g:\mathbb{N}\rightarrow\mathbb{N}$ is escaped by some $f$ in the set

• The smallest size of a set of measure-zero sets of reals whose union is $\mathbb{R}$

• The smallest size of a set of meager sets of reals whose union is $\mathbb{R}$

corresponds to $2^{\aleph_0}$, which under $\neg\mathsf{CH}$ is bigger than $\omega_1$. (Such descriptions are called "cardinal characteristics of the continuum.")